{"id":1043,"date":"2014-08-12T03:12:15","date_gmt":"2014-08-12T03:12:15","guid":{"rendered":"https:\/\/courses.candelalearning.com\/cheminter\/?post_type=chapter&#038;p=1043"},"modified":"2017-09-06T18:58:06","modified_gmt":"2017-09-06T18:58:06","slug":"the-behavior-of-gases","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/the-behavior-of-gases\/","title":{"raw":"The Behavior of Gases","rendered":"The Behavior of Gases"},"content":{"raw":"<h1 id=\"x-ck12-VGhlIEJlaGF2aW9yIG9mIEdhc2Vz-chapter\">The Behavior of Gases<\/h1>\r\n<div class=\"x-ck12-data\"><\/div>\r\n<h1 id=\"x-ck12-Q29tcHJlc3NpYmlsaXR5\">Compressibility<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-NmViYjAxYmNkMjlmNmYxNjk1NzM0NmNhMzI1YmU5YjQ.-czo\">\r\n \t<li>Define compressibility.<\/li>\r\n \t<li>Give examples of the uses of compressed gases.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-YzA0YTlmMzliMzM0YTZiNjgxYWVjM2YwZmY3YTEwNTY.-fl0\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212110\/20140811155522937478.jpeg\" alt=\"Compressing objects can help to squeeze them into small spaces\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-YzA0YTlmMzliMzM0YTZiNjgxYWVjM2YwZmY3YTEwNTY.-k01\"><strong> Will it all fit? <\/strong><\/p>\r\n<p id=\"x-ck12-OTI5ZDkwOTkxNDhjNDhiYzEzMGZiYjQzZGY1MmYyNWU.-ayx\">When we pack to go on vacation, there is always \u201cone more\u201d thing that we need to get in the suitcase. Maybe it\u2019s another bathing suit, pair of shoes, book \u2013 whatever the item, we need to get it in. Fortunately, we can squeeze things together somewhat. There is a little space between the folds of clothing, we can rearrange the shoes, and somehow we get that last thing in and close the suitcase.<\/p>\r\n\r\n<h3>Compressibility<\/h3>\r\n<p id=\"x-ck12-ZmM1NTllNjcyYTk5Y2MyNjMxNzQxNGRlNjMzMTFkZjA.-tda\">Scuba diving is a form of underwater diving in which a diver carries his own breathing gas, usually in the form of a tank of compressed air. The pressure in most commonly used scuba tanks ranges from 200 to 300 atmospheres. Gases are unlike other states of matter in that a gas expands to fill the shape and volume of its container. For this reason, gases can also be compressed so that a relatively large amount of gas can be forced into a small container. If the air in a typical scuba tank were transferred to a container at the standard pressure of 1 atm, the volume of that container would need to be about 2500 liters.<\/p>\r\n\r\n<div id=\"x-ck12-ODkzOWEyYzJjOGY0MGM3NGVkMzdiOGFmZTliOGJhNDA.-uak\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-bhf\"><img id=\"x-ck12-OTgwNDUtMTM2MzY4NDI2Mi00OC00LTI.\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212112\/20140811155523048566.jpeg\" alt=\"The pressure in a scuba tank is typically 200-300 atmospheres\" longdesc=\"Scuba%20diver.\" \/><\/p>\r\n<strong> Figure 14.1 <\/strong>\r\n<p id=\"x-ck12-Nzg4NDMyOTRmZmMyZjhhOTBkY2ZlYzdiZmMxNTg0Mzc.-lbs\">Scuba diver.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-YWNkZDRhYzZkNzE2ZjY4ZjM2NjY3N2VkOWU1Y2M3YWQ.-pwh\"><strong> Compressibility <\/strong> is the measure of how much a given volume of matter decreases when placed under pressure. If we put pressure on a solid or a liquid, there is essentially no change in volume. The atoms, ions, or molecules that make up the solid or liquid are very close together. There is no space between the individual particles, so they cannot pack together.<\/p>\r\n<p id=\"x-ck12-MTZiNzllZjc4YTI4ZDBjNGM3ZTlkN2I0Y2NkYmI0ZTA.-lqm\">The kinetic-molecular theory explains why gases are more compressible than either liquids or solids. Gases are compressible because most of the volume of a gas is composed of the large amounts of empty space between the gas particles. At room temperature and standard pressure, the average distance between gas molecules is about ten times the diameter of the molecules themselves. When a gas is compressed, as when the scuba tank is being filled, the gas particles are forced closer together.<\/p>\r\n<p id=\"x-ck12-NTdhNGE2MjNjNjZhZTFjN2YzOTAxNzQzNzE4OTM5YjE.-log\">Compressed gases are used in many situations. In hospitals, oxygen is often used for patients who have damaged lungs to help them breathe better. If a patient is having a major operation, the anesthesia that is administered will frequently be a compressed gas. Welding requires very hot flames produced by compresses acetylene and oxygen mixtures. Many summer barbeque grills are fueled by compressed propane.<\/p>\r\n\r\n<div id=\"x-ck12-NDlmNzMxMDBiMzZmNzdjNDg0OTA4N2RiMjAzZDM3Y2Q.-3db\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\r\n<p id=\"x-ck12-npe\"><img id=\"x-ck12-OTgwNDUtMTM2MzY4NDI4Ni04NC01My0z\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212113\/20140811155523191643.jpeg\" alt=\"Compressed gases such as oxygen are used for a wide variety of applications\" longdesc=\"Oxygen%20tank.\" \/><\/p>\r\n<strong> Figure 14.2 <\/strong>\r\n<p id=\"x-ck12-NzE5YTJmNmRlN2Q2MDhhMDk2ZWVjODkwNjEyNTRjMDI.-si5\">Oxygen tank.<\/p>\r\n\r\n<\/div>\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-YjM4YzZlN2ZiZDFjYjFkNTc0NTJhNDkzZDYxODQ3MTQ.-v8h\">\r\n \t<li>Gases will compress more easily that solids or liquids because here is so much space between the gas molecules.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-uaj\"><em> Questions <\/em><\/p>\r\n<p id=\"x-ck12-Y2JlMjQ5M2YzMTNmNmRjMzNmZTI0MTMzYzcwM2IzZmY.-8sp\">Use the link below to answer the following questions:<\/p>\r\n<p id=\"x-ck12-MmNmMDcwZTY1ZmFlZTE5NjRmMmZkMDE4MDJmNmZlOTM.-7as\"><a href=\"https:\/\/web.archive.org\/web\/20160110232445\/http:\/\/www.cdxetextbook.com\/engines\/motivePower\/4gasEng\/engcycle.html\" target=\"_blank\" rel=\"noopener\">4-stroke engine cycle<\/a><\/p>\r\n\r\n<ol id=\"x-ck12-M2JjODliZDZjODY4MmZlYTNmZTc3MDNlMzBiODdmYTM.-ibp\">\r\n \t<li>What brings the fuel-air mixture into the cylinder?<\/li>\r\n \t<li>What is the role of the compression cycle?<\/li>\r\n \t<li>Does the exhaust cycle compress the gases produced by ignition?<\/li>\r\n<\/ol>\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-4xl\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-MjAyYzI2YjllY2FmOWI4Zjg0MWZmMjU0MjkwNjc0YmE.-enh\">\r\n \t<li>Why is there no change in volume when pressure is applied to liquids and solids?<\/li>\r\n \t<li>Why do gases compress more easily than liquids and solids?<\/li>\r\n \t<li>List uses for compressed gases.<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-MzI4ZGI3ODM1YzVhMjEwMmM0ZTJhM2VkZDA0YWE5MGE.-j1l\">\r\n \t<li><strong> compressibility: <\/strong> The measure of how much a given volume of matter decreases when placed under pressure.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-RmFjdG9ycyBBZmZlY3RpbmcgR2FzIFByZXNzdXJl\">Factors Affecting Gas Pressure<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-NzQ2MjMzYzM5NzczNjEyNDY0M2QxYmFhZDJmNjM1YTU.-yku\">\r\n \t<li>List factors that affect gas pressure.<\/li>\r\n \t<li>Explain these effects in terms of the kinetic-molecular theory of gases.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-YTIzMmJkMTA4OGViNGViNDlhZjhhMzAxYjhiMjc5ZmU.-mkq\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212113\/20140811155523369990.jpeg\" alt=\"The pressure in a basketball in a game must be kept in a certain range\" width=\"200\" \/><\/span><\/p>\r\n<p id=\"x-ck12-NTYwMjJhZjEzNDNkYjhhZDc2NTUyYjYyMjUyNDhiYTA.-7kg\"><strong> How high does a basketball bounce? <\/strong><\/p>\r\n<p id=\"x-ck12-MmYzZjEzMTM3ZTFiOGRmOGExNTgyMDc5MWQ4ZjAwYTg.-1j3\">The pressure of the air in a basketball has to be adjusted so that the ball bounces to the correct height. Before a game, the officials check the ball by dropping it from shoulder height and seeing how far back up it bounces. What would the official do if the ball did not bounce up as far as it is supposed to? What would he do if it bounced too high?<\/p>\r\n<p id=\"x-ck12-OWFkMWE4YmY2YzY0ZmI3NmRjMzkwZDgwMGZiMjY0MDg.-8hg\">The pressure inside a container is dependent on the amount of gas inside the container. If the basketball does not bounce high enough, the official could remedy the situation by using a hand pump and adding more air to the ball. Conversely, if it bounces too high, he could let some air out of the ball.<\/p>\r\n\r\n<h3>Factors Affecting Gas Pressure<\/h3>\r\n<p id=\"x-ck12-MDIzM2FlNGQ0NGQwYTVjNjA2OWQ1OGU0ZGZkNjA1OTk.-cji\">Recall from the kinetic-molecular theory that gas particles move randomly and in straight lines until they elastically collide with either other gas particles or with one of the walls of the container. It is these collisions with the walls of the container that defines the pressure of the gas. Four variables are used to describe the condition of a gas. They are pressure <img id=\"x-ck12-MTQwMDYwODU5NTgzOA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212115\/1b290cf55a4792cc5dfa6917c4dd4c52.png\" alt=\"(P)\" width=\"26\" height=\"18\" \/> , volume <img id=\"x-ck12-MTQwMDYwODU5NTgzOQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212115\/beae5e0c3935dd8d8de532ebd274ad1c.png\" alt=\"(V)\" width=\"27\" height=\"18\" \/> , temperature <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212115\/ddb22c35d11ae2888dbca942bc72ede4.png\" alt=\"(T)\" width=\"25\" height=\"18\" \/> , and the amount of the gas as measured by the number moles <img id=\"x-ck12-MTQwMDYwODU5NTg0MA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211023\/4168b7a35bc24c982496aef5b556a0a5.png\" alt=\"(n)\" width=\"23\" height=\"18\" \/> . We will examine separately how the volume, temperature, and amount of gas each affect the pressure of an enclosed gas sample.<\/p>\r\n\r\n<h4>Amount of Gas<\/h4>\r\n<p id=\"x-ck12-OWY0YTc0OGNlYmNkZDhiMzIxZWI5YTkxY2E4MzZjMjQ.-sdx\">The <strong> Figure <\/strong> below shows what happens when air is added to a <strong> rigid container <\/strong> . A rigid container is one that is incapable of expanding or contracting. A steel canister is an example of a rigid container.<\/p>\r\n\r\n<div id=\"x-ck12-ODIyM2RlNzYzMGM5ZjFiNWM2ZTBjNDJiMTRkMTRkMzA.-tk2\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-jh6\"><img id=\"x-ck12-OTgwNDUtMTM2MzY4NDQ1OC02Ni05Ni0y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212115\/20140811155523544082.png\" alt=\"An increase in the number of gas particles causes an increase in the pressure of a gas\" longdesc=\"Increase%20in%20pressure%20with%20increase%20in%20number%20of%20gas%20particles.\" \/><\/p>\r\n<strong> Figure 14.3 <\/strong>\r\n<p id=\"x-ck12-NmUyZjZiNTExYTNmZWQzMGQwNzkyYmM2ZmRhMGRhODE.-xre\">Increase in pressure with increase in number of gas particles.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-OTNkODIzODI4NzgzMTIzYTdlNTUwNDQ2NjQ3YzZjODI.-5vi\">The canister on the left contains a gas at a certain pressure. The attached air pump is then used to double the amount of gas in the canister. Since the canister cannot expand, the increased number of air molecules will strike the inside walls of the canister twice as frequently as they did before. The result is that the pressure inside the canister doubles. As you might imagine, if more and more air is continually added to a rigid container, it may eventually burst. Reducing the number of molecules in a rigid container has the opposite effect and the pressure decreases.<\/p>\r\n\r\n<h4>Volume<\/h4>\r\n<p id=\"x-ck12-MmNmNTY1NGU3ZWY2MDc2MTA5MmZhOTgyY2E4YTZiZjY.-k0w\">Pressure is also affected by the volume of the container. If the volume of a container is decreased, the gas molecules have less space in which to move around. As a result, they will strike the walls of the container more often and the pressure increases.<\/p>\r\n<p id=\"x-ck12-YTk5OGIyOTA0MjFiOTVmMjY0MjI3MDU0N2E4MGM2MDI.-ult\"><strong> Figure <\/strong> below shows a cylinder of gas whose volume is controlled by an adjustable piston. On the left, the piston is pulled mostly out and the gauge reads a certain pressure. On the right, the piston has been pushed so that the volume of the enclosed portion of the container where the gas is located has been cut in half. The pressure of the gas doubles. Increasing the volume of the container would have the opposite effect and the pressure of the gas would decrease.<\/p>\r\n\r\n<div id=\"x-ck12-NTQ2YWQ2MTUyMmJmOWVkMmY4YzRhZmIzNTU5NjIzY2Y.-5yh\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-ext\"><img id=\"x-ck12-OTgwNDUtMTM2MzY4NTE4Ny0wLTk1LTQ.\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212116\/20140811155523730968.png\" alt=\"A decrease in volume causes an increase in pressure for a gas\" longdesc=\"Decrease%20in%20gas%20volume%20produced%20increase%20in%20gas%20pressure.\" \/><\/p>\r\n<strong> Figure 14.4 <\/strong>\r\n<p id=\"x-ck12-MTdhODhjMzIwZDA5OGI5ZTNjMzg2ZDVkZmRkOGU1ZjI.-0pj\">Decrease in gas volume produced increase in gas pressure.<\/p>\r\n\r\n<\/div>\r\n<h4>Temperature<\/h4>\r\n<p id=\"x-ck12-Njg3ODA5ZGY2NzUyZGNmNTZlODZhODEzZjA3NDBlOTg.-kfd\">It would be very unadvisable to place a can of soup over a campfire without venting the can. As the can heats up, it may explode. The kinetic-molecular theory explains why. The air inside the rigid can of soup is given more kinetic energy by the heat coming from the campfire. The kinetic energy causes the air molecules to move faster and they impact the container walls more frequently and with more force. The increase in pressure inside may eventually exceed the strength of the can and it will explode. An additional factor is that the soup may begin boiling which will then aid even more gas and more pressure to the inside of the can.<\/p>\r\n<p id=\"x-ck12-YzY3NmNhNzIyNDBiMjQyNDIwZDJmMDA2MmZkYmViYmE.-gdz\">Shown in the <strong> Figure <\/strong> below is a cylinder of gas on the left that is at room temperature (300 K). On the right, the cylinder has been heated until the Kelvin temperature has doubled to 600 K. The kinetic energy of the gas molecules increases, so collisions with the walls of the container are now more forceful than they were before. As a result, the pressure of the gas doubles. Decreasing the temperature would have the opposite effect, and the pressure of an enclosed gas would decrease.<\/p>\r\n\r\n<div id=\"x-ck12-Y2M3NDc1NmQ0OTUyOGY2MWNiMmJhODFjMDlhZjE1OTU.-aoa\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n\r\n<img id=\"x-ck12-OTgwNDUtMTM2MzY4NTA1OS0xLTI5LTM.\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212118\/20140811155523847727.png\" alt=\"An increase in temperature causes an increase in pressure for a gas\" longdesc=\"Increase%20in%20temperature%20produces%20increase%20in%20pressure.\" \/>\r\n\r\n<strong> Figure 14.5 <\/strong>\r\n<p id=\"x-ck12-MDk0NTYwOTE3ZDdmODg3ZDFlNGM2ZmJlYmJmODM0NzU.-r95\">Increase in temperature produces increase in pressure.<\/p>\r\n\r\n<\/div>\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-NzBiNzJlNTNmZmFiZDJkMzAzMzY5YmU1ZDI0MWJhZDY.-l9n\">\r\n \t<li>An increase in the number of gas molecules in the same volume container increases pressure.<\/li>\r\n \t<li>A decrease in container volume increases gas pressure.<\/li>\r\n \t<li>An increase in temperature of a gas in a rigid container increases the pressure.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-40k\"><em> Questions <\/em><\/p>\r\n<p id=\"x-ck12-YmMzM2NmMmUyYzQ1NmUxNTQ2YTNmYjA2MWZlY2FkODA.-qln\">Watch the video at the link below and answer the following questions:<\/p>\r\nhttp:\/\/www.youtube.com\/watch?v=0mVuWZ7nvcU\r\n<ol id=\"x-ck12-MjI4MTBhZjZmMjZiNWViYjk3ZjdjMDViZDI2NmQ1NzQ.-mdi\">\r\n \t<li>What causes pressure?<\/li>\r\n \t<li>What happens when you let gas out of the container?<\/li>\r\n \t<li>If you increase the temperature, what happens to the pressure?<\/li>\r\n \t<li>Why does the pressure drop when you increase the volume?<\/li>\r\n<\/ol>\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-1jz\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-MWRiZGM3YzE1MTFiZjE2NTViMGQwMmQzNWFlYmFlOTQ.-drd\">\r\n \t<li>What defines the pressure of a gas?<\/li>\r\n \t<li>Why does an increase in the number of molecules increase the pressure?<\/li>\r\n \t<li>Why does an increase in temperature increase the pressure?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-ZDBmZDcwYzRjMGI4ZDE4MDNkNzY1ZGQ4MGJlNzc5NTM.-son\">\r\n \t<li><strong> rigid container: <\/strong> One that is incapable of expanding or contracting.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-Qm95bGUncyBMYXc.\">Boyle's Law<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-ZWUxNDlmOWNhZGE0M2E0YjQ5ZGM4ZDQ0N2RjZTQ5NmM.-ll0\">\r\n \t<li>State Boyle\u2019s Law.<\/li>\r\n \t<li>Use Boyle\u2019s Law to calculate volume-pressure relationships.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-Y2JlMzlhNTViZGMyYjI1Nzk4YTEzMTNlNWQ4ZTFlYWE.-xmm\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212120\/20140811155523976471.jpeg\" alt=\"Weather balloons expand and eventually burst as they travel to higher altitudes\" width=\"200\" \/><\/span><\/p>\r\n<p id=\"x-ck12-ODMyMzg2Y2ZhNTg1YmM1NzMxZmY3MDA4NDJkYzRkYTY.-awf\"><strong> How important is it to check the weather? <\/strong><\/p>\r\n<p id=\"x-ck12-ZDJjY2FiOGFmMDRkM2Q4ODg3NjhmNTAzNWU3NjZmODc.-0pl\">Each day, hundreds of weather balloons are launched. Made of a synthetic rubber and carrying a box of instruments, the helium-filled balloon rises up into the sky. As it gains altitude, the atmospheric pressure becomes less and the balloon expands. At some point the balloon bursts due to the expansion, the instruments drop (aided by a parachute) to be retrieved and studied for information about the weather.<\/p>\r\n\r\n<h3>Boyle\u2019s Law<\/h3>\r\n<p id=\"x-ck12-ZWFiOGFlNzVhZDE4NWM2ODU2MjY0NmYwZjczZGJmNjQ.-hm4\">Robert Boyle (1627-1691), an English chemist, is widely considered to be one of the founders of the modern experimental science of chemistry. He discovered that doubling the pressure of an enclosed sample of gas while keeping its temperature constant caused the volume of the gas to be reduced by half. <strong> Boyle\u2019s law <\/strong> states that the volume of a given mass of gas varies inversely with the pressure when the temperature is kept constant. An inverse relationship is described in this way. As one variable increases in value, the other variable decreases.<\/p>\r\n<p id=\"x-ck12-OWNjNDQzZDM2YmNlODhiZjE2N2Y1MzQxOGI0OGU3MzE.-ucz\">Physically, what is happening? The gas molecules are moving and are a certain distance apart from one another. An increase in pressure pushes the molecules closer together, reducing the volume. If the pressure is decreased, the gases are free to move about in a larger volume.<\/p>\r\n\r\n<div id=\"x-ck12-NzQxZWNhZjg1YzQ0ZjAyOTYwODMwNmVjMTYwMzU5ODM.-tdq\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\r\n<p id=\"x-ck12-ndj\"><img id=\"x-ck12-OTgwNDUtMTM2MzY5MDU0Ny0yNi0yOC01\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212121\/20140811155524117221.jpeg\" alt=\"Robert Boyle states that PV is constant at a given temperature\" longdesc=\"Robert%20Boyle.\" \/><\/p>\r\n<strong> Figure 14.6 <\/strong>\r\n<p id=\"x-ck12-OWVmMWFlZmVkOTIzYTY5MzBiMjExNjZlNjc0MzNjMWY.-py9\">Robert Boyle.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-NGQyMzdhMzdmNTlhN2U1NGI2NDJhOTUwZjc5NWYwMzE.-wkb\">Mathematically, Boyle\u2019s law can be expressed by the equation:<\/p>\r\n<p id=\"x-ck12-uhx\"><img id=\"x-ck12-MTM2NjQzNzIwMjYwNw..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212122\/16476bd9759a44a54eeaf6837cecaa71.png\" alt=\"P times V = k\" width=\"85\" height=\"13\" \/><\/p>\r\n<p id=\"x-ck12-ZjczMjMwMzIxZGY5ODk0MGM4MDZhM2M0YjMwMWQyOTY.-who\">The\u00a0 <img id=\"x-ck12-MTM2NjQzNzIwMjYwOA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212122\/d301c6452e642017007fb7cd755ce08b.png\" alt=\"k\" width=\"9\" height=\"13\" \/> is a constant for a given sample of gas and depends only on the mass of the gas and the temperature. The <strong> Table <\/strong> below shows pressure and volume data for a set amount of gas at a constant temperature. The third column represents the value of the constant\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212122\/8c0cf0a5c7fa06aa679382f3c2076f4d.png\" alt=\"(k)\" width=\"22\" height=\"18\" \/> for this data and is always equal to the pressure multiplied by the volume. As one of the variables changes, the other changes in such a way that the product of\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212122\/ac34d0426fddd2c0ce37f74932678203.png\" alt=\"P times V\" width=\"51\" height=\"12\" \/> always remains the same. In this particular case, that constant is 500\u00a0atm\u00a0\u00b7\u00a0ml.<\/p>\r\n\r\n<table id=\"x-ck12-MzI3NTRlOWZlYmY5YjNlNWQ1NjUzNTUyZTQ5ODgxN2Q.-ggq\" class=\"x-ck12-nofloat\" border=\"1\"><caption>Pressure-Volume Data<\/caption>\r\n<tbody>\r\n<tr>\r\n<td><strong> Pressure (atm) <\/strong><\/td>\r\n<td><strong> Volume (mL) <\/strong><\/td>\r\n<td><img id=\"x-ck12-MTM2NjQzNzIwMjYxMA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212123\/7d6bf5e3739d48298c115ae53bb1315d.png\" alt=\"P times V = k (text{atm} cdot text{mL})\" width=\"168\" height=\"18\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0.5<\/td>\r\n<td>1000<\/td>\r\n<td>500<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>0.625<\/td>\r\n<td>800<\/td>\r\n<td>500<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1.0<\/td>\r\n<td>500<\/td>\r\n<td>\r\n<p id=\"x-ck12-Y2VlNjMxMTIxYzJlYzkyMzJmM2EyZjAyOGFkNWM4OWI.-luz\">500<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2.0<\/td>\r\n<td>250<\/td>\r\n<td>\r\n<p id=\"x-ck12-Y2VlNjMxMTIxYzJlYzkyMzJmM2EyZjAyOGFkNWM4OWI.-qrg\">500<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>5.0<\/td>\r\n<td>100<\/td>\r\n<td>500<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>8.0<\/td>\r\n<td>62.5<\/td>\r\n<td>500<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>10.0<\/td>\r\n<td>50<\/td>\r\n<td>500<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"x-ck12-ZmU5MDBmMDA5MTZiODZhYWEyYzFmNDRjMmY3YTMxYTU.-zol\">A graph of the data in the table further illustrates the inverse relationship nature of Boyle\u2019s Law (see <strong> Figure <\/strong> below ). Volume is plotted on the <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212123\/0f14991bebd0bc5ca2e82990e029d525.png\" alt=\"x\" width=\"10\" height=\"8\" \/> -axis, with the corresponding pressure on the <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212123\/b4a681cb4394e7a28b5b01d49cf52c9e.png\" alt=\"y\" width=\"9\" height=\"12\" \/> -axis.<\/p>\r\n\r\n<div id=\"x-ck12-ODViN2RhNmQ0ZDQxNGRhMThiMTlmNmM1ZjBlN2JiZGI.-qwa\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-5uo\"><img id=\"x-ck12-OTgwNDUtMTM2MzY5MDU5MS05MS01OC02\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212124\/20140811155524277894.png\" alt=\"The pressure of a gas decreases as its volume increases\" longdesc=\"The%20pressure%20of%20a%20gas%20decreases%20as%20the%20volume%20increases%2C%20making%20Boyle%E2%80%99s%20law%20an%20inverse%20relationship.\" \/><\/p>\r\n<strong> Figure 14.7 <\/strong>\r\n<p id=\"x-ck12-ODNmMzA3NDFmMDIyMTgxYTJlZjYyMTMxOTlkZTk4NzY.-fgt\">The pressure of a gas decreases as the volume increases, making Boyle\u2019s law an inverse relationship.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-NzBkODU5MGQxMjYxNGZmMTk5ZDZiYzY2MTU3MTJkYmQ.-ozb\">Boyle\u2019s Law can be used to compare changing conditions for a gas. We use\u00a0 <img id=\"x-ck12-MTM2NjQzNzIwMjYxMg..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/b95fd9f12605d93cbb24fd5fc1d83fbc.png\" alt=\"P_1\" width=\"18\" height=\"16\" \/> and\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/29c883aa1a216bc9157974e94c993965.png\" alt=\"V_1\" width=\"16\" height=\"16\" \/> to stand for the initial pressure and initial volume of a gas. After a change has been made,\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/f49ec791ef1ecb63d1c14840f9c2f983.png\" alt=\"P_2\" width=\"18\" height=\"15\" \/> and\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> stand for the final pressure and volume. The mathematical relationship of Boyle\u2019s Law becomes:<\/p>\r\n<p id=\"x-ck12-esr\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/f4c066bce5329c1006f6e21829e93d1a.png\" alt=\"P_1 times V_1=P_2 times V_2\" width=\"140\" height=\"16\" \/><\/p>\r\n<p id=\"x-ck12-YTI3NzZjYmQzMzBkYzFhNjhkYjJhNDA0ZjZmMDZkMmM.-xpx\">This equation can be used to calculate any one of the four quantities if the other three are known.<\/p>\r\n\r\n<h4>Sample Problem: Boyle\u2019s Law<\/h4>\r\n<p id=\"x-ck12-NTMyZTk3ODQ4MDdmOWMzNzcxZDJmMWRkMGUxZTNjOWQ.-ncg\">A sample of oxygen gas has a volume of 425 mL when the pressure is equal to 387 kPa. The gas is allowed to expand into a 1.75 L container. Calculate the new pressure of the gas.<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-o1i\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-6pr\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ln9\">\r\n \t<li><img id=\"x-ck12-MTQwMTM5OTE2NzcyNg..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/db5e81bc83ca81445200cbbc41291805.png\" alt=\"P_1=387 text{kPa}\" width=\"106\" height=\"17\" \/><\/li>\r\n \t<li><img id=\"x-ck12-MTQwMTM5OTE2NzcyNw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212127\/a39e628ca221147569ced5499f911de1.png\" alt=\"V_1=425 text{mL}\" width=\"101\" height=\"17\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212127\/41e797e168699fd6af53fda820e06914.png\" alt=\"V_2=1.75 text{L}=1750 text{mL}\" width=\"183\" height=\"16\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-uik\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-wqa\">\r\n \t<li><img id=\"x-ck12-MTQwMTM5OTE2NzcyOQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212128\/33dd88af6f64da7d9108a4e93ab30012.png\" alt=\"P_2=? text{kPa}\" width=\"82\" height=\"16\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-YmZmZDZiNjNmNzdmZDU1MjZiNDE1ZjAxMWIzMWZlYWY.-ahu\">Use Boyle\u2019s Law to solve for the unknown pressure <img id=\"x-ck12-MTM2NjQzNzIwMjYxMw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212128\/7309a9826c8e30de6d5ca051de922d91.png\" alt=\"(P_2)\" width=\"31\" height=\"18\" \/> . It is important that the two volumes ( <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/29c883aa1a216bc9157974e94c993965.png\" alt=\"V_1\" width=\"16\" height=\"16\" \/> and <img id=\"x-ck12-MTM2NjQzNzIwMjYxNA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> ) are expressed in the same units, so\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> has been converted to mL.<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-pqs\"><em> Step 2: Solve. <\/em><\/p>\r\n<p id=\"x-ck12-ZTZkYTlmMmZlMzNlOWE1MmY0OGVlNjA0MzgxYmM4ZTE.-txa\">First, rearrange the equation algebraically to solve for <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/f49ec791ef1ecb63d1c14840f9c2f983.png\" alt=\"P_2\" width=\"18\" height=\"15\" \/> .<\/p>\r\n<p id=\"x-ck12-3fb\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212128\/bff7f7fa27ec0593ea7b3c9219e71bb2.png\" alt=\"P_2=frac{P_1 times V_1}{V_2}\" width=\"103\" height=\"39\" \/><\/p>\r\n<p id=\"x-ck12-OTNiYzM1Nzk5YWMwYTdmNDZjM2RiOTlmODhhOGVkYTM.-5as\">Now substitute the known quantities into the equation and solve.<\/p>\r\n<p id=\"x-ck12-9zf\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212129\/7fb79e9ad9bcb40194be760d22e3cdf5.png\" alt=\"P_2=frac{387 text{ kPa} times 425 text{ mL}}{1750 text{ mL}}=94.0 text{ kPa}\" width=\"282\" height=\"39\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-nkb\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-Y2JlZjk5YTA0ZGY0NDNkZWZkZGU0YTZlZmU3OWYxOWU.-2qy\">The volume has increased to slightly over 4 times its original value and so the pressure is decreased by about <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212130\/c1689389012aed4922e2871602dab8bc.png\" alt=\"frac{1}{4}{th}\" width=\"24\" height=\"23\" \/> . The pressure is in kPa and the value has three significant figures. Note that any pressure or volume units can be used as long as they are consistent throughout the problem.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-NmZjNTA3ZTFmOTliNDgxOWRmNmIxMWU5NzMzMDIxZjc.-ler\">\r\n \t<li>The volume of a gas is inversely proportional to temperature.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-ZjRmZTgxMGFkNmYzZWFhZGViYzllOGZjOWIxZDdhZjQ.-5an\">Do the problems at the link below:<\/p>\r\n<p id=\"x-ck12-MDVhNjVlMmMyMjViMWU3N2U4OTA0ZDcxZTM5MzJiZWM.-wk0\"><a href=\"http:\/\/www.concord.org\/~ddamelin\/chemsite\/g_gasses\/handouts\/Boyle_Problems.pdf\"> http:\/\/www.concord.org\/~ddamelin\/chemsite\/g_gasses\/handouts\/Boyle_Problems.pdf <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-0em\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-N2U2ZDkxMWYwMjNjZTY4ZGUxNzg1ZDk2ZGVlOGU5Zjg.-xjk\">\r\n \t<li>What does \u201cinversely\u201d mean in this law?<\/li>\r\n \t<li>Explain Boyle\u2019s law in terms of the kinetic-molecular theory of gases.<\/li>\r\n \t<li>Does it matter what units are used?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-ZjQzYTMwY2U4MjgyNTBkZWZlYThlZmM2YjMwNWFlMWY.-bky\">\r\n \t<li><strong> Boyle\u2019s law: <\/strong> The volume of a given mass of gas varies inversely with the pressure when the temperature is kept constant.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-Q2hhcmxlcydzIExhdw..\">Charles's Law<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-YjAyNzVlYWE1MTQ0YWU4YjRmZWI4Zjk1YjMwZjY1NjU.-akf\">\r\n \t<li>State Charles\u2019 Law.<\/li>\r\n \t<li>Use this law to perform calculations involving volume-temperature relationships.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-N2Q4MmRkZTA4NjhmZjdlNmFiM2EwMWFkY2MwNDg1NGI.-wju\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212130\/20140811155524496248.jpeg\" alt=\"When bread is baked, the carbon dioxide inside expands\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-YTVjOWMzMzFlOTdkODBkYmNkYWQ4ZGVhMDQ1MTQ1ZjQ.-m5s\"><strong> How do you bake bread? <\/strong><\/p>\r\n<p id=\"x-ck12-MDU1MmViYjllZWVlNTM2NzkxNWViNmNjNTk1ZGY4NTU.-oy2\">Everybody enjoys the smell and taste of freshly-baked bread. It is light and fluffy as a result of the action of yeast on sugar. The yeast converts the sugar to carbon dioxide, which at high temperatures causes the dough to expand. The end-result is an enjoyable treat, especially when covered with melted butter.<\/p>\r\n\r\n<h3>Charles\u2019s Law<\/h3>\r\n<div id=\"x-ck12-YThhNjRlYWNjMGVjMjBmNzM3MzBmOGVlZGM5NzNiYjY.-y3m\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-tp3\"><img id=\"x-ck12-OTgwNDUtMTM2MzY5MjA3NS00MS0xOS0y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212132\/20140811155524707039.png\" alt=\"As a gas is heated at constant pressure, its volume increases\" longdesc=\"As%20a%20container%20of%20confined%20gas%20is%20heated%2C%20its%20molecules%20increase%20in%20kinetic%20energy%20and%20push%20the%20movable%20piston%20outward%2C%20resulting%20in%20an%20increase%20in%20volume.\" \/><\/p>\r\n<strong> Figure 14.8 <\/strong>\r\n<p id=\"x-ck12-MTUzMmUxNjQzMjZiY2Y4ZGQ3MThmNmE1YjExZmUyMTk.-jfu\">As a container of confined gas is heated, its molecules increase in kinetic energy and push the movable piston outward, resulting in an increase in volume.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-YWI3MzJmZDQzMDM1OTM4YjNjNDFlYjBlYWY1MTI0ODE.-zhd\">French physicist Jacques Charles (1746-1823) studied the effect of temperature on the volume of a gas at constant pressure. <strong> Charles\u2019s law <\/strong> states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stoppage of molecular motion.<\/p>\r\n<p id=\"x-ck12-Nzc2NzQ2ODk5OGUyZWM3ODM3NGYwYzc5MTAwYjhiZGE.-ma4\">Mathematically, the direct relationship of Charles\u2019s law can be represented by the following equation:<\/p>\r\n<p id=\"x-ck12-pvf\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212133\/68ec4151a7f8c52092668cc72efd0265.png\" alt=\"frac{V}{T}=k\" width=\"50\" height=\"37\" \/><\/p>\r\n<p id=\"x-ck12-NGE5NzZlOTVlNWFmMjM2ZmIwODQ1ZTFjOTMxYTFhNjc.-lc0\">As with Boyle\u2019s law, <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212122\/d301c6452e642017007fb7cd755ce08b.png\" alt=\"k\" width=\"9\" height=\"13\" \/> is constant only for a given gas sample. The <strong> Table <\/strong> below shows temperature and volume data for a set amount of gas at a constant pressure. The third column is the constant for this particular data set and is always equal to the volume divided by the Kelvin temperature.<\/p>\r\n\r\n<table id=\"x-ck12-MmJhZjZhNWU4OGVlMDlhMjg3MjhiZjUzZmE1ODE2NjI.-1ne\" class=\"x-ck12-nofloat\" border=\"1\"><caption>Temperature-Volume Data<\/caption>\r\n<tbody>\r\n<tr>\r\n<td><strong> Temperature (K) <\/strong><\/td>\r\n<td><strong> Volume (mL) <\/strong><\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212133\/de68e314bf5efae1e99b1782323d02a9.png\" alt=\"frac{V}{T}=kleft (frac{mL}{K}right)\" width=\"88\" height=\"24\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>50<\/td>\r\n<td>20<\/td>\r\n<td>0.40<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>100<\/td>\r\n<td>40<\/td>\r\n<td>0.40<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>150<\/td>\r\n<td>60<\/td>\r\n<td>0.40<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>200<\/td>\r\n<td>80<\/td>\r\n<td>0.40<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>300<\/td>\r\n<td>120<\/td>\r\n<td>0.40<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>500<\/td>\r\n<td>200<\/td>\r\n<td>0.40<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1000<\/td>\r\n<td>400<\/td>\r\n<td>0.40<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"x-ck12-NzYxZTI2Zjc0NTYxYzhmNDQ0NDI2ZDE4ZDEwMzMzY2U.-vmq\">When this data is graphed, the result is a straight line, indicative of a direct relationship, shown in <strong> Figure <\/strong> below .<\/p>\r\n\r\n<div id=\"x-ck12-NGFlYTMwM2RlMDUwNDI1M2Y2MTcyODQzZTBlMzlhY2Y.-vas\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-doy\"><img id=\"x-ck12-OTgwNDUtMTM2MzY5MjExMS0yLTMzLTM.\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212133\/20140811155524837584.png\" alt=\"The volume of a gas increases as the Kelvin temperature increases\" longdesc=\"The%20volume%20of%20a%20gas%20increases%20as%20the%20Kelvin%20temperature%20increases.\" \/><\/p>\r\n<strong> Figure 14.9 <\/strong>\r\n<p id=\"x-ck12-M2YyNDJmY2ZkNDQyYmQ0ZDI5ZjQ1ZDU0Y2IwMzJmYmU.-b6p\">The volume of a gas increases as the Kelvin temperature increases.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-MmQxZWEwODdiZTAyMTBhYmY5OWQ0YWZjZjEyNTllYTY.-fn5\">Notice that the line goes exactly toward the origin, meaning that as the absolute temperature of the gas approaches zero, its volume approaches zero. However, when a gas is brought to extremely cold temperatures, its molecules would eventually condense into the liquid state before reaching absolute zero. The temperature at which this change into the liquid state occurs varies for different gases.<\/p>\r\n<p id=\"x-ck12-NDJlYzE1YjgxZGE2ZTZiYTdiZTU4YTkzNjZiNmZjYjc.-ihc\">Charles\u2019s Law can also be used to compare changing conditions for a gas. Now we use\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/29c883aa1a216bc9157974e94c993965.png\" alt=\"V_1\" width=\"16\" height=\"16\" \/> and\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212135\/5ef2d3527b9eadc8e4f588dbe3795272.png\" alt=\"T_1\" width=\"16\" height=\"16\" \/> to stand for the initial volume and temperature of a gas, while <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> and\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212135\/d64b622498e5207084db58e1dea5b037.png\" alt=\"T_2\" width=\"16\" height=\"15\" \/> stand for the final volume and temperature. The mathematical relationship of Charles\u2019s Law becomes:<\/p>\r\n<p id=\"x-ck12-2kp\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212135\/6bb0892b51cef8a85ef101eb68a425e5.png\" alt=\"frac{V_1}{T_1}=frac{V_2}{T_2}\" width=\"63\" height=\"40\" \/><\/p>\r\n<p id=\"x-ck12-YmQ0ZThmMzUwZTFlMTI1ODcxZDEyMGZhY2EyMzBiNTg.-cm8\">This equation can be used to calculate any one of the four quantities if the other three are known. The direct relationship will only hold if the temperatures are expressed in Kelvin. Temperatures in Celsius will not work. Recall the relationship that K\u00a0=\u00a0\u00b0C\u00a0+\u00a0273.<\/p>\r\n\r\n<h4>Sample Problem: Charles\u2019s Law<\/h4>\r\n<p id=\"x-ck12-NDdkNmU2OGRiMTRmNzMwMmU0MTc5OTM5M2UxMWFkMjc.-inz\">A balloon is filled to a volume of 2.20 L at a temperature of 22\u00b0C. The balloon is then heated to a temperature of 71\u00b0C. Find the new volume of the balloon.<\/p>\r\n<p id=\"x-ck12-OGU2YWEyYzY0NzAwMDUxZjI4NjFjY2E4MjYyNmNhN2I.-ckj\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-ki5\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-s6x\">\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212135\/2f94b6d4455bd48ff87d466ed526654b.png\" alt=\"V_1=2.20 text{ L}\" width=\"91\" height=\"17\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212136\/b4d715ba982209d5f333b1baf3727991.png\" alt=\"T_1=22^circ text{C}=295 text{ K}\" width=\"151\" height=\"17\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212137\/b4893bb931993217110b5288da684114.png\" alt=\"T_2=71^circ text{C}=344 text{ K}\" width=\"151\" height=\"16\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-6jl\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-9qb\">\r\n \t<li><em> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212137\/bbda48a6df75d3defbe58c3b3ce2ebe6.png\" alt=\"V_2= ? text{ L}\" width=\"62\" height=\"16\" \/><\/em><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-NzRlZWZlMTBkM2Q2NmNkYjJiNGJlODZhMjg2ZDhmMGU.-mvb\">Use Charles\u2019s law to solve for the unknown volume <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212137\/0a4b20dd3f9257612d0c459edf0075c2.png\" alt=\"(V_2)\" width=\"30\" height=\"18\" \/> . The temperatures have first been converted to Kelvin.<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-pfi\"><em> Step 2: Solve. <\/em><\/p>\r\n<p id=\"x-ck12-Yzg1OGU4MDExNzQ5NGQ4OGU4NjY3ZWZlMTFmNDdhNzQ.-ebu\">First, rearrange the equation algebraically to solve for <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> .<\/p>\r\n<p id=\"x-ck12-ofi\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212138\/8866248b5555b9388de04b70a852ec31.png\" alt=\"V_2=frac{V_1 times T_2}{T_1}\" width=\"101\" height=\"40\" \/><\/p>\r\n<p id=\"x-ck12-OTNiYzM1Nzk5YWMwYTdmNDZjM2RiOTlmODhhOGVkYTM.-akc\">Now substitute the known quantities into the equation and solve.<\/p>\r\n<p id=\"x-ck12-2xd\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212138\/8d605d9b16320240278b22c3ad3b9919.png\" alt=\"V_2=frac{2.20 text{ L} times 344 text{ K}}{295 text{ K}}=2.57 text{ L}\" width=\"236\" height=\"39\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-rbu\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-YWRjNGQ0MGFjZWFlODYzN2NkMDI2OGYxMDFmNTRhY2Q.-n2j\">The volume increases as the temperature increases. The result has three significant figures.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-ZTMxNDFkOWExNWYwMGY4YTNhZTc4NjMwOTNiMDhkODg.-n2s\">\r\n \t<li>Increasing the temperature of a gas at constant pressure will produce and increase in the volume.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-MjJhNzc5ZTg5MmQxOGM2NjRhNjc1Mjk4MGFlODQ0NTQ.-spz\">Perform the calculations at the web site below:<\/p>\r\n<p id=\"x-ck12-Nzg4MWRiYzg2ZWZjMDM0MGVjNmUwZGIzNjU1MWJjOGI.-kxw\"><a href=\"http:\/\/mmsphyschem.com\/chuckL.pdf\"> http:\/\/mmsphyschem.com\/chuckL.pdf <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-lp0\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-YzQ0OTBlNGYwY2IwYmI0YTgyZmFiOWIyZmI1ZjAyNGI.-qka\">\r\n \t<li>Explain Charles\u2019 Law in terms of the kinetic molecular theory.<\/li>\r\n \t<li>Why does the temperature need to be in Kelvin?<\/li>\r\n \t<li>Does Charles\u2019 law hold when the gas becomes a liquid?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-Njg1MjU3NDcxYzdhN2JkMjg1NTI0OGNkZDRhOTU2NWQ.-6or\">\r\n \t<li><strong> Charles\u2019s law: <\/strong> The volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-R2F5LUx1c3NhYydzIExhdw..\">Gay-Lussac's Law<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-YzFjMWNmOTQ4Yzg5MGI2NTk5MzlhMDE1YTM5MGVjOGY.-xrq\">\r\n \t<li>State Gay-Lussac\u2019s law.<\/li>\r\n \t<li>Use this law to perform calculations involving pressure-temperature relationships.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-ZWJjODUzNTZiZDkwOWU5NTI4NjYzODRiZDJhNzdmNTg.-p8q\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212140\/20140811155524966201.jpeg\" alt=\"The temperature of a tank of gas needs to be taken into account when measuring its pressure\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-ODYwYjc2MmNkN2JjNzQ2ZjE1ZjI3ZTczYzI1OGYxZjU.-5ov\"><strong> How much propane is in the tank? <\/strong><\/p>\r\n<p id=\"x-ck12-NWY4ZjA5MDJmOTU1OGVhZDJjMmVkYjc3ODNiNjVkZTI.-087\">Propane tanks are widely used with barbeque grills. But it\u2019s not fun to find out half-way through your grilling that you\u2019ve run out of gas. You can buy gauges that measure the pressure inside the tank to see how much is left. The gauge measures pressure and will register a higher pressure on a hot day than it will on a cold day. So you need to take the air temperature into account when you decide whether or not to refill the tank before your next cook-out.<\/p>\r\n\r\n<h3>Gay-Lussac\u2019s Law<\/h3>\r\n<p id=\"x-ck12-Njk5MjJmODk5Y2UxOTE5YmM1NDM2MGMwNGEwMmNkNDg.-zqo\">When the temperature of a sample of gas in a rigid container is increased, the pressure of the gas increases as well. The increase in kinetic energy results in the molecules of gas striking the walls of the container with more force, resulting in a greater pressure. The French chemist Joseph Gay-Lussac (1778-1850) discovered the relationship between the pressure of a gas and its absolute temperature. <strong> Gay-Lussac\u2019s law <\/strong> states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant. Gay-Lussac\u2019s law is very similar to Charles\u2019s law, with the only difference being the type of container. Whereas the container in a Charles\u2019s law experiment is flexible, it is rigid in a Gay-Lussac\u2019s law experiment.<\/p>\r\n\r\n<div id=\"x-ck12-NDU2NjI2ZjZjZjVmYjA2ZjNiOTVmZThhNzQyOGU2NmM.-ryp\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\r\n<p id=\"x-ck12-oip\"><img id=\"x-ck12-OTgwNDUtMTM2MzY4ODcwMC03Mi04MS00LjIuNS4y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212142\/20140811155525097528.jpeg\" alt=\"In Gay-Lussac's Law, the pressure of a gas varies directly with the temperature of a gas\" longdesc=\"Joseph%20Louis%20Gay-Lussac.\" \/><\/p>\r\n<strong> Figure 14.10 <\/strong>\r\n<p id=\"x-ck12-NzU2YjdkMzFmNTk2NWNlODM4NjViYTM5NWMyMjg4Yjk.-kad\">Joseph Louis Gay-Lussac.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-ZDEwNGNmYjI1MWVmMGE2OGIxYjZlOWM3YTE4YjMyYTI.-r1t\">The mathematical expressions for Gay-Lussac\u2019s law are likewise similar to those of Charles\u2019s law:<\/p>\r\n<p id=\"x-ck12-YmU1ZDVkMzc1NDJkNzVmOTNhODcwOTQ0NTlmNzY2Nzg.-uxd\"><img id=\"x-ck12-MTM2NjYwNjgzMzcwMg..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212142\/5f3e7a33885242475c580acdf25b8f00.png\" alt=\"frac{P}{T}=k quad text{and} quad frac{P_1}{T_1}=frac{P_2}{T_2}\" width=\"182\" height=\"40\" \/><\/p>\r\n<p id=\"x-ck12-NTc3Mzc1YjczNDNjYjIwMTQxMjg5Y2E4OWI0MWI4ZTg.-ylo\">A graph of pressure vs. temperature also illustrates a direct relationship. As a gas is cooled at constant volume its pressure continually decreases until the gas condenses to a liquid.<\/p>\r\n\r\n<h4>Sample Problem: Gay-Lussac\u2019s Law<\/h4>\r\n<p id=\"x-ck12-MDhjZjhkNTNjOWYyODZmMWNhZWNkNjg4MzQyYjRjZWU.-tlw\">The gas in an aerosol can is under a pressure of 3.00 atm at a temperature of 25\u00b0C. It is dangerous to dispose of an aerosol can by incineration. What would the pressure in the aerosol can be at a temperature of 845\u00b0C?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-7eh\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-qtz\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-a6j\">\r\n \t<li><span class=\"x-ck12-underline\"> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212143\/d70729cd610db66b224f2f9beb8d97e8.png\" alt=\"P_1=3.00 text{atm}\" width=\"112\" height=\"16\" \/><\/span><\/li>\r\n \t<li><span class=\"x-ck12-underline\"> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212144\/7ea9fae24f1538e82b6f9bf895af1437.png\" alt=\"T_1=25^circ text{C}=298 text{ K}\" width=\"151\" height=\"17\" \/><\/span><\/li>\r\n \t<li><span class=\"x-ck12-underline\"> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212144\/7cd92943119aa0e329d3c6ae9162c438.png\" alt=\"T_2=845^circ text{C}=1118 text{ K}\" width=\"169\" height=\"16\" \/><\/span><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-v9l\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-dwb\">\r\n \t<li><em> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212145\/362eb69fcff4f8ce64140e9b9d372da6.png\" alt=\"P_2=? text{atm}\" width=\"83\" height=\"15\" \/><\/em><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-YjY3ZTEwNTY3MzY0ODc4NGMzMDM5ZTg1MjQ1ZmQ3Njg.-btn\">Use Gay-Lussac\u2019s law to solve for the unknown pressure <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212128\/7309a9826c8e30de6d5ca051de922d91.png\" alt=\"(P_2)\" width=\"31\" height=\"18\" \/> . The temperatures have first been converted to Kelvin.<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-fjb\"><em> Step 2: Solve. <\/em><\/p>\r\n<p id=\"x-ck12-Yzg1OGU4MDExNzQ5NGQ4OGU4NjY3ZWZlMTFmNDdhNzQ.-q4p\">First, rearrange the equation algebraically to solve for <img id=\"x-ck12-MTM2NjYwNjgzMzcwMw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> .<\/p>\r\n<p id=\"x-ck12-jaz\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212145\/2573f81dac377b98147ae4290c38664d.png\" alt=\"P_2=frac{P_1 times T_2}{T_1}\" width=\"103\" height=\"40\" \/><\/p>\r\n<p id=\"x-ck12-OTNiYzM1Nzk5YWMwYTdmNDZjM2RiOTlmODhhOGVkYTM.-b4c\">Now substitute the known quantities into the equation and solve.<\/p>\r\n<p id=\"x-ck12-mcc\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212146\/6658e7aac6eb46c39a3ae28767d7bf36.png\" alt=\"P_2=frac{3.00 text{ atm} times 1118 text{ K}}{298 text{ K}}=11.3 text{ atm}\" width=\"286\" height=\"39\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-neb\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-ZGIxYzgzMmIyNzc0YzY3NmJlZTA5NTkwYTNjYTcwYzY.-pqv\">The pressure increases dramatically due to large increase in temperature.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-MWIwMzllYzdmOTkyMDMzYjUyYTQ5Mzc0MjAxMTA0OWQ.-hgv\">\r\n \t<li>Pressure and temperature at constant volume are directly proportional.<\/li>\r\n<\/ul>\r\n<h4 id=\"x-ck12-OGMwMDRkN2UzYjhhNWE4NTBhYjI0NTljM2FmYzJmNTc.-wal_4-0rl\">Practice<\/h4>\r\n<p id=\"x-ck12-YjNkMmJlZTZmNzk4YzU3YTNjY2QzNWQ0NWY1YTM0MzM.-2mw\">Work on the problems found at the web site below:<\/p>\r\n<p id=\"x-ck12-YmRlYzMxYmYxY2VlNDBiMWI0ZTA4MGY5MWIyYTAzODU.-gku\"><a href=\"https:\/\/web.archive.org\/web\/20160302060512\/http:\/\/www.chemteam.info\/GasLaw\/WS-Gay-Lussac.html\" target=\"_blank\" rel=\"noopener\">Gay Lussac's Law<\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-znu\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-ODVmNzk2ZDg5NTExYjYzYmI1Yzk5N2M3NDc4ODUxMWU.-0wn\">\r\n \t<li>Explain Gay-Lussac\u2019s Law in terms of the kinetic-molecular theory.<\/li>\r\n \t<li>What would a graph of pressure vs. temperature show us?<\/li>\r\n \t<li>What is the difference in containers in Charles\u2019 Law and Gay-Lussac\u2019s Law?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-ZGYwMjA5OTQxMTEzM2Q4NzMxNGU2ZmNhNWU2ZjUwMjY.-6xc\">\r\n \t<li><strong> Gay-Lussac\u2019s law: <\/strong> The pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-Q29tYmluZWQgR2FzIExhdw..\">Combined Gas Law<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-MTkwN2NhNmM3MDgwY2M4NTQ4MmY2NmM4NjJmMzBhMzA.-nyu\">\r\n \t<li>State the combined gas law.<\/li>\r\n \t<li>Use the law to calculate parameters in general gas problems.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-OTdiZjMzYjQzMWVmNjJjZmQzMjBjNDBjNzhhODZjMGI.-w5q\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212147\/20140811155525236852.jpeg\" alt=\"A refrigerator uses the combined gas law to dissipate heat\" width=\"200\" \/><\/span><\/p>\r\n<p id=\"x-ck12-NjE2OTEwMDhmZmNiZTRhZjc3NDFhYTUzNWRkMmFhZTY.-xdz\"><strong> What keeps things cold? <\/strong><\/p>\r\n<p id=\"x-ck12-MzhkNTk3MjNlZDg5MGFmMzU1YTU1Y2M5MmExZDk3YWE.-bkw\">The modern refrigerator takes advantage of the gas laws to remove heat from a system. Compressed gas in the coils (see above) is allowed to expand. This expansion lowers the temperature of the gas and transfers heat energy from the material in the refrigerator to the gas. As the gas is pumped through the coils, the pressure on the gas compresses it and raises the gas temperature. This heat is then dissipated through the coils into the outside air. As the compressed gas is pumped through the system again, the process repeats itself.<\/p>\r\n\r\n<h3>Combined Gas Law<\/h3>\r\n<p id=\"x-ck12-YTg4YTM1YjZmNWIyNjcyOTU2NmU1OGM2YWQ1YTlkZjQ.-5wq\">To this point, we have examined the relationships between any two of the variables of <img id=\"x-ck12-MTM2NjYwNzg1Mjk4Ng..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/dc2a557e4775924079e5de1061889778.png\" alt=\"P\" width=\"14\" height=\"12\" \/> , <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/adb347871c296be723ff3085c0023a28.png\" alt=\"V\" width=\"14\" height=\"12\" \/> , and <img id=\"x-ck12-MTM2NjYwNzg1Mjk4Nw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/a7284843d100e640c0e00ae83ab66a90.png\" alt=\"T\" width=\"13\" height=\"12\" \/> , while the third variable is held constant. However, situations arise where all three variables change. The <strong> combined gas law <\/strong> expresses the relationship between the pressure, volume, and absolute temperature of a fixed amount of gas. For a combined gas law problem, only the amount of gas is held constant.<\/p>\r\n<p id=\"x-ck12-YmU1ZDVkMzc1NDJkNzVmOTNhODcwOTQ0NTlmNzY2Nzg.-rzv\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212149\/15e5e85314846399443271ebb3f9bcb0.png\" alt=\"frac{P times V}{T}=k quad text{and} quad frac{P_1 times V_1}{T_1}=frac{P_2 times V_2}{T_2}\" width=\"299\" height=\"40\" \/><\/p>\r\n\r\n<h4>Sample Problem: Combined Gas Law<\/h4>\r\n<p id=\"x-ck12-OGI0ZTEyNDJlOTdiZWM1OWQ0N2ViMmU4YmQ1N2I4ODg.-r4p\">2.00 L of a gas at 35\u00b0C and 0.833 atm is brought to standard temperature and pressure (STP). What will be the new gas volume?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-8tv\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-tvg\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-n4j\">\r\n \t<li><em> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212149\/064576e16ae72453b8d008e85e54c48d.png\" alt=\"P_1=0.833 text{ atm}\" width=\"121\" height=\"16\" \/><\/em><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212150\/604efeb1aaaf796728459e3e8d41ca05.png\" alt=\"V_1=2.00 text{ L}\" width=\"91\" height=\"17\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212150\/999d8e43bf34ed87892a395f377a853c.png\" alt=\"T_1=35^circ text{C}=308 text{ K}\" width=\"151\" height=\"17\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212151\/5d24e643f58e553426043959bc772fc0.png\" alt=\"P_2=1.00 text{ atm}\" width=\"112\" height=\"15\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212151\/bf67d672df5dea053630c457b15ee84b.png\" alt=\"T_2=0^circ text{C}=273 text{ K}\" width=\"142\" height=\"16\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-uwj\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-heb\">\r\n \t<li><em> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212152\/51e1944bfa6d5ec4a5ee7088baf64635.png\" alt=\"V_2=? text{ L}\" width=\"62\" height=\"16\" \/><\/em><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-YWFmZDAyZWY3MjlmNTAzMzc5YjVlMjE3MmZlZDg0ZDU.-abl\">Use the combined gas law to solve for the unknown volume <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212137\/0a4b20dd3f9257612d0c459edf0075c2.png\" alt=\"(V_2)\" width=\"30\" height=\"18\" \/> . STP is 273 K and 1 atm. The temperatures have been converted to Kelvin.<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-ex2\"><em> Step 2: Solve <\/em><\/p>\r\n<p id=\"x-ck12-MDBkZWE2Yzk1YTE0NzIyMmExMGE0NmZmYzU1OWI1YjI.-kjc\">First, rearrange the equation algebraically to solve for <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> .<\/p>\r\n<p id=\"x-ck12-9jl\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212152\/d2ebee1bf37edce72fd61d9dc57d6902.png\" alt=\"V_2=frac{P_1 times V_1 times T_2}{P_2 times T_1}\" width=\"143\" height=\"40\" \/><\/p>\r\n<p id=\"x-ck12-OTNiYzM1Nzk5YWMwYTdmNDZjM2RiOTlmODhhOGVkYTM.-zpr\">Now substitute the known quantities into the equation and solve.<\/p>\r\n<img id=\"x-ck12-MTM2NjYwNzg1Mjk4OA..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212152\/f6b309b8131990189c8e66d2820fb70e.png\" alt=\"V_2=frac{0.833 text{ atm} times 2.00 text{ L} times 273 text{ K}}{1.00 text{ atm} times 308 text{ K}}=1.48 text{ L}\" width=\"336\" height=\"39\" \/>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-fsp\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-M2FlNGFlZjk0OTg4NjIwZDczYjg0NzA3MzVjMjI4ZmU.-g7k\">Both the increase in pressure and the decrease in temperature cause the volume of the gas sample to decrease.\u00a0 Since both changes are relatively small, the volume does not decrease dramatically.<\/p>\r\n<p id=\"x-ck12-OWVlY2JmZGU4MTcwYmExOTY1NDIwMDEwYzAzZGUxZmE.-rkt\">It may seem challenging to remember all the different gas laws introduced so far. Fortunately, Boyle\u2019s, Charles\u2019s, and Gay-Lussac\u2019s laws can all be easily derived from the combined gas law. For example, consider a situation where a change occurs in the volume and pressure of a gas while the temperature is being held constant. In that case, it can be said that\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212153\/ad0be0d3264b5c6c9cabf30715cc5e0b.png\" alt=\"T_1 = T_2\" width=\"58\" height=\"16\" \/> . Look at the combined gas law and cancel the\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/a7284843d100e640c0e00ae83ab66a90.png\" alt=\"T\" width=\"13\" height=\"12\" \/> variable out from both sides of the equation. What is left over is Boyle\u2019s law:<\/p>\r\n<p id=\"x-ck12-ZDkwZDdmM2E4ODAyMDBmZTRiOWFiMDUxOGY4MzAwYzg.-wto\"><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212154\/9db530c3c955711edeb1d9ebd41a60ee.png\" alt=\"P_1 times V_1 = P_2 times V_2\" width=\"140\" height=\"16\" \/> . Likewise, if the pressure is constant, then\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212154\/4d1c4ecefed32beff2939628cb3a3ab9.png\" alt=\"P_1 = P_2\" width=\"61\" height=\"16\" \/> and canceling\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/dc2a557e4775924079e5de1061889778.png\" alt=\"P\" width=\"14\" height=\"12\" \/> out of the equation leaves Charles\u2019s law. If the volume is constant, then\u00a0 <img id=\"x-ck12-MTM2NjYwNzg1Mjk4OQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212154\/de70d3338a1bb694e07d1b31ad411ad4.png\" alt=\"V_1 = V_2\" width=\"58\" height=\"16\" \/> and canceling\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/adb347871c296be723ff3085c0023a28.png\" alt=\"V\" width=\"14\" height=\"12\" \/> out of the equation leaves Gay-Lussac\u2019s law.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-YTVhNmMyZDEyZDA3ZGJlYWEzMWQwYzEwZmZlNmMzNzI.-imk\">\r\n \t<li>The combined gas law shows the relationships among temperature, volume, and pressure.<\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212155\/6568f29084a4b11bb32a0ff5d5e40b8f.png\" alt=\"frac{P_1V_1}{T_1} = frac{P_2V_2}{T_2}\" width=\"85\" height=\"25\" \/><\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-ZGIyN2QxMzI1ODI1MWRlNWVlMjE5YTlhNmM5NDJlN2M.-fqy\">Work on the problems at the link below:<\/p>\r\n<p id=\"x-ck12-YzNhMDI4NGExMjlmY2QzODE1ZGJiMzA4OTEwNDQ5MWE.-o8v\"><a href=\"http:\/\/misterguch.brinkster.net\/WKS001_007_146637.pdf\"> http:\/\/misterguch.brinkster.net\/WKS001_007_146637.pdf <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-b7i\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-NDZmM2ZiYmMzYzljYjVlMGY0OWUxMzBhYzZiNmE4OGY.-dgp\">\r\n \t<li>What is the only thing held constant in a combined gas law problem?<\/li>\r\n \t<li>If you want to solve for the volume of a gas\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212137\/0a4b20dd3f9257612d0c459edf0075c2.png\" alt=\"(V_2)\" width=\"30\" height=\"18\" \/> and\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/b95fd9f12605d93cbb24fd5fc1d83fbc.png\" alt=\"P_1\" width=\"18\" height=\"16\" \/> is greater than <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/f49ec791ef1ecb63d1c14840f9c2f983.png\" alt=\"P_2\" width=\"18\" height=\"15\" \/> , would you expect\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> to be larger or smaller than <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/29c883aa1a216bc9157974e94c993965.png\" alt=\"V_1\" width=\"16\" height=\"16\" \/> ?<\/li>\r\n \t<li>What would be the equation for finding\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/f49ec791ef1ecb63d1c14840f9c2f983.png\" alt=\"P_2\" width=\"18\" height=\"15\" \/> given all the other parameters?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-ZTU5Y2NmZmU0MTZkODQ5Y2M1ZDQ3OGYwNzA2Y2IzMTk.-vsu\">\r\n \t<li><strong> combined gas law: <\/strong> Expresses the relationship between the pressure, volume, and absolute temperature of a fixed amount of gas.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-QXZvZ2Fkcm8ncyBMYXc.\">Avogadro's Law<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-ZWQ0Y2E1ZjgwYzFmZDQ1MzM4YzdmMmU1MzhlYTBlYTc.-pde\">\r\n \t<li>State Avogadro\u2019s Law.<\/li>\r\n \t<li>Use this law to perform calculations involving quantities of gases.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-MGM2Zjg3ZGYyNjVmZTk4Yjc4NDBjZTAxMzNiZmFjMWE.-du0\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212155\/20140811155525460045.jpeg\" alt=\"Avogadro's Law tells us that putting gas into a tire increases its pressure\" width=\"200\" \/><\/span><\/p>\r\n<p id=\"x-ck12-ZjFjODM1NmVlOGVkY2M3MGUwOWRiZTYyZTMwZTE0Zjg.-p6v\"><strong> How much air do you put into a tire? <\/strong><\/p>\r\n<p id=\"x-ck12-NTZhOWYyNDA0YzNiYzBhMGEyZTI0ZjQwNTdhNWM5MDE.-5qy\">A flat tire is not very useful. It does not cushion the rim of the wheel and creates a very uncomfortable ride. When air is added to the tire, the pressure increases as more molecules of gas are forced into the rigid tire. How much air should be put into a tire depends on the pressure rating for that tire. Too little pressure and the tire will not hold its shape. Too much pressure and the tire could burst.<\/p>\r\n\r\n<h3>Avogadro\u2019s Law<\/h3>\r\n<p id=\"x-ck12-YTlmZDk4ZTcxNjViYzY2OGExYjkxYWNhZTQxMjA1Nzk.-mjv\">You have learned about Avogadro\u2019s hypothesis: equal volumes of any gas at the same temperature and pressure contain the same number of molecules. It follows that the volume of a gas is directly proportional to the number of moles of gas present in the sample. <strong> Avogadro\u2019s law <\/strong> states that the volume of a gas is directly proportional to the number of moles of gas when the temperature and pressure are held constant. The mathematical expression of Avogadro\u2019s law is<\/p>\r\n<p id=\"x-ck12-YmU1ZDVkMzc1NDJkNzVmOTNhODcwOTQ0NTlmNzY2Nzg.-ea0\"><img id=\"x-ck12-MTM2NjYwODI3MzEzNw..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212156\/4557cd2f339a4ea7c168510243bd9c8a.png\" alt=\"V=k times n quad text{and} quad frac{V_1}{n_1}=frac{V_2}{n_2}\" width=\"211\" height=\"40\" \/><\/p>\r\n<p id=\"x-ck12-NzVhMWE4YmNlYmRhZDIzNDM0NTI3OWRlYzE3YWU5Njk.-loh\">where\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211002\/c73c4e5aa4eb7c39977bbf98a8cd212a.png\" alt=\"n\" width=\"11\" height=\"8\" \/> is the number of moles of gas and\u00a0 <img id=\"x-ck12-MTM2NjYwODI3MzEzOA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212122\/d301c6452e642017007fb7cd755ce08b.png\" alt=\"k\" width=\"9\" height=\"13\" \/> is a constant. Avogadro\u2019s law is in evidence whenever you blow up a balloon. The volume of the balloon increases as you add moles of gas to the balloon by blowing it up.<\/p>\r\n<p id=\"x-ck12-OTBmODhmYWJjNDJmYTEwYmVhNmRmZTM2ZmQyYmQzZTQ.-x6g\">If the container holding the gas is rigid rather than flexible, pressure can be substituted for volume in Avogadro\u2019s law. Adding gas to a rigid container makes the pressure increase.<\/p>\r\n\r\n<h4>Sample Problem: Avogadro\u2019s Law<\/h4>\r\n<p id=\"x-ck12-YTI1YjgyMmIxMTkyZDFlNWQwYWFmZjdmY2ViZWYzOTA.-2ab\">A balloon has been filled to a volume of 1.90 L with 0.0920 mol of helium gas. If 0.0210 mol of additional helium is added to the balloon while the temperature and pressure are held constant, what is the new volume of the balloon?<\/p>\r\n<p id=\"x-ck12-OGU2YWEyYzY0NzAwMDUxZjI4NjFjY2E4MjYyNmNhN2I.-ydl\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-iyb\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-dey\">\r\n \t<li><em> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212157\/b18c8945f2039f0506b418c4c5664e8d.png\" alt=\"V_1=1.90 text{ L}\" width=\"91\" height=\"17\" \/><\/em><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212157\/b9491744c2ff6024c2a1498e31574164.png\" alt=\"n_1=0.0920 text{ mol}\" width=\"127\" height=\"17\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212158\/f3d75b46fb576905ae99c18e03cde536.png\" alt=\"n_2=0.0920+0.0210=0.1130 text{ mol}\" width=\"272\" height=\"16\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-hot\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-1iu\">\r\n \t<li><em> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212152\/51e1944bfa6d5ec4a5ee7088baf64635.png\" alt=\"V_2=? text{ L}\" width=\"62\" height=\"16\" \/><\/em><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ZTMyNDVhMzcwNTgwYzQ3ZjExY2IwNDhlNGY2OTQ2N2I.-jw1\">Note that the final number of moles has to be calculated by adding the original number of moles to the moles of added helium. Use Avogadro\u2019s law to solve for the final volume.<\/p>\r\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-fpw\"><em> Step 2: Solve. <\/em><\/p>\r\n<p id=\"x-ck12-Yzg1OGU4MDExNzQ5NGQ4OGU4NjY3ZWZlMTFmNDdhNzQ.-nxw\">First, rearrange the equation algebraically to solve for <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> .<\/p>\r\n<p id=\"x-ck12-8pd\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212158\/b1eab62cefeb445a02b827e2c1fcffee.png\" alt=\"V_2=frac{V_1 times n_2}{n_1}\" width=\"101\" height=\"40\" \/><\/p>\r\n<p id=\"x-ck12-OTNiYzM1Nzk5YWMwYTdmNDZjM2RiOTlmODhhOGVkYTM.-ghy\">Now substitute the known quantities into the equation and solve.<\/p>\r\n<p id=\"x-ck12-mkt\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212159\/ec3e3cd202cf649f2a46e280d3d59cca.png\" alt=\"V_2=frac{1.90 text{ L} times 0.1130 text{ mol}}{0.0920 text{ mol}}=2.33 text{ L}\" width=\"274\" height=\"39\" \/><\/p>\r\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-wu4\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-NDAxYzUyZmMxYmFlYWQyMTI3MGVmYTc1MWI5N2JlYWQ.-2jc\">Since a relatively small amount of additional helium was added to the balloon, its volume increases slightly.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-ODJkMGZkZWU1YWM0ZjVmNGNiYzFkMTlkNmYyZTA1MDc.-msg\">\r\n \t<li>Calculations are shown for relationships between volume and number of moles of a gas.<\/li>\r\n<\/ul>\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-idx\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-NWE1NzA1NzUzYmNiNjE0N2UwNWU5NjgxZmZhMjg4MDY.-daw\">\r\n \t<li>What is held constant in the Avogadro\u2019s Law relationship?<\/li>\r\n \t<li>What happens if you add gas to a rigid container?<\/li>\r\n \t<li>Why does a balloon expand when you add air to it?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-MDhhMmZiMjM4OTI1ZDM2N2I4ZjBhNTI4YjVkMTY1OTU.-tiw\">\r\n \t<li><strong> Avogadro\u2019s law: <\/strong> The volume of a gas is directly proportional to the number of moles of gas when the temperature and pressure are held constant.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-SWRlYWwgR2FzIExhdw..\">Ideal Gas Law<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-MGVlOTkwZTY4MjU5NzExYTBmZjQxYmVjYTZiNjNiYjQ.-8xy\">\r\n \t<li>Derive the ideal gas law from the combined gas law and Avogadro\u2019s law.<\/li>\r\n \t<li>Calculate the value of the ideal gas constant.<\/li>\r\n \t<li>Use the ideal gas law to calculate parameters for ideal gases.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-OTRiNjdiZWVlZjZjMGZhOTljMDczNjViNjUyOTRjMWY.-3v1\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212200\/20140811155525625008.jpeg\" alt=\"The ideal gas law can be used to determine the amount of gas present in a system\" width=\"100\" \/><\/span><\/p>\r\n<p id=\"x-ck12-MmIyZDljNjI4YmRlZWQwYTZhZjc1YTMzYjBlZmQ4YzA.-yxd\"><strong> What chemical reactions require ammonia? <\/strong><\/p>\r\n<p id=\"x-ck12-MmI1M2Q5YjRlYzE4MGY4MjJiYTZlYWE3OGQ5M2YwZmU.-wvt\">There are a number of chemical reactions that require ammonia. In order to carry out the reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system.<\/p>\r\n\r\n<h3>Ideal Gas Law<\/h3>\r\n<p id=\"x-ck12-ZTg3NzkxMzViZWJhNDIxNDhiZTcyNjFiMDJjMDdiMTA.-ed4\">The combined gas law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro\u2019s law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these together leaves us with the following equation:<\/p>\r\n<p id=\"x-ck12-wr3\"><img id=\"x-ck12-MTM2NjYwOTEyODc4OQ..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212203\/b18bf265809dd0c3b00092aeae02977d.png\" alt=\"frac{P_1 times V_1}{T_1 times n_1}=frac{P_2 times V_2}{T_2 times n_2}\" width=\"145\" height=\"40\" \/><\/p>\r\n<p id=\"x-ck12-ZDlmYjBhZTQ3MmRiOWRiYmNjZDFjMTAxYmUwNjBmNzk.-0fr\">As with the other gas laws, we can also say that\u00a0 <img id=\"x-ck12-MTM2NjYwOTEyODc5MA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212203\/426d96bd9cee0ae6f889fd1cee17c81a.png\" alt=\"frac{left(P times V right)}{left(T times n right)}\" width=\"41\" height=\"29\" \/> is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal.<\/p>\r\n<p id=\"x-ck12-NTcyNzU3YTgwN2UxMmQ1NjllNzcyMzZkMTQ1OGEyNmU.-5wf\">The <strong> ideal gas law <\/strong> is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> for the constant, the equation becomes:<\/p>\r\n<p id=\"x-ck12-7qo\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/1a80d20b13965449f86ff0bd84c737d8.png\" alt=\"frac{P times V}{T times n}=R\" width=\"91\" height=\"37\" \/><\/p>\r\n<p id=\"x-ck12-MGIwODRkZDIxMzlmZWMyYTM1MmI0YzBiYTRhNzQ5MmY.-ht0\">The ideal gas law is conventionally rearranged to look this way, with the multiplication signs omitted:<\/p>\r\n<p id=\"x-ck12-13m\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/11cf69a276a7ad759359c56f0512e141.png\" alt=\"PV=nRT\" width=\"90\" height=\"12\" \/><\/p>\r\n<p id=\"x-ck12-NTZmMmI4YzU0M2RjNmY0YzE4MDFiOGExMTlmYmEyNjE.-z5o\">The variable\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> in the equation is called the <strong> ideal gas constant <\/strong> .<\/p>\r\n\r\n<h4>Evaluating the Ideal Gas Constant<\/h4>\r\n<p id=\"x-ck12-NDU1YzQ4NjBlZjA3ZDE4MzI0ZWU3ODgyNTUwNjc1NDY.-fwa\">The value of <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> , the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. However, pressure is commonly measured in one of three units: kPa, atm, or mmHg. Therefore,\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> can have three different values.<\/p>\r\n<p id=\"x-ck12-ZTk0N2FlYWJkYmRmZTNmZjM4ZmE5NjgxOTExMTc3N2E.-p0x\">We will demonstrate how\u00a0 <img id=\"x-ck12-MTM2NjYwOTEyODc5MQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> is calculated when the pressure is measured in kPa. Recall that the volume of 1.00 mol of any gas at STP is measured to be 22.414 L. We can substitute 101.325 kPa for pressure, 22.414 L for volume, and 273.15 K for temperature into the ideal gas equation and solve for <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> .<\/p>\r\n<img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212205\/c0037c63a8c77edb80a7778550aee404.png\" alt=\"R=frac{PV}{nT}=frac{101.325 text{kPa} times 22.414 text{ L}}{1.000 text{mol} times 273.15 text{ K}}=8.314 text{kPa} cdot text{L\/K} cdot text{mol}\" width=\"472\" height=\"39\" \/>\r\n<p id=\"x-ck12-YmUyYzJhZWZkODU3MTA4NGUxOTcwNGVlNzYyMzI0ODI.-tgm\">This is the value of\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> that is to be used in the ideal gas equation when the pressure is given in kPa. The <strong> Table <\/strong> below shows a summary of this and the other possible values of <img id=\"x-ck12-MTM2NjYwOTEyODc5Mg..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> . It is important to choose the correct value of\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> to use for a given problem.<\/p>\r\n\r\n<table id=\"x-ck12-ZmVlMThiZTFmNTVkMjg4ZGYzY2UxNGE5NTQyYjllNzU.-br6\" class=\"x-ck12-nofloat\" border=\"1\"><caption>Values of the Ideal Gas Constant<\/caption>\r\n<tbody>\r\n<tr>\r\n<td><strong> Unit of <\/strong> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/dc2a557e4775924079e5de1061889778.png\" alt=\"P\" width=\"14\" height=\"12\" \/><strong>\r\n<\/strong><\/td>\r\n<td><strong> Unit of <\/strong> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/adb347871c296be723ff3085c0023a28.png\" alt=\"V\" width=\"14\" height=\"12\" \/><\/td>\r\n<td><strong> Unit of <\/strong> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211002\/c73c4e5aa4eb7c39977bbf98a8cd212a.png\" alt=\"n\" width=\"11\" height=\"8\" \/><strong>\r\n<\/strong><\/td>\r\n<td><strong> Unit of <\/strong> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/a7284843d100e640c0e00ae83ab66a90.png\" alt=\"T\" width=\"13\" height=\"12\" \/><\/td>\r\n<td><strong> Value and unit of <\/strong> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>kPa<\/td>\r\n<td>L<\/td>\r\n<td>mol<\/td>\r\n<td>K<\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212207\/d9e6f5f6e76170eb98b5b1f0eaed65ee.png\" alt=\"8.314 text{J\/K} cdot text{mol}\" width=\"121\" height=\"18\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>atm<\/td>\r\n<td>L<\/td>\r\n<td>mol<\/td>\r\n<td>K<\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212208\/5f341d284dec12741f5809ebf5477684.png\" alt=\"0.08206 text{L} cdot text{atm\/K} cdot text{mol}\" width=\"185\" height=\"18\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>mmHg<\/td>\r\n<td>L<\/td>\r\n<td>mol<\/td>\r\n<td>K<\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212208\/65451fb51aff398ea23a71bd4256f85d.png\" alt=\"62.36 text{ L} cdot text{mmHg\/K} cdot text{mol}\" width=\"188\" height=\"18\" \/><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"x-ck12-Mjg1N2M4YzMwZTNiYzhjNjI3Yjg0MWZkODFlZGM1Njg.-xv1\">Notice that the unit for\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> when the pressure is in kPa has been changed to J\/K\u00a0\u2022\u00a0mol. A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule (J).<\/p>\r\n\r\n<h4>Sample Problem: Ideal Gas Law<\/h4>\r\n<p id=\"x-ck12-MTQyMTY0NDY2ZmRkMjFiMDY2OTNhYWY1YjEzYmMxODQ.-tlt\">What volume is occupied by 3.760 g of oxygen gas at a pressure of 88.4 kPa and a temperature of 19\u00b0C? Assume the oxygen is ideal.<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-nct\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-fhn\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-cy0\">\r\n \t<li><em> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212209\/55b6d82cc42bc408a485e0f528f06d6f.png\" alt=\"P = 88.4 text{kPa}\" width=\"106\" height=\"14\" \/><\/em><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212210\/05e91eb9d11d1c8050952111f84da1c7.png\" alt=\"T = 19^circ text{C} = 292 text{ K}\" width=\"146\" height=\"14\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212210\/443c32746f0a9dbd4aac11fda6a45aa2.png\" alt=\"text{mass} O_2 = 3.760 text{ g}\" width=\"145\" height=\"16\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212211\/5c54eeabf3c70df08cb455f8a8907ef7.png\" alt=\"O_2 = 32.00 text{ g\/mol}\" width=\"139\" height=\"18\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212211\/4707dcbdc7b4220c7b11dc57c8d97f93.png\" alt=\"R = 8.314 text{ J\/K} cdot text{mol}\" width=\"159\" height=\"18\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-bkm\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ysq\">\r\n \t<li><em> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212212\/0657dfe2c2c1e6586aa37f082316afef.png\" alt=\"V = ? text{ L}\" width=\"59\" height=\"13\" \/><\/em><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-MGE4NmVlNDJlMGFkMTQ4Y2I4NjRkMTEzY2VmY2JlZGI.-iae\">In order to use the ideal gas law, the number of moles of O <sub> 2 <\/sub> \u00a0 <img id=\"x-ck12-MTM2NjYwOTEyODc5Mw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211023\/4168b7a35bc24c982496aef5b556a0a5.png\" alt=\"(n)\" width=\"23\" height=\"18\" \/> must be found from the given mass and the molar mass. Then, use\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212212\/b5a071ec470437e7416a7e999a24b3fc.png\" alt=\"PV = nRT\" width=\"90\" height=\"12\" \/> to solve for the volume of oxygen.<\/p>\r\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-afs\"><em> Step 2: Solve <\/em> .<\/p>\r\n<p id=\"x-ck12-sdp\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212213\/a7682991a7f0112a32d04072b051fdaf.png\" alt=\"3.760 text{ g} times frac{1 text{mol} O_2}{32.00 text{ g} O_2}=0.1175 text{mol} O_2\" width=\"299\" height=\"41\" \/><\/p>\r\n<p id=\"x-ck12-MDZhOTZjMmFlOTE5MGFjMTE0Y2UxYjQ3OTYyN2RkZmY.-pqw\">Rearrange the ideal gas law and solve for <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/adb347871c296be723ff3085c0023a28.png\" alt=\"V\" width=\"14\" height=\"12\" \/> .<\/p>\r\n<p id=\"x-ck12-wh2\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212214\/a9fa443dfe97998bf81bc7d37b0788f8.png\" alt=\"V=frac{nRT}{P}=frac{0.1175 text{mol} times 8.314 text{ J\/K} cdot text{mol} times 292 text{ K}}{88.4 text{kPa}}=3.23 text{ L } O_2\" width=\"503\" height=\"40\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-vm9\"><em> Step 3: Think about your result <\/em><\/p>\r\n<p id=\"x-ck12-YWNmNTQwMjBmOGQxNjJhN2NmYjQxZmJkOTk3ZTNlNzg.-bie\">The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume (22.4 L\/mol) since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/a7284843d100e640c0e00ae83ab66a90.png\" alt=\"T\" width=\"13\" height=\"12\" \/> and <img id=\"x-ck12-MTM2NjYwOTEyODc5NA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/dc2a557e4775924079e5de1061889778.png\" alt=\"P\" width=\"14\" height=\"12\" \/> . Since a joule (J)\u00a0=\u00a0kPa\u00a0\u2022\u00a0L, the units cancel correctly, leaving a volume in liters.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-ZTk5OTRiY2YyZmNiYTAzODY5MzhjNjI1NThlOWNmZDc.-h8x\">\r\n \t<li>The ideal gas constant is calculated.<\/li>\r\n \t<li>An example of calculations using the ideal gas law is shown.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-ZGIyN2QxMzI1ODI1MWRlNWVlMjE5YTlhNmM5NDJlN2M.-blq\">Work on the problems at the link below:<\/p>\r\n<p id=\"x-ck12-NmY2MDNmYTFhOGM4N2Y4MjJiNmU5MjczMTFhYzU0NDk.-qpg\"><a href=\"http:\/\/chemsite.lsrhs.net\/gasses\/handouts\/Ideal_Problems.pdf\"> http:\/\/chemsite.lsrhs.net\/gasses\/handouts\/Ideal_Problems.pdf <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-dmh\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-MDBkYTM5MGMzOWJmOGVmNGZjNGQxNWRkOTA5M2JhMzQ.-654\">\r\n \t<li>Which value of\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> will you use if the pressure is given in atm?<\/li>\r\n \t<li>You are doing a calculation where the pressure is given in mm Hg. You select 8.314\u00a0J\/K\u00a0\u2022\u00a0mol as your value for <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> . Will you get a correct answer?<\/li>\r\n \t<li>How would you check that you have chosen the correct value of\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> for your problem?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-ZTYwZTU0MTExMTZhOGUzNDQ0NzkwNmJlOGRkMTE5N2M.-hkb\">\r\n \t<li><strong> ideal gas constant: <\/strong> The variable\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212215\/e096e3ca730e7ebae2c2660b5801dea8.png\" alt=\"R\" width=\"14\" height=\"12\" \/> in the ideal gas law equation.<\/li>\r\n \t<li><strong> ideal gas law: <\/strong> A single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-Q2FsY3VsYXRpbmcgdGhlIE1vbGFyIE1hc3Mgb2YgYSBHYXM.\">Calculating the Molar Mass of a Gas<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-Y2E4MjBhYmZjMTllOGEwNTIwNWQ5Nzg2MDFkNDEyMTk.-zy4\">\r\n \t<li>Calculate the molar mass of a gas.<\/li>\r\n \t<li>Calculate the density of a gas.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-YWExZWY3ZGZiYjViZGY3YWEwOWUwYWYxMmUxYzliOGM.-9uq\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212215\/20140811155525773654.jpeg\" alt=\"Gases with low molar masses such as hydrogen and helium can be used to float balloons\" width=\"200\" \/><\/span><\/p>\r\n<p id=\"x-ck12-ODZiYWMwYjZjOWViMDUzN2QyMWVmYWViZjk2NjA4MTU.-gio\"><strong> What makes it float? <\/strong><\/p>\r\n<p id=\"x-ck12-ZTQwNTI2NzQ2YjQ2MmYzYmZkMWVkODg1ZmVkYTJlODM.-s8i\">Helium has long been used in balloons and blimps. Since it is much less dense than air, it will float above the ground. We can buy small balloons filled with helium at stores, but large ones (such as the balloon seen above) are much more expensive and take up a lot more helium.<\/p>\r\n\r\n<h3>Calculating Molar Mass and Density of a Gas<\/h3>\r\n<p id=\"x-ck12-M2ZjZTlhNGQzNzU5YmIyMmU3ZmU0YmExMDQ3YzYyODg.-yuh\">A chemical reaction, which produces a gas, is performed. The produced gas is then collected and its mass and volume are determined. The molar mass of the unknown gas can be found using the ideal gas law, provided the temperature and pressure of the gas are also known.<\/p>\r\n\r\n<h4>Sample Problem: Molar Mass and the Ideal Gas Law<\/h4>\r\n<p id=\"x-ck12-MjkyOTkzODk5MGJjMjg5ZmEwZjkwMTY0YjRmMDUyZTQ.-sed\">A certain reaction occurs, producing an oxide of nitrogen as a gas. The gas has a mass of 1.211 g and occupies a volume of 677 mL. The temperature in the laboratory is 23\u00b0C and the air pressure is 0.987 atm. Calculate the molar mass of the gas and deduce its formula. Assume the gas is ideal.<\/p>\r\n<p id=\"x-ck12-OGU2YWEyYzY0NzAwMDUxZjI4NjFjY2E4MjYyNmNhN2I.-vxd\"><em> Step 1: List the known quantities and plan the problem <\/em> .<\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-em0\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-u3m\">\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212217\/15ee7ca6617ec89c8e9b6af4c30f8368.png\" alt=\"text{mass}=1.211text{ g}\" width=\"118\" height=\"15\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212218\/0938b476f561ff3e3eb8105649e1de84.png\" alt=\"V = 677 text{ ml} = 0.677 text{ L}\" width=\"173\" height=\"14\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212218\/3bab40e282f77f715cc5a084ba006150.png\" alt=\"T= 23^circ text{C}=296 text{ K}\" width=\"146\" height=\"14\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212219\/fcc8aab2b57db7f9d6d576ef7ff81f0f.png\" alt=\"P =0.987 text{atm}\" width=\"116\" height=\"14\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212219\/0338a5ae6c992b4ab8e04603d609fee2.png\" alt=\"R =0.08206 text{ L} cdot text{atm\/K} cdot text{mol}\" width=\"223\" height=\"18\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-rcu\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-3dm\">\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212220\/c528206210b32c4e36374e35fe2432f1.png\" alt=\"n= ? text{mol}\" width=\"73\" height=\"14\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212221\/25947cbe6aff86043fd47cc71379fe36.png\" alt=\"text{molar mass} = ? text{g\/mol}\" width=\"169\" height=\"18\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-OTc0Yzk5MTVlYmQ4YWViMjAxOWYzZWQwZmY2NTMxNzg.-mlj\">First the ideal gas law will be used to solve for the moles of unknown gas <img id=\"x-ck12-MTM2NjYxMDQ5NjY2OA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211023\/4168b7a35bc24c982496aef5b556a0a5.png\" alt=\"(n)\" width=\"23\" height=\"18\" \/> . Then the mass of the gas divided by the moles will give the molar mass.<\/p>\r\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-njd\"><em> Step 2: Solve <\/em> .<\/p>\r\n<p id=\"x-ck12-3wi\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212221\/c03079f46a2fff832953b85b039e9ee8.png\" alt=\"n=frac{PV}{RT}=frac{0.987 text{atm} times 0.677 text{ L}}{0.08206 text{ L} cdot text{atm\/K} cdot text{mol} times 296 text{ K}}=0.0275 text{mol}\" width=\"457\" height=\"42\" \/><\/p>\r\n<p id=\"x-ck12-NzdlN2U4YjljYjdkYjdjZThkMDgwN2VlY2NjOTBmMTE.-6om\">Now divide g by mol to get the molar mass.<\/p>\r\n<p id=\"x-ck12-gau\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212223\/6f092e645d73d411fbd30dffab7aa0ee.png\" alt=\"text{molar mass}=frac{1.211 text{ g}}{0.0275 text{mol}}=44.0 text{g\/mol}\" width=\"309\" height=\"38\" \/><\/p>\r\n<p id=\"x-ck12-NzlmMDhlMGI3NDhiODcyY2FlNTNjYzJmNTI1Zjc4Mjc.-7sg\">Since N has a molar mass of 14 g\/mol and O has a molar mass of 16 g\/mol, the formula N <sub> 2 <\/sub> O would produce the correct molar mass.<\/p>\r\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-jqn\"><em> Step 3: Think about your result <\/em><\/p>\r\n<p id=\"x-ck12-Yjg1MWQ1YzEwMWEyZDI5NTgxMjZkMTY4MjYzYzljMmI.-ghd\">The\u00a0 <img id=\"x-ck12-MTM2NjYxMDQ5NjY2OQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> value that corresponds to a pressure in atm was chosen for this problem. The calculated molar mass gives a reasonable formula for dinitrogen monoxide.<\/p>\r\n\r\n<h4>Calculating Density of a Gas<\/h4>\r\n<p id=\"x-ck12-NTJjODQ4YjIzOGRiYTdlMzZmODcxZTU0NTM4NjIyN2M.-xz1\">The ideal gas law can be used to find the density of a gas at conditions that are not standard. For example, we will determine the density of ammonia gas (NH <sub> 3 <\/sub> ) at 0.913 atm and 20\u00b0C, assuming the ammonia is ideal. First, the molar mass of ammonia is calculated to be 17.04 g\/mol. Next, assume exactly 1 mol of ammonia\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212224\/a357460ec50f481e8810da93911a02f7.png\" alt=\"(n = 1)\" width=\"56\" height=\"18\" \/> and calculate the volume that such an amount would occupy at the given temperature and pressure.<\/p>\r\n<p id=\"x-ck12-7v6\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212224\/bae49338d0e06b594a1d1fb981dcc822.png\" alt=\"V = frac{ nRT}{P}=frac{1.00 text{mol} times 0.08206 text{ L} cdot text{atm\/K}cdot text{mol} times 293 text{ K}}{0.913 text{atm}}=26.3 text{ L}\" width=\"522\" height=\"40\" \/><\/p>\r\n<p id=\"x-ck12-NDQ1OWIzM2EzYzgyMTc4MjRkMGRhN2M5NzhiNTIxZDA.-dtf\">Now the density can be calculated by dividing the mass of one mole of ammonia by the volume above.<\/p>\r\n<p id=\"x-ck12-mnz\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212226\/89d5d4933d53b9bc38998e4cccf6822f.png\" alt=\"text{density}=frac{17.04 text{ g}}{26.3 text{ L}}=0.647 text{g\/L}\" width=\"239\" height=\"38\" \/><\/p>\r\n<p id=\"x-ck12-MmVjYTcxZjNjYTU1NDc3ZTJjYjAwZDcwYTI0ZGQwOTc.-nlm\">As a point of comparison, this density is slightly less than the density of ammonia at STP, which is equal to <img id=\"x-ck12-MTM2NjYxMDQ5NjY3Ng..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212227\/8001f341f27809461141f496b63aa091.png\" alt=\"frac{(17.04 text{g\/mol})}{(22.4 text{L\/mol})} = 0.761 text{g\/L}\" width=\"181\" height=\"29\" \/> . It makes sense that the density should be lower compared to that at STP since both the increase in temperature (from 0\u00b0C to 20\u00b0C) and the decrease in pressure (from 1 atm to 0.913 atm) would cause the NH <sub> 3 <\/sub> molecules to spread out a bit further from one another.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-ZmFkMmQzNTEyNjIyZGM0NWE1OGQxMmJiOGEyZTUxZjQ.-ldq\">\r\n \t<li>Calculations of molar mass and density of an ideal gas are described.<\/li>\r\n<\/ul>\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-ijl\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-MjhiZWM4OWM1MTIzZTFiNDgwMzBmNGM3ODU4ZGUwMGQ.-b6d\">\r\n \t<li>Why do you need the volume, temperature, and pressure of the gas to calculate molar mass?<\/li>\r\n \t<li>What assumption about the gas is made in all these calculations?<\/li>\r\n \t<li>Why do you need the mass of the gas to calculate the molar mass?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\"><\/div>\r\n<h1 id=\"x-ck12-R2FzIFN0b2ljaGlvbWV0cnk.\">Gas Stoichiometry<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-YjIwODVmZTdjMmMyZDBjNzMyNDdlZWQzNmU4ZjM3MzU.-t5r\">\r\n \t<li>Use the ideal gas law to calculate stoichiometry problems for gases.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-NzhiODljMzE5YzczYzUzMjdjNDU2NzMyNzkyNDFiMDU.-qrz\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212227\/20140811155525942661.jpeg\" alt=\"Knowing the amount of a gas needed is important for many applications\" width=\"400\" \/><\/span><\/p>\r\n\r\n<h4><span class=\"x-ck12-img-inline\"> How is fertilizer produced?\r\n<\/span><\/h4>\r\n<p id=\"x-ck12-OWI3NjFjMTI3NmNhZWExMzYyMTc3OWNlN2MxYWE3YWE.-fdi\">The Haber cycle reaction of gaseous nitrogen and hydrogen to form ammonia is a critical step in the production of fertilizer from ammonia. It is important to have an excess of the starting materials so that a maximum yield of ammonia can be achieved. By knowing how much ammonia is needed for manufacture of a batch of fertilizer, the proper amounts of nitrogen and hydrogen gases can be incorporated into the process.<\/p>\r\n\r\n<h3>Gas Stoichiometry<\/h3>\r\n<p id=\"x-ck12-ZGNjYmJhM2NlMTZhOWYyYjc4NjAzNjgzMDczOGI3ODU.-c5l\">You have learned how to use molar volume to solve stoichiometry problems for chemical reactions involving one or more gases at STP. Now, we can use the ideal gas law to expand our treatment of chemical reactions to solve stoichiometry problems for reactions that occur at any temperature and pressure.<\/p>\r\n\r\n<h4>Sample Problem: Gas Stoichiometry and the Ideal Gas Law<\/h4>\r\n<p id=\"x-ck12-OWJiMTg0MGY5ZWQ3MWVkMGI4ZmY4Y2QxYTNmY2ExZGY.-b5o\">What volume of carbon dioxide is produced by the combustion of 25.21 g of ethanol (C <sub> 2 <\/sub> H <sub> 5 <\/sub> OH)\u00a0at 54\u00b0C and 728 mmHg? Assume the gas is ideal.<\/p>\r\n<p id=\"x-ck12-ODg1ODBlZjhiMTIwZjJjM2Q5NmJmODc4MDNlZDRhYTA.-ls2\">Before using the ideal gas law, it is necessary to write and balance the chemical equation. Recall that most combustion reactions, the given substance reacts with O <sub> 2 <\/sub> to form CO <sub> 2 <\/sub> and H <sub> 2 <\/sub> O. Here is the balanced equation for the combustion of ethanol.<\/p>\r\n<img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212229\/da8532c253d207d89961bb44eda89513.png\" alt=\"text{C}_2text{H}_5text{OH}(l)+3text{O}_2(g) rightarrow 2text{CO}_2(g)+3text{H}_2text{O}(l)\" width=\"342\" height=\"18\" \/>\r\n<p id=\"x-ck12-ZWRlZGNhY2I2ZjdkODlhOWVlMDkxNTRmNzlmMjUxZDI.-uvx\"><em> Step 1: List the known quantities and solve the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-sbi\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-hup\">\r\n \t<li><span class=\"x-ck12-underline\"> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212230\/8895b9609b51970be97e9f8aedb863dc.png\" alt=\"text{mass} text{C}_2text{H}_5text{OH}=25.21 text{ g}\" width=\"192\" height=\"16\" \/><\/span><\/li>\r\n \t<li><span class=\"x-ck12-underline\"> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212231\/8c8f86b472bd9d1c10b3d7e3bda54cee.png\" alt=\"text{molar mass} text{C}_2text{H}_5text{OH}=46.08 text{ g\/mol}\" width=\"281\" height=\"18\" \/><\/span><\/li>\r\n \t<li><span class=\"x-ck12-underline\"> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212232\/bdd8d0aab73462de83ee60a33c299f81.png\" alt=\"P=728 text{ mmHg}\" width=\"123\" height=\"16\" \/><\/span><\/li>\r\n \t<li><span class=\"x-ck12-underline\"> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212233\/816a80e184ef6694c5135dea0d517288.png\" alt=\"T=54^circ text{C}=327 text{ K}\" width=\"146\" height=\"14\" \/><\/span><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-bx5\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-nts\">\r\n \t<li><em> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212233\/a8c85ab5d338790007e8d3caf51b2348.png\" alt=\"text{Volume} text{CO}_2=? text{ L}\" width=\"143\" height=\"16\" \/><\/em><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-MDRjYzYyOTJlNjA3MThjOTllNjk0ZjczMjNkMDEzNzc.-6mw\">The number of moles of carbon dioxide gas is first calculated by stoichiometry. Then the ideal gas law is used to calculate the volume of CO <sub> 2 <\/sub> produced.<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-d7z\"><em> Step 2: Solve. <\/em><\/p>\r\n<p id=\"x-ck12-i9w\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212234\/98be15fbc832bb992b6eaa3261013639.png\" alt=\"25.21 text{ g } text{C}_2text{H}_5text{OH} times frac{1 text{ mol } text{C}_2text{H}_5text{OH}}{46.08 text{ g } text{C}_2text{H}_5text{OH}} times frac{2 text{ mol } text{CO}_2}{1 text{ mol } text{C}_2text{H}_5text{OH}}=1.094 text{ mol } text{C}_2text{H}_5text{OH}\" width=\"603\" height=\"41\" \/><\/p>\r\n<p id=\"x-ck12-M2ZkOTE2ZDVhM2Y1OTZmYmRjNTQwNGY0Yzc0ODA1ZTg.-ase\">The moles of ethanol\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211023\/4168b7a35bc24c982496aef5b556a0a5.png\" alt=\"(n)\" width=\"23\" height=\"18\" \/> is now substituted into\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212235\/fb2df3ecf87eaa43f2bdf8fd25a59ab0.png\" alt=\"PV=nRT\" width=\"90\" height=\"12\" \/> to solve for the volume.<\/p>\r\n<p id=\"x-ck12-6lx\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212236\/f803c72bae9ad2aa8a0b1cb827650230.png\" alt=\"V=frac{nRT}{P}=frac{1.094 text{ mol} times 62.36 text{ L} cdot text{mmHg\/K} cdot text{mol} times 327 text{ K}}{728 text{ mmHg}}=30.6 text{ L}\" width=\"535\" height=\"42\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-g2z\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-ZWM3ZjEzYjUxM2ZkMDVlYzg3OTZjMzM1M2IwYWY1ZWQ.-dgx\">The mass of ethanol is slightly more than one half mole, meaning that the mole ratio results in slightly more than one mole of carbon dioxide being produced. Because of the elevated temperature and reduced pressure compared to STP, the resulting volume is larger than 22.4 L.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-Mjg4NjJiNDVjZGFhYzQxNDc4ZjI3MzhkOTdkODA5ZTA.-ixb\">\r\n \t<li>The ideal gas law is used to calculate stoichiometry problems for gases.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-MGIxMTE2NjA5NmRjNjU4NzY3YjYxMDkzY2YxYTFjOTA.-iv8\">Solve the problems on the worksheet at this site:<\/p>\r\n<p id=\"x-ck12-ODc1ZjJlNmYwMDhiNTA5YzRhZDNiOWNiMDEzZTYzMzA.-thm\"><a href=\"http:\/\/misterguch.brinkster.net\/PRA036.pdf\"> http:\/\/misterguch.brinkster.net\/PRA036.pdf <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-ns9\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-NDgwOGNmYWFiN2ZkYjYxZWQ4ODA4OWM0NDYxYzMyYmU.-ubv\">\r\n \t<li>Do we need gas conditions to be at STP to calculate stoichiometry problems?<\/li>\r\n \t<li>Why do we want to determine the stoichiometry of these reactions?<\/li>\r\n \t<li>What assumption are we making about the gases involved?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\"><\/div>\r\n<h1 id=\"x-ck12-UmVhbCBhbmQgSWRlYWwgR2FzZXM.\">Real and Ideal Gases<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-NTE4NzU5NjA1ZDQzYjBkOGViMTRiNGRjODg5ODQwNmU.-jyp\">\r\n \t<li>Define a real gas.<\/li>\r\n \t<li>Describe differences between real gases and ideal gases.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-NDk0NjE3M2VlOGFlMDg1ZDgyN2E5YjE1YzIyNGYzODY.-q1i\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212237\/20140811155526027455.png\" alt=\"Intermolecular forces can cause deviations from ideality in gases\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-NjRjZGRhODE5M2NiY2FhNzE3ODY1OGYzM2JmZjM4ODk.-vtw\"><strong> Location, Location, Location <\/strong><\/p>\r\n<p id=\"x-ck12-NDRhOTA5YjRkYzJjMWE2ZGQ3ZDJjMDBkYmM3OGE1NTc.-1av\">The behavior of a molecule depends a lot on its structure. We can have two compounds with the same number of atoms and yet they act very differently. Ethanol (C <sub> 2 <\/sub> H <sub> 5 <\/sub> OH) is a clear liquid that has a boiling point of about 79\u00b0C. Dimethylether (CH <sub> 3 <\/sub> OCH <sub> 3 <\/sub> ) has the same number of carbons, hydrogens, and oxygens, but boils at a much lower temperature (-25\u00b0C). The difference lies in the amount of intermolecular interaction (strong H-bonds for ethanol, weak van der Waals force for the ether).<\/p>\r\n\r\n<h3>Real and Ideal Gases<\/h3>\r\n<p id=\"x-ck12-ZDkyZDcwOTI3NmU1NTg1ZGFiNjA0MmY2NjFiMWQwMGI.-upf\">An ideal gas is one that follows the gas laws at all conditions of temperature and pressure. To do so, the gas would need to completely abide by the kinetic-molecular theory. The gas particles would need to occupy zero volume and they would need to exhibit no attractive forces what so ever toward each other. Since neither of those conditions can be true, there is no such thing as an ideal gas. A <strong> real gas <\/strong> is a gas that does not behave according to the assumptions of the kinetic-molecular theory. Fortunately, at the conditions of temperature and pressure that are normally encountered in a laboratory, real gases tend to behave very much like ideal gases.<\/p>\r\n<p id=\"x-ck12-NDZmN2QyOTExNWJkNDcwZDdmOWE0MDYzOTQyNzA1YWI.-e8n\">Under what conditions then, do gases behave least ideally? When a gas is put under high pressure, its molecules are forced closer together as the empty space between the particles is diminished. A decrease in the empty space means that the assumption that the volume of the particles themselves is negligible is less valid. When a gas is cooled, the decrease in kinetic energy of the particles causes them to slow down. If the particles are moving at slower speeds, the attractive forces between them are more prominent. Another way to view it is that continued cooling the gas will eventually turn it into a liquid and a liquid is certainly not an ideal gas anymore (see liquid nitrogen in the <strong> Figure <\/strong> below ). In summary, a real gas deviates most from an ideal gas at low temperatures and high pressures. Gases are most ideal at high temperature and low pressure.<\/p>\r\n\r\n<div id=\"x-ck12-NTAyNjRkMDljODM3NWJiZDc3NmM2ZmJhNzgyYWVmZGU.-2cf\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\r\n<p id=\"x-ck12-tyd\"><img id=\"x-ck12-OTgwNDUtMTM2Mzc1NTg3Ni0zNy01Ny0z\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212238\/20140811155526151540.jpeg\" alt=\"Liquid nitrogen is at such low temperatures that it is no longer a gas\" longdesc=\"Nitrogen%20gas%20that%20has%20been%20cooled%20to%2077%20K%20has%20turned%20to%20a%20liquid%20and%20must%20be%20stored%20in%20a%20vacuum%20insulated%20container%20to%20prevent%20it%20from%20rapidly%20vaporizing.\" \/><\/p>\r\n<strong> Figure 14.11 <\/strong>\r\n<p id=\"x-ck12-OTk5ZTM3MWRiNDhjM2U5M2Y5MTdjNjgzNGU4YjAyYzM.-cjm\">Nitrogen gas that has been cooled to 77 K has turned to a liquid and must be stored in a vacuum insulated container to prevent it from rapidly vaporizing.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-Mzg3Yjg1ZWVkMjQ3MjJjOWU5YzQ2ZGM1OWNmOWY0Mjc.-x5r\">The <strong> Figure <\/strong> below shows a graph of\u00a0 <img id=\"x-ck12-MTM2NjYxMjE2Njg1OA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212239\/f044bae598a4b75a3862b89f6c3a056c.png\" alt=\"frac{PV}{RT}\" width=\"21\" height=\"23\" \/> plotted against pressure for 1 mol of a gas at three different temperatures - 200 K, 500 K, and 1000 K. An ideal gas would have a value of 1 for that ratio at all temperatures and pressures and the graph would simply be a horizontal line. As can be seen, deviations from an ideal gas occur. As the pressure begins to rise, the attractive forces cause the volume of the gas to be less than expected and the value of\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212239\/f044bae598a4b75a3862b89f6c3a056c.png\" alt=\"frac{PV}{RT}\" width=\"21\" height=\"23\" \/> drops under 1. Continued pressure increase results in the volume of the particles to become significant and the value of\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212239\/f044bae598a4b75a3862b89f6c3a056c.png\" alt=\"frac{PV}{RT}\" width=\"21\" height=\"23\" \/> rises to greater than 1. Notice, that the magnitude of the deviations from ideality is greatest for the gas at 200 K and least for the gas at 1000 K.<\/p>\r\n\r\n<div id=\"x-ck12-OWI3YmI5NWJmNTNmMWI4NjcyNGE1YTM5Yzg0MjQxYzk.-wbi\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-mrc\"><img id=\"x-ck12-OTgwNDUtMTM2Mzc1NTgwOS02Mi01OS01\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212239\/20140811155526233028.png\" alt=\"Real gases deviate from ideal gases at high pressures and at low temperatures\" longdesc=\"Real%20gases%20deviate%20from%20ideal%20gases%20at%20high%20pressures%20and%20at%20low%20temperatures.\" \/><\/p>\r\n<strong> Figure 14.12 <\/strong>\r\n<p id=\"x-ck12-ZDBmNzgyODBjZDEwZmIwNWQ2MDRiOTU2NjM2ZDAzNzY.-o17\">Real gases deviate from ideal gases at high pressures and at low temperatures.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-NDE3Njg1Mjg4MTE2ZTU3OWNkNmQyY2ZiOTBiYWRiNWE.-zdw\">The ideality of a gas also depends on the strength and type of intermolecular attractive forces that exist between the particles. Gases whose attractive forces are weak are more ideal than those with strong attractive forces. At the same temperature and pressure, neon is more ideal than water vapor because neon\u2019s atoms are only attracted by weak dispersion forces, while water vapor\u2019s molecules are attracted by relatively stronger hydrogen bonds. Helium is a more ideal gas than neon because its smaller number of electrons means that helium\u2019s dispersion forces are even weaker than those of neon.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-NWNkMTllNWU1NzUzMzFkMWQ0ZDJlOWIzYjAyMmI1NWE.-7tn\">\r\n \t<li>The properties of real gases and their deviations from ideality are described.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-msh\"><em> Questions <\/em><\/p>\r\n<p id=\"x-ck12-Y2JlMjQ5M2YzMTNmNmRjMzNmZTI0MTMzYzcwM2IzZmY.-ab2\">Use the link below to answer the following questions:<\/p>\r\n<p id=\"x-ck12-ZGU5NWE1N2U5NGJjYjI3ODFiYmRhOWFlMTJlMmE5MzI.-bvl\"><a href=\"http:\/\/www.adichemistry.com\/physical\/gaseous\/deviation\/van-der-waals-equation.html\"> http:\/\/www.adichemistry.com\/physical\/gaseous\/deviation\/van-der-waals-equation.html <\/a><\/p>\r\n\r\n<ol id=\"x-ck12-ZDk2NGZmMTg4OWVjNzhlZWNhZmNkNzgzZjMxMGQ4Zjk.-vok\">\r\n \t<li>What is the compressibility factor for a perfect (ideal) gas?<\/li>\r\n \t<li>What does it mean if <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212241\/8befac7dd5d20767941551d93fee0ed7.png\" alt=\"Z&gt;1\" width=\"45\" height=\"13\" \/> ?<\/li>\r\n \t<li>What does it mean if <img id=\"x-ck12-MTM2NjYxMjE2Njg1OQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212241\/982265d061076cef029721fba31dab33.png\" alt=\"Z&lt;1\" width=\"45\" height=\"13\" \/> ?<\/li>\r\n<\/ol>\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-pxa\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-Y2NmZWU1ODY4ZGE1YjA2YzljOGI0ZTJkNzE5NzEzMDg.-2o0\">\r\n \t<li>What becomes more significant as the pressure increases?<\/li>\r\n \t<li>Do the attractive forces between gas particles become more prominent at higher or lower temperatures?<\/li>\r\n \t<li>Would HCl gas be more or less ideal than helium?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-MmMyMjgyZmVmMmRmZmE3MDYzZTNlZTA4YTUyNDlmNWM.-zfb\">\r\n \t<li><strong> real gas: <\/strong> A gas that does not behave according to the assumptions of the kinetic-molecular theory.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-RGFsdG9uJ3MgTGF3IG9mIFBhcnRpYWwgUHJlc3N1cmVz\">Dalton's Law of Partial Pressures<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-MDY3MDVkOTBhNDJkNWVhN2JhNzYzMWE1Y2QxYWE5MTE.-bxj\">\r\n \t<li>Define partial pressure.<\/li>\r\n \t<li>State Dalton\u2019s law of partial pressures.<\/li>\r\n \t<li>Use this law to calculate pressures of gas mixtures.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-M2JmNzk2NzlmMmY3NGIwNGRlNWY3NjdlOGY0OGI4YjE.-x3j\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212241\/20140811155526314531.jpeg\" alt=\"Venus has a high partial pressure of carbon dioxide and nitrogen\" width=\"250\" \/><\/span><\/p>\r\n\r\n<h4 id=\"x-ck12-NWVlNTZlMjQyMmMyOGQ2MzIwZDhkZTVlYjU2ZGEwZTY.-4ly_11-ojp\">Is there oxygen available on Venus?<\/h4>\r\n<p id=\"x-ck12-OWJhOTI4ZDYxMjExYjZlZDk4MjM2ZDVmZjRlNThmZjQ.-g0t\">The atmosphere of Venus is markedly different from that of Earth. The gases in the Venusian atmosphere are 96.5% carbon dioxide and 3% nitrogen. The atmospheric pressure on Venus is roughly 92 times that of Earth, so the amount of nitrogen on Venus would contribute a pressure well over 2700 mm Hg. And there is no oxygen present, so we couldn\u2019t breathe there. Not that we would want to go to Venus \u2013 the surface temperature is usually over 460\u00b0C.<\/p>\r\n\r\n<h3>Dalton\u2019s Law of Partial Pressures<\/h3>\r\n<p id=\"x-ck12-ZDYzYzQ3MDU4MjUxNDFiZmFhOTUyYmIxYjk4ODFhYTE.-w4c\">Gas pressure results from collisions between gas particles and the inside walls of their container. If more gas is added to a rigid container, the gas pressure increases. The identities of the two gases do not matter. John Dalton, the English chemist who proposed the atomic theory, also studied mixtures of gases. He found that each gas in a mixture exerts a pressure independently of every other gas in the mixture. For example, our atmosphere is composed of about 78% nitrogen and 21% oxygen, with smaller amounts of several other gases making up the rest. Since nitrogen makes up 78% of the gas particles in a given sample of air, it exerts 78% of the pressure. If the overall atmospheric pressure is 1.00 atm, then the pressure of just the nitrogen in the air is 0.78 atm. The pressure of the oxygen in the air is 0.21 atm.<\/p>\r\n<p id=\"x-ck12-NWNjZTIxYWM1OWIwZTBiNjg2Njk0ZDEzODJiYjQxM2U.-3q5\">The <strong> partial pressure <\/strong> of a gas is the contribution that gas makes to the total pressure when the gas is part of a mixture. The partial pressure of a gas is indicated by a\u00a0 <img id=\"x-ck12-MTM2NjYxMjgyNjQ2Mw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/dc2a557e4775924079e5de1061889778.png\" alt=\"P\" width=\"14\" height=\"12\" \/> with a subscript that is the symbol or formula of that gas. The partial pressure of nitrogen is represented by\u00a0 <img id=\"x-ck12-MTM2NjYxMjgyNjQ2NA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212243\/28beb3b275887c063f1f720929c422b2.png\" alt=\"P_{N_2}\" width=\"27\" height=\"17\" \/> . <strong> Dalton\u2019s law of partial pressures <\/strong> states that the total pressure of a mixture of gases is equal to the sum of all of the partial pressures of the component gases. Dalton\u2019s law can be expressed with the following equation:<\/p>\r\n<p id=\"x-ck12-ysv\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212243\/dc796502abd45ae9ca6892fa10b9b1fe.png\" alt=\"P_{text{total}}=P_1+P_2+P_3+ldots\" width=\"205\" height=\"16\" \/><\/p>\r\n<p id=\"x-ck12-ZjEyMmZiNmI2NjIyMDI3NmEyMzRjNmQ4MmVjZGM1OTI.-6sh\">The <strong> Figure <\/strong> below shows two gases that are in separate, equal-sized containers at the same temperature and pressure. Each exerts a different pressure,\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/b95fd9f12605d93cbb24fd5fc1d83fbc.png\" alt=\"P_1\" width=\"18\" height=\"16\" \/> and <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/f49ec791ef1ecb63d1c14840f9c2f983.png\" alt=\"P_2\" width=\"18\" height=\"15\" \/> , reflective of the number of particles in the container. On the right, the two gases are combined into the same container, with no volume change. The total pressure of the gas mixture is equal to the sum of the individual pressures. If\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212244\/273489be1c6bb8b9fbd5c8559112efa8.png\" alt=\"P_1 = 300 text{ mmHg}\" width=\"128\" height=\"17\" \/> and <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212244\/daa4b3a9457f11fff4ff03fa2aea9830.png\" alt=\"P_2 = 500 text{ mmHg}\" width=\"128\" height=\"16\" \/> , then <img id=\"x-ck12-MTM2NjYxMjgyNjQ2NQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212245\/512977e45cf92f9bca54108929ca5623.png\" alt=\"P_{text{Total}}=800 text{ mmHg}\" width=\"151\" height=\"17\" \/> .<\/p>\r\n\r\n<div id=\"x-ck12-N2M0MmU0NDIxMjg3MTk5ZTEwYjRjMjBkYTFjYTlkOTU.-usn\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-dxk\"><img id=\"x-ck12-OTgwNDUtMTM2Mzc1ODExNi0xNS0xMS0xMA..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212246\/20140811155526465849.png\" alt=\"Dalton's law states that the pressure of a gas mixture is equal to the partial pressure of the components\" longdesc=\"Dalton%E2%80%99s%20law%20says%20that%20the%20pressure%20of%20a%20gas%20mixture%20is%20equal%20to%20the%20partial%20pressures%20of%20the%20combining%20gases.\" \/><\/p>\r\n<strong> Figure 14.13 <\/strong>\r\n<p id=\"x-ck12-YzE1MjkzNzQ0NWRjYjNhODNmZWY3NjcxM2ZhMTAwZWI.-acd\">Dalton\u2019s law says that the pressure of a gas mixture is equal to the partial pressures of the combining gases.<\/p>\r\n\r\n<\/div>\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-MDA2YmNlNThmNjljZDNhYzVkZTA0YjIzMDZjMWNjMzM.-j0h\">\r\n \t<li>The total pressure in a system is equal to the sums of the partial pressures of the gases present.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-NDU5MThkNzZhZDRiYmFiYmM1ZGM1ZmRiNjI1Y2QyYzA.-tvi\">Review the concepts at the link below and work the sample problems:<\/p>\r\n<p id=\"x-ck12-NDgxMjFkMzA0NTNmN2E0MDRlMDA3ZTM0MDY5YTk5ZDQ.-t1g\"><a href=\"http:\/\/www.kentchemistry.com\/links\/GasLaws\/dalton.htm\"> http:\/\/www.kentchemistry.com\/links\/GasLaws\/dalton.htm <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-eeu\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-ZWJjZmYxZjgzNDVlYjYyODljNzcxZTFhMTA3YWJmMzc.-bmk\">\r\n \t<li>What is the foundation for Dalton\u2019s law?<\/li>\r\n \t<li>Argon makes up about 0.93% of our atmosphere. If the atmospheric pressure is 760 mm Hg, what is the pressure contributed by argon?<\/li>\r\n \t<li>On a given day, the water vapor in the air is 2.5%. If the partial pressure of the vapor is 19.4 mm Hg, what is the atmospheric pressure?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-ZjJhNzE4MGMxNTdkZGYwMDdlNGQ0OTcyMDM0ZGIxODI.-sd2\">\r\n \t<li><strong> Partial pressure: <\/strong> The contribution that gas makes to the total pressure when the gas is part of a mixture.<\/li>\r\n \t<li><strong> Dalton\u2019s law of partial pressures: <\/strong> The total pressure of a mixture of gases is equal to the sum of all of the partial pressures of the component gases.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-TW9sZSBGcmFjdGlvbg..\">Mole Fraction<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-OWFkOTlkY2Q1MzM4ZTllNGUxZDg1Y2VjZmFhYTU3ZDg.-ter\">\r\n \t<li>Define mole fraction.<\/li>\r\n \t<li>Perform calculations involving mole fractions.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-YTcwYmVhNGMxODY3NjVjMWM0N2Q3ODlhNzcyYWM0MTY.-rve\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212247\/20140811155526612018.jpeg\" alt=\"The changing mole fraction of sulfur dioxide is a mixed blessing\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-ZDU1ZWJkMGNmOTc0MWNiNmUxYzljYjAyYTIwMmNjMjc.-3is\"><strong> The mixed blessing of sulfur dioxide <\/strong><\/p>\r\n<p id=\"x-ck12-ZWNlYTg2MTlmNjkzNzlkOThhNmIwYTk2ODFkMTA0MGU.-erg\">Sulfur dioxide is a by-product of many processes, both natural and human-made. Massive amounts of this gas are released during volcanic eruptions such as the one seen above on the Big Island (Hawaii). Humans produce sulfur dioxide by burning coal. The gas has a cooling effect when in the atmosphere by reflecting sunlight back away from the earth. However, sulfur dioxide is also a component of smog and acid rain, both of which are harmful to the environment. Many efforts have been made to reduce SO <sub> 2 <\/sub> levels to lower acid rain production. An unforeseen complication: as we lower the concentration of this gas in the atmosphere, we lower its ability to cool and then we have global warming concerns.<\/p>\r\n\r\n<h3>Mole Fraction<\/h3>\r\n<p id=\"x-ck12-MDQyZjA4ZTJlNjk2NDE4ODk2NjM5NzQ3MDFhYzY4Njk.-yrq\">One way to express relative amounts of substances in a mixture is with the mole fraction. <strong> Mole fraction <\/strong> \u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212249\/0124750beaa395c3119a87d9c19f2789.png\" alt=\"(X)\" width=\"28\" height=\"18\" \/> is the ratio of moles of one substance in a mixture to the total number of moles of all substances. For a mixture of two substances,\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> and <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/00025e2ee78e7b712fbb42f74f2d6cb7.png\" alt=\"B\" width=\"14\" height=\"12\" \/> , the mole fractions of each would be written as follows:<\/p>\r\n<p id=\"x-ck12-YmU1ZDVkMzc1NDJkNzVmOTNhODcwOTQ0NTlmNzY2Nzg.-zx9\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212249\/3c708d707a41f03b77c141794a18fb87.png\" alt=\"X_A= frac{text{mol} A}{text{mol} A+text{mol} B} quad text{and} quad X_B=frac{text{mol} B}{text{mol} A+text{mol} B}\" width=\"409\" height=\"39\" \/><\/p>\r\n<p id=\"x-ck12-MzhlNDU0M2Y0MmZlYzA1MDI4NzYwZmYyYWJmOTRiMmI.-xr8\">If a mixture consists of 0.50 mol\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> and 1.00 mol <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/00025e2ee78e7b712fbb42f74f2d6cb7.png\" alt=\"B\" width=\"14\" height=\"12\" \/> , then the mole fraction of\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> would be <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212250\/28245fb4dc2eab5bad3743911a8f1571.png\" alt=\"X_A=frac{0.5}{1.5} = 0.33\" width=\"126\" height=\"24\" \/> .\u00a0 Similarly, the mole fraction of\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/00025e2ee78e7b712fbb42f74f2d6cb7.png\" alt=\"B\" width=\"14\" height=\"12\" \/> would be <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212251\/00d6fe0bf594a44daa1ff7af4a776b56.png\" alt=\"X_B =frac{1.0}{1.5} = 0.67\" width=\"126\" height=\"24\" \/> .<\/p>\r\n<p id=\"x-ck12-Y2M5M2FiMTlmYWZiZGY5NjAxMzIyMWJkMTE0NDY0ZjU.-m8y\">Mole fraction is a useful quantity for analyzing gas mixtures in conjunction with Dalton\u2019s law of partial pressures. Consider the following situation: A 20.0 liter vessel contains 1.0 mol of hydrogen gas at a pressure of 600 mmHg. Another 20.0 liter vessel contains 3.0 mol of helium at a pressure of 1800 mmHg. These two gases are mixed together in an identical 20.0 liter vessel. Because each will exert its own pressure according to Dalton\u2019s law, we can express the partial pressures as follows:<\/p>\r\n<p id=\"x-ck12-YmU1ZDVkMzc1NDJkNzVmOTNhODcwOTQ0NTlmNzY2Nzg.-qox\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212251\/09e954a86ef27a64811ada0fe15d0537.png\" alt=\"P_{H_2}=X_{H_2} times P_{text{Total}} quad text{and} quad P_{He}=X_{He} times P_{text{Total}}\" width=\"365\" height=\"17\" \/><\/p>\r\n<p id=\"x-ck12-NzBjM2NiODE0OTQwYmRmM2YzZjhkNDVkZTE1ZTA0OTA.-ewj\">The partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. For our mixture of hydrogen and helium:<\/p>\r\n<p id=\"x-ck12-YmU1ZDVkMzc1NDJkNzVmOTNhODcwOTQ0NTlmNzY2Nzg.-jum\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212253\/0c0019ee084d9c55406b47b1d7e2cf9b.png\" alt=\"X_{H_2}=frac{1.0 text{mol}}{1.0 text{mol}+3.0 text{mol}}=0.25 quad text{and} quad X_{He}=frac{3.0 text{mol}}{1.0 text{mol} + 3.0 text{mol}}=0.75\" width=\"572\" height=\"39\" \/><\/p>\r\n<p id=\"x-ck12-MjQ4ODRlNzAzNTMzYTU3MzIzYTk0OWI3ZDYwNDIwNzg.-dsc\">The total pressure according to Dalton\u2019s law is <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212254\/a35110bf11c36dac7a069303648e986c.png\" alt=\"600 text{ mmHg} + 1800 text{ mmHg} = 2400 text{ mmHg}\" width=\"319\" height=\"16\" \/> . So, each partial pressure will be:<\/p>\r\n<p id=\"x-ck12-fwz\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212255\/cf654c3d8ff44acf9656dcfbcdc889cd.png\" alt=\"&amp; P_{H_2}=0.25 times 2400 text{ mmHg}=600 text{ mmHg} \\&amp; P_{He}=0.75 times 2400 text{ mmHg}=1800 text{ mmHg}\" width=\"320\" height=\"42\" \/><\/p>\r\n<p id=\"x-ck12-ODBlMTNhYjlkOWQ0YTE2OGFlMjRjZDg4OWQ4MWEzNWE.-5rp\">The partial pressures of each gas in the mixture don\u2019t change since they were mixed into the same size vessel and the temperature was not changed.<\/p>\r\n\r\n<h4>Sample Problem: Dalton\u2019s Law<\/h4>\r\n<p id=\"x-ck12-NjY0NmIwNDFhZmM1NjQ2MDkyMTkyYWYwZTBiZDdlMjg.-cop\">A flask contains a mixture of 1.24 moles of hydrogen gas and 2.91 moles of oxygen gas. If the total pressure is 104 kPa, what is the partial pressure of each gas?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-utm\"><em> Step 1: List the known quantities and plan the problem <\/em> .<\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-0ju\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ZjRlMTgwMTJjM2RhNGYxMzIxZjQxYjMxMmIwODhlY2Q.-j0s\">\r\n \t<li>1.24 mol H <sub> 2 <\/sub><\/li>\r\n \t<li>2.91 mol O <sub> 2 <\/sub><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212256\/fe4864d1f5f23f08708b43e2ab118bd2.png\" alt=\"P_{text{Total}}=104 text{kPa}\" width=\"129\" height=\"17\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-mux\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-6qx\">\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212257\/128358483d347569abcf233b1977c445.png\" alt=\"P_{H_2}=? text{kPa}\" width=\"93\" height=\"18\" \/><\/li>\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212257\/5ad22f50a3cb357fc8f14ceab55a2d6e.png\" alt=\"P_{O_2}=? text{kPa}\" width=\"92\" height=\"18\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ZWZjNzRkYTQzYTVhNjBlMTFiMDlhM2I2YzFmYjBkNTQ.-mds\">First, the mole fraction of each gas can be determined. Then, the partial pressure can be calculated by multiplying the mole fraction by the total pressure.<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-hsv\"><em> Step 2: Solve <\/em> .<\/p>\r\n<p id=\"x-ck12-6ik\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212258\/4a7e9cd4699c81f04274bc8325e98a43.png\" alt=\"&amp; X_{H_2}=frac{1.24 text{mol}}{1.24 text{mol} + 2.91 text{mol}}=0.299 &amp;&amp; X_{O_2}=frac{2.91 text{mol}}{1.24 text{mol} + 2.91 text{mol}}=0.701 \\&amp; P_{H_2}=0.299 times 104 text{ kPa}=31.1 text{ kPa} &amp;&amp; P_{O_2}=0.701 times 104 text{ kPa}=72.9 text{ kPa}\" width=\"574\" height=\"65\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-s3f\"><em> Step 3: Think about your result <\/em> .<\/p>\r\n<p id=\"x-ck12-YzZhZjZiZGFmMmIyZWVhMjZjODExYWM3NTc0MWMyOTk.-lwh\">The hydrogen is slightly less than one third of the mixture, so it exerts slightly less than one third of the total pressure.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-ZmIwNmY3MjRiNjFjNTYzYjVjMjliOTk0Mzk3ZDM2MmY.-afk\">\r\n \t<li>Use of the mole fraction allows calculation to be made for mixtures of gases.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-pc8\"><em> Questions <\/em><\/p>\r\n<p id=\"x-ck12-YmMzM2NmMmUyYzQ1NmUxNTQ2YTNmYjA2MWZlY2FkODA.-odu\">Watch the video at the link below and answer the following questions:<\/p>\r\nhttp:\/\/www.youtube.com\/watch?v=7BaX__s-4Ls\r\n<ol id=\"x-ck12-ZWJhNDUwMTNlZmM3Y2E3NWZjYjJmNDQ4ZGM1NjczMDY.-grk\">\r\n \t<li>What is mole percent?<\/li>\r\n \t<li>Do the mole fractions add up to 1.00?<\/li>\r\n \t<li>What other way could you calculate the mole fraction of oxygen once you have the mole fraction of nitrogen?<\/li>\r\n<\/ol>\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-ebi\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-YmFmMjgyNmNlMDkwZDM5ODBmMjQwMGRhZmJjZjI1MmM.-c02\">\r\n \t<li>What is mole fraction?<\/li>\r\n \t<li>How do you determine partial pressure of a gas when given the mole fraction and the total pressure?<\/li>\r\n \t<li>In a gas mixture containing equal numbers of moles of two gases, what can you say about the partial pressures of each gas?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-MjMzYjRhNDk1ZDFhYWQ2NGNhZDA1MDBhNjZkZDBhYjc.-njd\">\r\n \t<li><strong> Mole fraction <\/strong> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212249\/0124750beaa395c3119a87d9c19f2789.png\" alt=\"(X)\" width=\"28\" height=\"18\" \/><strong> : <\/strong> The ratio of moles of one substance in a mixture to the total number of moles of all substances.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-R2FzIENvbGxlY3Rpb24gYnkgV2F0ZXIgRGlzcGxhY2VtZW50\">Gas Collection by Water Displacement<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-NGE2NzM5OTk5MmRhYjQxMDIwMzBjZDk5YjQxMGRjMmM.-hqb\">\r\n \t<li>Calculate volumes of dry gases obtained after collecting over water.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-NThiNjgwMzQ2NTQzNjVhYmE3OGZmNzllNzhlN2FkMTU.-jwe\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212300\/20140811155526744059.png\" alt=\"The pressure of gases collected over water can be determined by using the atmospheric pressure in the room\" width=\"400\" \/><\/span><\/p>\r\n<p id=\"x-ck12-NzBjNDNhMTBkM2VkMzE3MjY1YmVmMmEwNGUyNjU5NmQ.-5mw\"><strong> What is the pressure? <\/strong><\/p>\r\n<p id=\"x-ck12-MjFjZjViMzQ3M2M3NGRjMjgwYWY3MTRhYTdiMGM2YWE.-3lk\">You need to do a lab experiment where hydrogen gas is generated. In order to calculate the yield of gas, you have to know the pressure inside the tube where the gas is collected. But how can you get a barometer in there? Very simple: you don\u2019t. All you need is the atmospheric pressure in the room. As the gas pushed out the water, it is pushing against the atmosphere, so the pressure inside is equal to the pressure outside.<\/p>\r\n\r\n<h3>Gas Collection by Water Displacement<\/h3>\r\n<p id=\"x-ck12-NjdiMDQ2N2I1MWExMWRlYTQ0NDBhYmZlNDlhMjQ2ODg.-glt\">Gases that are produced in laboratory experiments are often collected by a technique called <strong> water displacement <\/strong> (see <strong> Figure <\/strong> below ). A bottle is filled with water and placed upside-down in a pan of water. The reaction flask is fitted with rubber tubing which is then fed under the bottle of water. As the gas is produced in the reaction flask, it exits through the rubber tubing and displaces the water in the bottle. When the bottle is full of the gas, it can be sealed with a lid.<\/p>\r\n\r\n<div id=\"x-ck12-MDM5MDYzN2ZhZmJlZTk4ODllZjFhNzA4YzAwNzFhYzk.-eff\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n<p id=\"x-ck12-1tt\"><img id=\"x-ck12-OTgwNDUtMTM2Mzc1ODQ1Ny0zNy0xNi0y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212301\/20140811155526874362.png\" alt=\"A gas produced in a chemical reaction can be collected by water displacement.\" longdesc=\"A%20gas%20produced%20in%20a%20chemical%20reaction%20can%20be%20collected%20by%20water%20displacement.\" \/><\/p>\r\n<strong> Figure 14.14 <\/strong>\r\n<p id=\"x-ck12-NzVjZDljODZhZDQwOGFkZmIxY2U1NmY3NmNiM2JiNDE.-tqt\">A gas produced in a chemical reaction can be collected by water displacement.<\/p>\r\n\r\n<\/div>\r\n<p id=\"x-ck12-YTgzZmRiNzE3YjkzYThkMzE3N2QzZWI2NTA4MzYyYzA.-yvf\">Because the gas is collected over water, it is not pure but is mixed with vapor from the evaporation of the water. Dalton\u2019s law can be used to calculate the amount of the desired gas by subtracting the contribution of the water vapor.<\/p>\r\n<p id=\"x-ck12-gbt\"><img id=\"x-ck12-MTQwMDYxMzQzODg4MQ..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212303\/7bf208c74a09188dc83bb4d88013e23c.png\" alt=\"P_{text{Total}} &amp;=P_g+P_{H_2 O} qquad P_g text{ is the pressure of the desired gas}\\P_g &amp;=P_{text{Total}}- P_{H_2 O}\" width=\"462\" height=\"45\" \/><\/p>\r\n<p id=\"x-ck12-Yjg0MGMyMjBjNjBmMGJmZjM0MTM3ZTdhNGRjODEyOGE.-zrl\">In order to solve a problem, it is necessary to know the vapor pressure of water at the temperature of the reaction (see <strong> Table <\/strong> below). The sample problem illustrates the use of Dalton\u2019s law when a gas is collected over water.<\/p>\r\n\r\n<table>\r\n<thead>\r\n<tr>\r\n<th colspan=\"4\">Vapor Pressure of Water (mmHg) at Selected Temperatures (\u00b0C)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Temperature (\u00b0C)<\/th>\r\n<th>Vapor Pressure (mmHg)<\/th>\r\n<th>Temperature (\u00b0C)<\/th>\r\n<th>Vapor Pressure (mmHg)<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>0<\/td>\r\n<td>4.58<\/td>\r\n<td>40<\/td>\r\n<td>55.32<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>5<\/td>\r\n<td>6.54<\/td>\r\n<td>45<\/td>\r\n<td>71.88<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>10<\/td>\r\n<td>9.21<\/td>\r\n<td>50<\/td>\r\n<td>92.51<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>15<\/td>\r\n<td>12.79<\/td>\r\n<td>55<\/td>\r\n<td>118.04<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>20<\/td>\r\n<td>17.54<\/td>\r\n<td>60<\/td>\r\n<td>149.38<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>25<\/td>\r\n<td>23.76<\/td>\r\n<td>65<\/td>\r\n<td>187.54<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>30<\/td>\r\n<td>31.82<\/td>\r\n<td>70<\/td>\r\n<td>233.7<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>35<\/td>\r\n<td>42.18<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Sample Problem: Gas Collected by Water Displacement<\/h4>\r\n<p id=\"x-ck12-NDI4YzllZTZmOWEyNzBjY2I2NWQ1ZjQzYjg4ZjI3ODg.-3sz\">A certain experiment generates 2.58 L of hydrogen gas, which is collected over water. The temperature is 20\u00b0C and the atmospheric pressure is 98.60 kPa. Find the volume that the dry hydrogen would occupy at STP.<\/p>\r\n<p id=\"x-ck12-dkc\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-ykk\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-cy3\">\r\n \t<li><span class=\"x-ck12-underline\"> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212304\/006c3d6caa49cb5e202603952dba195f.png\" alt=\"V_{text{Total}} =2.58 text{ L}\" width=\"114\" height=\"17\" \/><\/span><\/li>\r\n \t<li><img id=\"x-ck12-MTQwMDYxMzQzODg4Mw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212304\/dc11e1e2cbee5090672917d93547ea32.png\" alt=\"T=20^ circ text{C}=293 text{ K}\" width=\"146\" height=\"14\" \/><\/li>\r\n \t<li><span class=\"x-ck12-underline\"> <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212305\/37bd71ce8cb54b5b5aa31b7d661d6160.png\" alt=\"P_{text{Total}} =98.60 text{ kPa}=739.7 text{ mmHg}\" width=\"266\" height=\"17\" \/><\/span><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-ykd\"><span class=\"x-ck12-underline\"> Unknown <em>\r\n<\/em> <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ifv\">\r\n \t<li><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212306\/2f65228bce106ab00ed628005b1f47fa.png\" alt=\"V_{H_2} text{at} STP= ? text{ L}\" width=\"140\" height=\"18\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-Y2VkOTFjYmIxYjRhYmVhZjQzYmQ2YzJhZjI2NzBiNWU.-kgq\">The atmospheric pressure is converted from kPa to mmHg in order to match units with the table. The sum of the pressures of the hydrogen and the water vapor is equal to the atmospheric pressure. The pressure of the hydrogen is found by subtraction. Then, the volume of the gas at STP can be calculated by using the combined gas law.<\/p>\r\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-4yd\"><em> Step 2: Solve. <\/em><\/p>\r\n<p id=\"x-ck12-3jx\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212306\/6c113d922b4e57a5765d16193c0ce8c2.png\" alt=\"P_{H_2}=P_{text{Total}}-P_{H_2 O}=739.7 text{ mmHg} -17.54 text{ mmHg}=722.2 text{ mmHg}\" width=\"524\" height=\"17\" \/><\/p>\r\n<p id=\"x-ck12-MWU2OGE5MzZmNTk5NmMxNmI5YjJhMDIyYzlhOWIzMmM.-dgz\">Now the combined gas law is used, solving for <img id=\"x-ck12-MTQwMDYxMzQzODg4NA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> , the volume of hydrogen at STP.<\/p>\r\n<p id=\"x-ck12-vft\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212307\/38c5ef825b1c1b16462d721800e47177.png\" alt=\"V_2=frac{P_1 times V_1 times T_2}{P_2 times T_1}=frac{722.2 text{ mmHg} times 2.58 text{ L} times 273 text{ K}}{760 text{ mmHg} times 293 text{ K}}=2.28 text{ L } H_2\" width=\"511\" height=\"41\" \/><\/p>\r\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-pv0\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-MWZhNjY4NGY5OTUwN2RjNWVkNjQ2MDY2NGYzMDE0YTA.-dqo\">If the hydrogen gas were to be collected at STP and without the presence of the water vapor, its volume would be 2.28 L. This is less than the actual collected volume because some of that is water vapor. The conversion using STP is useful for stoichiometry purposes.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-ZjMyMzQ1NjQxODdjODQ0NzhjNWMwOWQ4MmEzM2I0MTI.-8sw\">\r\n \t<li>The vapor pressure due to water in a sample can be corrected for in order to get the true value for the pressure of the gas.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-teg\"><em> Questions <\/em><\/p>\r\n<p id=\"x-ck12-YmMzM2NmMmUyYzQ1NmUxNTQ2YTNmYjA2MWZlY2FkODA.-afu\">Watch the video at the link below and answer the following questions:<\/p>\r\nhttp:\/\/www.youtube.com\/watch?v=xmL2Pax4yUQ\r\n<ol id=\"x-ck12-MTg0ZjQ4MWJmYmQ5NTU1OTFhOWM5ZDUxZDEzMjMxNDY.-jip\">\r\n \t<li>What was the thistle tube used for?<\/li>\r\n \t<li>How did the instructor tests for oxygen?<\/li>\r\n \t<li>Did you observe any unsafe lab practices in the video?<\/li>\r\n \t<li>What would have happened to the splint if carbon dioxide had been collected?<\/li>\r\n<\/ol>\r\n<h4>Review<\/h4>\r\n<em> Questions <\/em>\r\n<ol id=\"x-ck12-ZWEyYWIwN2U4ZjllZTVlYjcxM2Q3YTIxNGRiMzY2M2I.-ygo\">\r\n \t<li>Why is gas collected over water not pure?<\/li>\r\n \t<li>Why would we want to correct for water vapor?<\/li>\r\n \t<li>A student wants to collect his gas over diethyl ether (vapor pressure of 530 mm Hg at 25\u00b0C). Is this a good idea? Explain your answer.<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-MzNkNDQ2MmYxZjA0ZTg1ZjEyYjdmOTJhZDhmOTRmMTY.-9vg\">\r\n \t<li><strong> water displacement: <\/strong> Collection of a gas over water.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1 id=\"x-ck12-RGlmZnVzaW9uIGFuZCBFZmZ1c2lvbiBhbmQgR3JhaGFtJ3MgTGF3\">Diffusion and Effusion and Graham's Law<\/h1>\r\n<div class=\"x-ck12-data-objectives\">\r\n<ul id=\"x-ck12-MDc3OWVlN2RiZDc0OTU3Y2Y1Y2VkZTI2NDc2MDFlZWI.-ub9\">\r\n \t<li>Define diffusion and effusion.<\/li>\r\n \t<li>State Graham\u2019s law.<\/li>\r\n \t<li>Use Graham\u2019s law to perform calculations involving movement of gases.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"x-ck12-MmE4Yzc2MzJhZWQyMDE4NTcyMTRjNzAyY2Y3YWI3MjE.-vle\"><span class=\"x-ck12-img-inline\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212310\/20140811155526959330.png\" alt=\"A classic experiment to find the rate of diffusion for gases uses hydrochloric acid and ammonia\" width=\"450\" \/><\/span><\/p>\r\n<p id=\"x-ck12-ODYwMDBlYTZmMjBiNmZlOWVmMzdiMGYwZmIzODUyYTU.-tvq\"><strong> How do we know how fast a gas moves? <\/strong><\/p>\r\n<p id=\"x-ck12-Njg5MWRkOWIzOGI2YWM5NDljZjI1OTY3M2ZlOTRkNDY.-ypb\">We usually cannot see gases, so we need ways to detect their movements indirectly. The relative rates of diffusion of ammonia to hydrogen chloride can be observed in a simple experiment. Cotton balls are soaked with solutions of ammonia and hydrogen chloride (hydrochloric acid) and attached to two different rubber stoppers. These are simultaneously plugged into either end of a long glass tube. The vapors of each travel down the tube at different rates. Where the vapors meet, they react to form ammonium chloride (NH <sub> 4 <\/sub> Cl), a white solid that appears in the glass tube as a ring.<\/p>\r\n\r\n<h3>Graham\u2019s Law<\/h3>\r\n<p id=\"x-ck12-ODA4ZGIxYTRiNjA1MTQ4YzY5OGQ0YWYzOGIwYzNjYjA.-ehk\">When a person opens a bottle of perfume in one corner of a large room, it doesn\u2019t take very long for the scent to spread throughout the entire room. Molecules of the perfume evaporate and the vapor spreads out to fill the entire space. <strong> Diffusion <\/strong> is the tendency of molecules to move from an area of high concentration to an area of low concentration until the concentration is uniform. While gases diffuse rather quickly, liquids diffuse much more slowly. Solids essentially do not diffuse.<\/p>\r\n<p id=\"x-ck12-OGNhODJkZWQ1YzBiMTQ3MzVkNGI1Mjc2NjQ1OTVlYWU.-2px\">Video of bromine diffusion:<\/p>\r\nhttp:\/\/www.youtube.com\/watch?v=R_xDe004oTQ\r\n<p id=\"x-ck12-ZjUyZmQ0ZjJiM2IwYmFhNDY2YTA0OTgzMTk1NzBhMTQ.-d8t\">A related process to diffusion is the effusion. <strong> Effusion <\/strong> is the process of a confined gas escaping through a tiny hole in its container. Effusion can be observed by the fact that a helium-filled balloon will stop floating and sink to the floor after a day or so. This is because the helium gas effuses through tiny pores in the balloon. Both diffusion and effusion are related to the speed at which various gas molecules move. Gases that have a lower molar mass effuse and diffuse at a faster rate than gases that have a higher molar mass.<\/p>\r\n<p id=\"x-ck12-Mjk5ZDVhMDA5NzFkOWI5ZDMzZDlkMzdlZDg0ZmZiOTk.-lgz\">Scottish chemist Thomas Graham (1805-1869) studied the rates of effusion and diffusion of gases. <strong> Graham\u2019s law <\/strong> states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. Graham\u2019s law can be understood by comparing two gases ( <img id=\"x-ck12-MTM2NjYxOTU4ODEwMw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> and <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/00025e2ee78e7b712fbb42f74f2d6cb7.png\" alt=\"B\" width=\"14\" height=\"12\" \/> ) at the same temperature, meaning the gases have the same kinetic energy. The kinetic energy of a moving object is given by the equation <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212312\/50ba5f09928eb42fb31d9b0d2f6e4854.png\" alt=\"KE =frac{1}{2}mv^2\" width=\"96\" height=\"23\" \/> ,<\/p>\r\n<p id=\"x-ck12-NDJiZjk4ZjQ2MzlhZDY4MjIzYTEwYjc0MjlhZWMyZGE.-ckf\">where\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211009\/de65f173073697541b369ca6047387a8.png\" alt=\"m\" width=\"16\" height=\"8\" \/> is mass and\u00a0 <img id=\"x-ck12-MTM2NjYxOTU4ODEwNA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19210944\/ab157ad37ef6efcf3559476aa5ec6133.png\" alt=\"v\" width=\"9\" height=\"8\" \/> is velocity. Setting the kinetic energies of the two gases equal to one another gives:<\/p>\r\n<p id=\"x-ck12-qnu\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212313\/b3bc5c29051f9d8156106e18b7e5adeb.png\" alt=\"frac{1}{2}m_Av^2_A=frac{1}{2}m_Bv^2_B\" width=\"138\" height=\"37\" \/><\/p>\r\n<p id=\"x-ck12-YmMzY2MxZmEzNzg3YzNmZmJjZGU1MjRjYmQ3OGVlNzE.-0mk\">The equation can be rearranged to solve for the ratio of the velocity of gas\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> to the velocity of gas\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212313\/cddfbc8003d4628a6a87301133d54c3f.png\" alt=\"Bleft(frac{v_A}{v_B}right)\" width=\"56\" height=\"34\" \/> .<\/p>\r\n<p id=\"x-ck12-ODAyYmE5MzhjOTc5YWE5MzAyOTFkOTY0Njk4YWMzYmY.-fbq\"><img id=\"x-ck12-MTM2NjYxOTU4ODEwNQ..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212313\/07a84d6027d74eedc76b326f7240482c.png\" alt=\"frac{v^2_A}{v^2_B}=frac{m_B}{m_A} text{ which becomes } frac{v_A}{v_B}=sqrt{frac{m_B}{m_A}}\" width=\"301\" height=\"47\" \/><\/p>\r\n<p id=\"x-ck12-N2FiOWVlMTNiMGYxNTZiZGFiNTc1NjI0Mjg3OGMyOWM.-wn8\">For the purposes of comparing the rates of effusion or diffusion of two gases at the same temperature, the molar masses of each gas can be used in the equation for <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211009\/de65f173073697541b369ca6047387a8.png\" alt=\"m\" width=\"16\" height=\"8\" \/> .<\/p>\r\n\r\n<h4>Sample Problem: Graham\u2019s Law<\/h4>\r\n<p id=\"x-ck12-ZDUzYjg0NGQxMzZmMjZkNTJkZDhhN2FmYzlkOTNlYTc.-yrt\">Calculate the ratio of diffusion rates of ammonia gas (NH <sub> 3 <\/sub> ) to hydrogen chloride (HCl) at the same temperature and pressure.<\/p>\r\n<p id=\"x-ck12-OGU2YWEyYzY0NzAwMDUxZjI4NjFjY2E4MjYyNmNhN2I.-tqe\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-kdy\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-M2Q3ZjVmZTM3YjFlOTU3OGQ0ZjMxMTk1M2FmOGIzNjE.-qvq\">\r\n \t<li>molar mass NH <sub> 3 <\/sub> = 17.04 g\/mol<\/li>\r\n \t<li>molar mass HCl = 36.46 g\/mol<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-bqu\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ZGU5MzNiY2VmMTI1ZWNlY2YxNmE2OTMwNzUwNTE3YjE.-sqs\">\r\n \t<li>velcoity ratio\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212314\/1dcf1549ad77a06a51f20268d8f21e59.png\" alt=\"dfrac{v_{text{NH}_3}}{v_{text{HCl}}}\" width=\"36\" height=\"37\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ZTFiZjBkMGM5ODZhYjBkZWY3YmNjYzRkZjQ4Yjc2MDU.-0zg\">Substitute the molar masses of the gases into Graham\u2019s law and solve for the ratio.<\/p>\r\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-cnq\"><em> Step 2: Solve. <\/em><\/p>\r\n<p id=\"x-ck12-tnf\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212315\/dddb559fd0c6a5257db93b5b44578c43.png\" alt=\"frac{v_{NH_3}}{v_{HCl}}=sqrt{frac{36.46 text{ g\/mol}}{17.04 text{ g\/mol}}}=1.46\" width=\"235\" height=\"56\" \/><\/p>\r\n<p id=\"x-ck12-YmYwNmQwN2NiZGZhZjNmZTRhYjQzODE4NmU1MDIxNWI.-ak8\">The rate of diffusion of ammonia is 1.46 times faster than the rate of diffusion of hydrogen chloride.<\/p>\r\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-ork\"><em> Step 3: Think about your result <\/em><\/p>\r\n<p id=\"x-ck12-NTk2ZDY1NDhjMjI1MTA4OTY2MzllM2Q1YzU5MGZjYWE.-bdx\">Since ammonia has a smaller molar mass than hydrogen chloride, the velocity of its molecules is greater and the velocity ratio is larger than 1.<\/p>\r\n\r\n<h4>Summary<\/h4>\r\n<ul id=\"x-ck12-Mzg3OWM5YWMyYTRiNmE4YmQ1MmYwOWY5ZjE1ZTRjZGU.-7kb\">\r\n \t<li>The processes of gas diffusion and effusion are described.<\/li>\r\n \t<li>Graham\u2019s law relates the molecular mass of a gas to its rate of diffusion or effusion.<\/li>\r\n<\/ul>\r\n<h4>Practice<\/h4>\r\n<p id=\"x-ck12-Yjc3YzRjOTEyYmFmY2ZkMTdhOGMxM2VhZDRiMzQyMjU.-n25\">Read the material on the link below and do the practice problems:<\/p>\r\n<p id=\"x-ck12-ZGFjY2IzM2Q1MGUxMTdmNjc1NDkwY2U1YjcxMmYwNzE.-odl\"><a href=\"http:\/\/www.kentchemistry.com\/links\/GasLaws\/GrahamsLaw.htm\"> http:\/\/www.kentchemistry.com\/links\/GasLaws\/GrahamsLaw.htm <\/a><\/p>\r\n\r\n<h4>Review<\/h4>\r\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-orq\"><em> Questions <\/em><\/p>\r\n\r\n<ol id=\"x-ck12-ZDU1MWJiY2FmNWQ2Nzg1MmViOTA0ZjRkNTQ5ZWU2ODY.-tfs\">\r\n \t<li>Why can you smell food cooking when you are in the next room?<\/li>\r\n \t<li>Why does a helium-filled balloon gradually sink?<\/li>\r\n \t<li>What does temperature have to do with gas kinetic energies?<\/li>\r\n<\/ol>\r\n<div class=\"x-ck12-data-problem-set\"><\/div>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-MzFjY2RjZmMwYTY2YjY5YjM5NTJiNDdmYzZlNWNhMjM.-mnw\">\r\n \t<li><strong> diffusion: <\/strong> The tendency of molecules to move from an area of high concentration to an area of low concentration until the concentration is uniform.<\/li>\r\n \t<li><strong> effusion: <\/strong> The process of a confined gas escaping through a tiny hole in its container.<\/li>\r\n \t<li><strong> Graham\u2019s law: <\/strong> The rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas.<\/li>\r\n<\/ul>\r\n<\/div>\r\n[reveal-answer q=\"836080\"]Show References[\/reveal-answer]\r\n[hidden-answer a=\"836080\"]\r\n<h2>References<\/h2>\r\n<ol>\r\n \t<li>Image copyright Jandrie Lombard, 2014. <a href=\"http:\/\/www.shutterstock.com\"> http:\/\/www.shutterstock.com <\/a> .<\/li>\r\n \t<li>Ian Myles. <a href=\"http:\/\/www.flickr.com\/photos\/imphotography\/3754144111\/\"> http:\/\/www.flickr.com\/photos\/imphotography\/3754144111\/ <\/a> .<\/li>\r\n \t<li>Flickr: rick. <a href=\"http:\/\/www.flickr.com\/photos\/spine\/309730216\/\"> http:\/\/www.flickr.com\/photos\/spine\/309730216\/ <\/a> .<\/li>\r\n \t<li>Courtesy of US Navy Photographer's Mate 2nd Class Damon J. Moritz. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Basketball_game.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Basketball_game.jpg <\/a> .<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung. .<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung. .<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung. .<\/li>\r\n \t<li>Courtesy of NOAA Photo Library, NOAA Central Library; OAR\/ERL\/National Severe Storms Laboratory (NSSL). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Nssl0020.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Nssl0020.jpg <\/a> .<\/li>\r\n \t<li>Johann Kerseboom. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Robert_Boyle_0001.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Robert_Boyle_0001.jpg <\/a> .<\/li>\r\n \t<li>CK-12 Foundation - Wade Baxter. CK-12 Foundation .<\/li>\r\n \t<li>Courtesy of Petty Officer 3rd Class Charles Oki, US Navy. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Navy_baking_bread.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Navy_baking_bread.jpg <\/a> .<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung. .<\/li>\r\n \t<li>CK-12 Foundation - Wade Baxter. CK-12 Foundation .<\/li>\r\n \t<li>Courtesy of Robert Kaufmann, FEMA. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:FEMA_-_22585_-_Photograph_by_Robert_Kaufmann_taken_on_02-27-2006_in_Louisiana.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:FEMA_-_22585_-_Photograph_by_Robert_Kaufmann_taken_on_02-27-2006_in_Louisiana.jpg <\/a> .<\/li>\r\n \t<li>. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Joseph_louis_gay-lussac.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Joseph_louis_gay-lussac.jpg <\/a> .<\/li>\r\n \t<li>User:Calipper\/It.Wikipedia. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Frigorifero_Ignis_componenti.JPG\"> http:\/\/commons.wikimedia.org\/wiki\/File:Frigorifero_Ignis_componenti.JPG <\/a> .<\/li>\r\n \t<li>User:Martijn\/Nl.Wikipedia. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Fietspomp.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Fietspomp.jpg <\/a> .<\/li>\r\n \t<li>User:Masur\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Gas_cylinder_ammonia.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Gas_cylinder_ammonia.jpg <\/a> .<\/li>\r\n \t<li>User:GeorgHH\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:HighFlyer_Hamburg02.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:HighFlyer_Hamburg02.jpg <\/a> .<\/li>\r\n \t<li>User:J\u00fc\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Isomere_Ethanol_Dimethylether.png\"> http:\/\/commons.wikimedia.org\/wiki\/File:Isomere_Ethanol_Dimethylether.png <\/a> .<\/li>\r\n \t<li>Flickr: andrechinn. <a href=\"http:\/\/www.flickr.com\/photos\/andrec\/2699842079\/\"> http:\/\/www.flickr.com\/photos\/andrec\/2699842079\/ <\/a> .<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung. .<\/li>\r\n \t<li>Courtesy of NASA\/JPL. <a href=\"http:\/\/photojournal.jpl.nasa.gov\/catalog\/pia00124\"> http:\/\/photojournal.jpl.nasa.gov\/catalog\/pia00124 <\/a> .<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung. .<\/li>\r\n \t<li>Courtesy of J. D. Griggs, US Geological Survey. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Pahoeoe_fountain_sharpen.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Pahoeoe_fountain_sharpen.jpg <\/a> .<\/li>\r\n \t<li>Laura Guerin. CK-12 Foundation .<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung. .<\/li>\r\n \t<li>CK-12 Foundation - Christopher Auyeung. .<\/li>\r\n<\/ol>\r\n[\/hidden-answer]","rendered":"<h1 id=\"x-ck12-VGhlIEJlaGF2aW9yIG9mIEdhc2Vz-chapter\">The Behavior of Gases<\/h1>\n<div class=\"x-ck12-data\"><\/div>\n<h1 id=\"x-ck12-Q29tcHJlc3NpYmlsaXR5\">Compressibility<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-NmViYjAxYmNkMjlmNmYxNjk1NzM0NmNhMzI1YmU5YjQ.-czo\">\n<li>Define compressibility.<\/li>\n<li>Give examples of the uses of compressed gases.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-YzA0YTlmMzliMzM0YTZiNjgxYWVjM2YwZmY3YTEwNTY.-fl0\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212110\/20140811155522937478.jpeg\" alt=\"Compressing objects can help to squeeze them into small spaces\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-YzA0YTlmMzliMzM0YTZiNjgxYWVjM2YwZmY3YTEwNTY.-k01\"><strong> Will it all fit? <\/strong><\/p>\n<p id=\"x-ck12-OTI5ZDkwOTkxNDhjNDhiYzEzMGZiYjQzZGY1MmYyNWU.-ayx\">When we pack to go on vacation, there is always \u201cone more\u201d thing that we need to get in the suitcase. Maybe it\u2019s another bathing suit, pair of shoes, book \u2013 whatever the item, we need to get it in. Fortunately, we can squeeze things together somewhat. There is a little space between the folds of clothing, we can rearrange the shoes, and somehow we get that last thing in and close the suitcase.<\/p>\n<h3>Compressibility<\/h3>\n<p id=\"x-ck12-ZmM1NTllNjcyYTk5Y2MyNjMxNzQxNGRlNjMzMTFkZjA.-tda\">Scuba diving is a form of underwater diving in which a diver carries his own breathing gas, usually in the form of a tank of compressed air. The pressure in most commonly used scuba tanks ranges from 200 to 300 atmospheres. Gases are unlike other states of matter in that a gas expands to fill the shape and volume of its container. For this reason, gases can also be compressed so that a relatively large amount of gas can be forced into a small container. If the air in a typical scuba tank were transferred to a container at the standard pressure of 1 atm, the volume of that container would need to be about 2500 liters.<\/p>\n<div id=\"x-ck12-ODkzOWEyYzJjOGY0MGM3NGVkMzdiOGFmZTliOGJhNDA.-uak\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-bhf\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MzY4NDI2Mi00OC00LTI.\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212112\/20140811155523048566.jpeg\" alt=\"The pressure in a scuba tank is typically 200-300 atmospheres\" longdesc=\"Scuba%20diver.\" \/><\/p>\n<p><strong> Figure 14.1 <\/strong><\/p>\n<p id=\"x-ck12-Nzg4NDMyOTRmZmMyZjhhOTBkY2ZlYzdiZmMxNTg0Mzc.-lbs\">Scuba diver.<\/p>\n<\/div>\n<p id=\"x-ck12-YWNkZDRhYzZkNzE2ZjY4ZjM2NjY3N2VkOWU1Y2M3YWQ.-pwh\"><strong> Compressibility <\/strong> is the measure of how much a given volume of matter decreases when placed under pressure. If we put pressure on a solid or a liquid, there is essentially no change in volume. The atoms, ions, or molecules that make up the solid or liquid are very close together. There is no space between the individual particles, so they cannot pack together.<\/p>\n<p id=\"x-ck12-MTZiNzllZjc4YTI4ZDBjNGM3ZTlkN2I0Y2NkYmI0ZTA.-lqm\">The kinetic-molecular theory explains why gases are more compressible than either liquids or solids. Gases are compressible because most of the volume of a gas is composed of the large amounts of empty space between the gas particles. At room temperature and standard pressure, the average distance between gas molecules is about ten times the diameter of the molecules themselves. When a gas is compressed, as when the scuba tank is being filled, the gas particles are forced closer together.<\/p>\n<p id=\"x-ck12-NTdhNGE2MjNjNjZhZTFjN2YzOTAxNzQzNzE4OTM5YjE.-log\">Compressed gases are used in many situations. In hospitals, oxygen is often used for patients who have damaged lungs to help them breathe better. If a patient is having a major operation, the anesthesia that is administered will frequently be a compressed gas. Welding requires very hot flames produced by compresses acetylene and oxygen mixtures. Many summer barbeque grills are fueled by compressed propane.<\/p>\n<div id=\"x-ck12-NDlmNzMxMDBiMzZmNzdjNDg0OTA4N2RiMjAzZDM3Y2Q.-3db\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\n<p id=\"x-ck12-npe\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MzY4NDI4Ni04NC01My0z\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212113\/20140811155523191643.jpeg\" alt=\"Compressed gases such as oxygen are used for a wide variety of applications\" longdesc=\"Oxygen%20tank.\" \/><\/p>\n<p><strong> Figure 14.2 <\/strong><\/p>\n<p id=\"x-ck12-NzE5YTJmNmRlN2Q2MDhhMDk2ZWVjODkwNjEyNTRjMDI.-si5\">Oxygen tank.<\/p>\n<\/div>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-YjM4YzZlN2ZiZDFjYjFkNTc0NTJhNDkzZDYxODQ3MTQ.-v8h\">\n<li>Gases will compress more easily that solids or liquids because here is so much space between the gas molecules.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-uaj\"><em> Questions <\/em><\/p>\n<p id=\"x-ck12-Y2JlMjQ5M2YzMTNmNmRjMzNmZTI0MTMzYzcwM2IzZmY.-8sp\">Use the link below to answer the following questions:<\/p>\n<p id=\"x-ck12-MmNmMDcwZTY1ZmFlZTE5NjRmMmZkMDE4MDJmNmZlOTM.-7as\"><a href=\"https:\/\/web.archive.org\/web\/20160110232445\/http:\/\/www.cdxetextbook.com\/engines\/motivePower\/4gasEng\/engcycle.html\" target=\"_blank\" rel=\"noopener\">4-stroke engine cycle<\/a><\/p>\n<ol id=\"x-ck12-M2JjODliZDZjODY4MmZlYTNmZTc3MDNlMzBiODdmYTM.-ibp\">\n<li>What brings the fuel-air mixture into the cylinder?<\/li>\n<li>What is the role of the compression cycle?<\/li>\n<li>Does the exhaust cycle compress the gases produced by ignition?<\/li>\n<\/ol>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-4xl\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-MjAyYzI2YjllY2FmOWI4Zjg0MWZmMjU0MjkwNjc0YmE.-enh\">\n<li>Why is there no change in volume when pressure is applied to liquids and solids?<\/li>\n<li>Why do gases compress more easily than liquids and solids?<\/li>\n<li>List uses for compressed gases.<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-MzI4ZGI3ODM1YzVhMjEwMmM0ZTJhM2VkZDA0YWE5MGE.-j1l\">\n<li><strong> compressibility: <\/strong> The measure of how much a given volume of matter decreases when placed under pressure.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-RmFjdG9ycyBBZmZlY3RpbmcgR2FzIFByZXNzdXJl\">Factors Affecting Gas Pressure<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-NzQ2MjMzYzM5NzczNjEyNDY0M2QxYmFhZDJmNjM1YTU.-yku\">\n<li>List factors that affect gas pressure.<\/li>\n<li>Explain these effects in terms of the kinetic-molecular theory of gases.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-YTIzMmJkMTA4OGViNGViNDlhZjhhMzAxYjhiMjc5ZmU.-mkq\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212113\/20140811155523369990.jpeg\" alt=\"The pressure in a basketball in a game must be kept in a certain range\" width=\"200\" \/><\/span><\/p>\n<p id=\"x-ck12-NTYwMjJhZjEzNDNkYjhhZDc2NTUyYjYyMjUyNDhiYTA.-7kg\"><strong> How high does a basketball bounce? <\/strong><\/p>\n<p id=\"x-ck12-MmYzZjEzMTM3ZTFiOGRmOGExNTgyMDc5MWQ4ZjAwYTg.-1j3\">The pressure of the air in a basketball has to be adjusted so that the ball bounces to the correct height. Before a game, the officials check the ball by dropping it from shoulder height and seeing how far back up it bounces. What would the official do if the ball did not bounce up as far as it is supposed to? What would he do if it bounced too high?<\/p>\n<p id=\"x-ck12-OWFkMWE4YmY2YzY0ZmI3NmRjMzkwZDgwMGZiMjY0MDg.-8hg\">The pressure inside a container is dependent on the amount of gas inside the container. If the basketball does not bounce high enough, the official could remedy the situation by using a hand pump and adding more air to the ball. Conversely, if it bounces too high, he could let some air out of the ball.<\/p>\n<h3>Factors Affecting Gas Pressure<\/h3>\n<p id=\"x-ck12-MDIzM2FlNGQ0NGQwYTVjNjA2OWQ1OGU0ZGZkNjA1OTk.-cji\">Recall from the kinetic-molecular theory that gas particles move randomly and in straight lines until they elastically collide with either other gas particles or with one of the walls of the container. It is these collisions with the walls of the container that defines the pressure of the gas. Four variables are used to describe the condition of a gas. They are pressure <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTQwMDYwODU5NTgzOA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212115\/1b290cf55a4792cc5dfa6917c4dd4c52.png\" alt=\"(P)\" width=\"26\" height=\"18\" \/> , volume <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTQwMDYwODU5NTgzOQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212115\/beae5e0c3935dd8d8de532ebd274ad1c.png\" alt=\"(V)\" width=\"27\" height=\"18\" \/> , temperature <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212115\/ddb22c35d11ae2888dbca942bc72ede4.png\" alt=\"(T)\" width=\"25\" height=\"18\" \/> , and the amount of the gas as measured by the number moles <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTQwMDYwODU5NTg0MA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211023\/4168b7a35bc24c982496aef5b556a0a5.png\" alt=\"(n)\" width=\"23\" height=\"18\" \/> . We will examine separately how the volume, temperature, and amount of gas each affect the pressure of an enclosed gas sample.<\/p>\n<h4>Amount of Gas<\/h4>\n<p id=\"x-ck12-OWY0YTc0OGNlYmNkZDhiMzIxZWI5YTkxY2E4MzZjMjQ.-sdx\">The <strong> Figure <\/strong> below shows what happens when air is added to a <strong> rigid container <\/strong> . A rigid container is one that is incapable of expanding or contracting. A steel canister is an example of a rigid container.<\/p>\n<div id=\"x-ck12-ODIyM2RlNzYzMGM5ZjFiNWM2ZTBjNDJiMTRkMTRkMzA.-tk2\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-jh6\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MzY4NDQ1OC02Ni05Ni0y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212115\/20140811155523544082.png\" alt=\"An increase in the number of gas particles causes an increase in the pressure of a gas\" longdesc=\"Increase%20in%20pressure%20with%20increase%20in%20number%20of%20gas%20particles.\" \/><\/p>\n<p><strong> Figure 14.3 <\/strong><\/p>\n<p id=\"x-ck12-NmUyZjZiNTExYTNmZWQzMGQwNzkyYmM2ZmRhMGRhODE.-xre\">Increase in pressure with increase in number of gas particles.<\/p>\n<\/div>\n<p id=\"x-ck12-OTNkODIzODI4NzgzMTIzYTdlNTUwNDQ2NjQ3YzZjODI.-5vi\">The canister on the left contains a gas at a certain pressure. The attached air pump is then used to double the amount of gas in the canister. Since the canister cannot expand, the increased number of air molecules will strike the inside walls of the canister twice as frequently as they did before. The result is that the pressure inside the canister doubles. As you might imagine, if more and more air is continually added to a rigid container, it may eventually burst. Reducing the number of molecules in a rigid container has the opposite effect and the pressure decreases.<\/p>\n<h4>Volume<\/h4>\n<p id=\"x-ck12-MmNmNTY1NGU3ZWY2MDc2MTA5MmZhOTgyY2E4YTZiZjY.-k0w\">Pressure is also affected by the volume of the container. If the volume of a container is decreased, the gas molecules have less space in which to move around. As a result, they will strike the walls of the container more often and the pressure increases.<\/p>\n<p id=\"x-ck12-YTk5OGIyOTA0MjFiOTVmMjY0MjI3MDU0N2E4MGM2MDI.-ult\"><strong> Figure <\/strong> below shows a cylinder of gas whose volume is controlled by an adjustable piston. On the left, the piston is pulled mostly out and the gauge reads a certain pressure. On the right, the piston has been pushed so that the volume of the enclosed portion of the container where the gas is located has been cut in half. The pressure of the gas doubles. Increasing the volume of the container would have the opposite effect and the pressure of the gas would decrease.<\/p>\n<div id=\"x-ck12-NTQ2YWQ2MTUyMmJmOWVkMmY4YzRhZmIzNTU5NjIzY2Y.-5yh\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-ext\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MzY4NTE4Ny0wLTk1LTQ.\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212116\/20140811155523730968.png\" alt=\"A decrease in volume causes an increase in pressure for a gas\" longdesc=\"Decrease%20in%20gas%20volume%20produced%20increase%20in%20gas%20pressure.\" \/><\/p>\n<p><strong> Figure 14.4 <\/strong><\/p>\n<p id=\"x-ck12-MTdhODhjMzIwZDA5OGI5ZTNjMzg2ZDVkZmRkOGU1ZjI.-0pj\">Decrease in gas volume produced increase in gas pressure.<\/p>\n<\/div>\n<h4>Temperature<\/h4>\n<p id=\"x-ck12-Njg3ODA5ZGY2NzUyZGNmNTZlODZhODEzZjA3NDBlOTg.-kfd\">It would be very unadvisable to place a can of soup over a campfire without venting the can. As the can heats up, it may explode. The kinetic-molecular theory explains why. The air inside the rigid can of soup is given more kinetic energy by the heat coming from the campfire. The kinetic energy causes the air molecules to move faster and they impact the container walls more frequently and with more force. The increase in pressure inside may eventually exceed the strength of the can and it will explode. An additional factor is that the soup may begin boiling which will then aid even more gas and more pressure to the inside of the can.<\/p>\n<p id=\"x-ck12-YzY3NmNhNzIyNDBiMjQyNDIwZDJmMDA2MmZkYmViYmE.-gdz\">Shown in the <strong> Figure <\/strong> below is a cylinder of gas on the left that is at room temperature (300 K). On the right, the cylinder has been heated until the Kelvin temperature has doubled to 600 K. The kinetic energy of the gas molecules increases, so collisions with the walls of the container are now more forceful than they were before. As a result, the pressure of the gas doubles. Decreasing the temperature would have the opposite effect, and the pressure of an enclosed gas would decrease.<\/p>\n<div id=\"x-ck12-Y2M3NDc1NmQ0OTUyOGY2MWNiMmJhODFjMDlhZjE1OTU.-aoa\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MzY4NTA1OS0xLTI5LTM.\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212118\/20140811155523847727.png\" alt=\"An increase in temperature causes an increase in pressure for a gas\" longdesc=\"Increase%20in%20temperature%20produces%20increase%20in%20pressure.\" \/><\/p>\n<p><strong> Figure 14.5 <\/strong><\/p>\n<p id=\"x-ck12-MDk0NTYwOTE3ZDdmODg3ZDFlNGM2ZmJlYmJmODM0NzU.-r95\">Increase in temperature produces increase in pressure.<\/p>\n<\/div>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-NzBiNzJlNTNmZmFiZDJkMzAzMzY5YmU1ZDI0MWJhZDY.-l9n\">\n<li>An increase in the number of gas molecules in the same volume container increases pressure.<\/li>\n<li>A decrease in container volume increases gas pressure.<\/li>\n<li>An increase in temperature of a gas in a rigid container increases the pressure.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-40k\"><em> Questions <\/em><\/p>\n<p id=\"x-ck12-YmMzM2NmMmUyYzQ1NmUxNTQ2YTNmYjA2MWZlY2FkODA.-qln\">Watch the video at the link below and answer the following questions:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Factors That Influence Gas Pressure\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/0mVuWZ7nvcU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<ol id=\"x-ck12-MjI4MTBhZjZmMjZiNWViYjk3ZjdjMDViZDI2NmQ1NzQ.-mdi\">\n<li>What causes pressure?<\/li>\n<li>What happens when you let gas out of the container?<\/li>\n<li>If you increase the temperature, what happens to the pressure?<\/li>\n<li>Why does the pressure drop when you increase the volume?<\/li>\n<\/ol>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-1jz\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-MWRiZGM3YzE1MTFiZjE2NTViMGQwMmQzNWFlYmFlOTQ.-drd\">\n<li>What defines the pressure of a gas?<\/li>\n<li>Why does an increase in the number of molecules increase the pressure?<\/li>\n<li>Why does an increase in temperature increase the pressure?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-ZDBmZDcwYzRjMGI4ZDE4MDNkNzY1ZGQ4MGJlNzc5NTM.-son\">\n<li><strong> rigid container: <\/strong> One that is incapable of expanding or contracting.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-Qm95bGUncyBMYXc.\">Boyle&#8217;s Law<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-ZWUxNDlmOWNhZGE0M2E0YjQ5ZGM4ZDQ0N2RjZTQ5NmM.-ll0\">\n<li>State Boyle\u2019s Law.<\/li>\n<li>Use Boyle\u2019s Law to calculate volume-pressure relationships.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-Y2JlMzlhNTViZGMyYjI1Nzk4YTEzMTNlNWQ4ZTFlYWE.-xmm\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212120\/20140811155523976471.jpeg\" alt=\"Weather balloons expand and eventually burst as they travel to higher altitudes\" width=\"200\" \/><\/span><\/p>\n<p id=\"x-ck12-ODMyMzg2Y2ZhNTg1YmM1NzMxZmY3MDA4NDJkYzRkYTY.-awf\"><strong> How important is it to check the weather? <\/strong><\/p>\n<p id=\"x-ck12-ZDJjY2FiOGFmMDRkM2Q4ODg3NjhmNTAzNWU3NjZmODc.-0pl\">Each day, hundreds of weather balloons are launched. Made of a synthetic rubber and carrying a box of instruments, the helium-filled balloon rises up into the sky. As it gains altitude, the atmospheric pressure becomes less and the balloon expands. At some point the balloon bursts due to the expansion, the instruments drop (aided by a parachute) to be retrieved and studied for information about the weather.<\/p>\n<h3>Boyle\u2019s Law<\/h3>\n<p id=\"x-ck12-ZWFiOGFlNzVhZDE4NWM2ODU2MjY0NmYwZjczZGJmNjQ.-hm4\">Robert Boyle (1627-1691), an English chemist, is widely considered to be one of the founders of the modern experimental science of chemistry. He discovered that doubling the pressure of an enclosed sample of gas while keeping its temperature constant caused the volume of the gas to be reduced by half. <strong> Boyle\u2019s law <\/strong> states that the volume of a given mass of gas varies inversely with the pressure when the temperature is kept constant. An inverse relationship is described in this way. As one variable increases in value, the other variable decreases.<\/p>\n<p id=\"x-ck12-OWNjNDQzZDM2YmNlODhiZjE2N2Y1MzQxOGI0OGU3MzE.-ucz\">Physically, what is happening? The gas molecules are moving and are a certain distance apart from one another. An increase in pressure pushes the molecules closer together, reducing the volume. If the pressure is decreased, the gases are free to move about in a larger volume.<\/p>\n<div id=\"x-ck12-NzQxZWNhZjg1YzQ0ZjAyOTYwODMwNmVjMTYwMzU5ODM.-tdq\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\n<p id=\"x-ck12-ndj\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MzY5MDU0Ny0yNi0yOC01\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212121\/20140811155524117221.jpeg\" alt=\"Robert Boyle states that PV is constant at a given temperature\" longdesc=\"Robert%20Boyle.\" \/><\/p>\n<p><strong> Figure 14.6 <\/strong><\/p>\n<p id=\"x-ck12-OWVmMWFlZmVkOTIzYTY5MzBiMjExNjZlNjc0MzNjMWY.-py9\">Robert Boyle.<\/p>\n<\/div>\n<p id=\"x-ck12-NGQyMzdhMzdmNTlhN2U1NGI2NDJhOTUwZjc5NWYwMzE.-wkb\">Mathematically, Boyle\u2019s law can be expressed by the equation:<\/p>\n<p id=\"x-ck12-uhx\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjQzNzIwMjYwNw..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212122\/16476bd9759a44a54eeaf6837cecaa71.png\" alt=\"P times V = k\" width=\"85\" height=\"13\" \/><\/p>\n<p id=\"x-ck12-ZjczMjMwMzIxZGY5ODk0MGM4MDZhM2M0YjMwMWQyOTY.-who\">The\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjQzNzIwMjYwOA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212122\/d301c6452e642017007fb7cd755ce08b.png\" alt=\"k\" width=\"9\" height=\"13\" \/> is a constant for a given sample of gas and depends only on the mass of the gas and the temperature. The <strong> Table <\/strong> below shows pressure and volume data for a set amount of gas at a constant temperature. The third column represents the value of the constant\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212122\/8c0cf0a5c7fa06aa679382f3c2076f4d.png\" alt=\"(k)\" width=\"22\" height=\"18\" \/> for this data and is always equal to the pressure multiplied by the volume. As one of the variables changes, the other changes in such a way that the product of\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212122\/ac34d0426fddd2c0ce37f74932678203.png\" alt=\"P times V\" width=\"51\" height=\"12\" \/> always remains the same. In this particular case, that constant is 500\u00a0atm\u00a0\u00b7\u00a0ml.<\/p>\n<table id=\"x-ck12-MzI3NTRlOWZlYmY5YjNlNWQ1NjUzNTUyZTQ5ODgxN2Q.-ggq\" class=\"x-ck12-nofloat\">\n<caption>Pressure-Volume Data<\/caption>\n<tbody>\n<tr>\n<td><strong> Pressure (atm) <\/strong><\/td>\n<td><strong> Volume (mL) <\/strong><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjQzNzIwMjYxMA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212123\/7d6bf5e3739d48298c115ae53bb1315d.png\" alt=\"P times V = k (text{atm} cdot text{mL})\" width=\"168\" height=\"18\" \/><\/td>\n<\/tr>\n<tr>\n<td>0.5<\/td>\n<td>1000<\/td>\n<td>500<\/td>\n<\/tr>\n<tr>\n<td>0.625<\/td>\n<td>800<\/td>\n<td>500<\/td>\n<\/tr>\n<tr>\n<td>1.0<\/td>\n<td>500<\/td>\n<td>\n<p id=\"x-ck12-Y2VlNjMxMTIxYzJlYzkyMzJmM2EyZjAyOGFkNWM4OWI.-luz\">500<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>2.0<\/td>\n<td>250<\/td>\n<td>\n<p id=\"x-ck12-Y2VlNjMxMTIxYzJlYzkyMzJmM2EyZjAyOGFkNWM4OWI.-qrg\">500<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td>5.0<\/td>\n<td>100<\/td>\n<td>500<\/td>\n<\/tr>\n<tr>\n<td>8.0<\/td>\n<td>62.5<\/td>\n<td>500<\/td>\n<\/tr>\n<tr>\n<td>10.0<\/td>\n<td>50<\/td>\n<td>500<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"x-ck12-ZmU5MDBmMDA5MTZiODZhYWEyYzFmNDRjMmY3YTMxYTU.-zol\">A graph of the data in the table further illustrates the inverse relationship nature of Boyle\u2019s Law (see <strong> Figure <\/strong> below ). Volume is plotted on the <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212123\/0f14991bebd0bc5ca2e82990e029d525.png\" alt=\"x\" width=\"10\" height=\"8\" \/> -axis, with the corresponding pressure on the <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212123\/b4a681cb4394e7a28b5b01d49cf52c9e.png\" alt=\"y\" width=\"9\" height=\"12\" \/> -axis.<\/p>\n<div id=\"x-ck12-ODViN2RhNmQ0ZDQxNGRhMThiMTlmNmM1ZjBlN2JiZGI.-qwa\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-5uo\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MzY5MDU5MS05MS01OC02\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212124\/20140811155524277894.png\" alt=\"The pressure of a gas decreases as its volume increases\" longdesc=\"The%20pressure%20of%20a%20gas%20decreases%20as%20the%20volume%20increases%2C%20making%20Boyle%E2%80%99s%20law%20an%20inverse%20relationship.\" \/><\/p>\n<p><strong> Figure 14.7 <\/strong><\/p>\n<p id=\"x-ck12-ODNmMzA3NDFmMDIyMTgxYTJlZjYyMTMxOTlkZTk4NzY.-fgt\">The pressure of a gas decreases as the volume increases, making Boyle\u2019s law an inverse relationship.<\/p>\n<\/div>\n<p id=\"x-ck12-NzBkODU5MGQxMjYxNGZmMTk5ZDZiYzY2MTU3MTJkYmQ.-ozb\">Boyle\u2019s Law can be used to compare changing conditions for a gas. We use\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjQzNzIwMjYxMg..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/b95fd9f12605d93cbb24fd5fc1d83fbc.png\" alt=\"P_1\" width=\"18\" height=\"16\" \/> and\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/29c883aa1a216bc9157974e94c993965.png\" alt=\"V_1\" width=\"16\" height=\"16\" \/> to stand for the initial pressure and initial volume of a gas. After a change has been made,\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/f49ec791ef1ecb63d1c14840f9c2f983.png\" alt=\"P_2\" width=\"18\" height=\"15\" \/> and\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> stand for the final pressure and volume. The mathematical relationship of Boyle\u2019s Law becomes:<\/p>\n<p id=\"x-ck12-esr\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/f4c066bce5329c1006f6e21829e93d1a.png\" alt=\"P_1 times V_1=P_2 times V_2\" width=\"140\" height=\"16\" \/><\/p>\n<p id=\"x-ck12-YTI3NzZjYmQzMzBkYzFhNjhkYjJhNDA0ZjZmMDZkMmM.-xpx\">This equation can be used to calculate any one of the four quantities if the other three are known.<\/p>\n<h4>Sample Problem: Boyle\u2019s Law<\/h4>\n<p id=\"x-ck12-NTMyZTk3ODQ4MDdmOWMzNzcxZDJmMWRkMGUxZTNjOWQ.-ncg\">A sample of oxygen gas has a volume of 425 mL when the pressure is equal to 387 kPa. The gas is allowed to expand into a 1.75 L container. Calculate the new pressure of the gas.<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-o1i\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-6pr\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-ln9\">\n<li><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTQwMTM5OTE2NzcyNg..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/db5e81bc83ca81445200cbbc41291805.png\" alt=\"P_1=387 text{kPa}\" width=\"106\" height=\"17\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTQwMTM5OTE2NzcyNw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212127\/a39e628ca221147569ced5499f911de1.png\" alt=\"V_1=425 text{mL}\" width=\"101\" height=\"17\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212127\/41e797e168699fd6af53fda820e06914.png\" alt=\"V_2=1.75 text{L}=1750 text{mL}\" width=\"183\" height=\"16\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-uik\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-wqa\">\n<li><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTQwMTM5OTE2NzcyOQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212128\/33dd88af6f64da7d9108a4e93ab30012.png\" alt=\"P_2=? text{kPa}\" width=\"82\" height=\"16\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-YmZmZDZiNjNmNzdmZDU1MjZiNDE1ZjAxMWIzMWZlYWY.-ahu\">Use Boyle\u2019s Law to solve for the unknown pressure <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjQzNzIwMjYxMw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212128\/7309a9826c8e30de6d5ca051de922d91.png\" alt=\"(P_2)\" width=\"31\" height=\"18\" \/> . It is important that the two volumes ( <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/29c883aa1a216bc9157974e94c993965.png\" alt=\"V_1\" width=\"16\" height=\"16\" \/> and <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjQzNzIwMjYxNA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> ) are expressed in the same units, so\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> has been converted to mL.<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-pqs\"><em> Step 2: Solve. <\/em><\/p>\n<p id=\"x-ck12-ZTZkYTlmMmZlMzNlOWE1MmY0OGVlNjA0MzgxYmM4ZTE.-txa\">First, rearrange the equation algebraically to solve for <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/f49ec791ef1ecb63d1c14840f9c2f983.png\" alt=\"P_2\" width=\"18\" height=\"15\" \/> .<\/p>\n<p id=\"x-ck12-3fb\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212128\/bff7f7fa27ec0593ea7b3c9219e71bb2.png\" alt=\"P_2=frac{P_1 times V_1}{V_2}\" width=\"103\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-OTNiYzM1Nzk5YWMwYTdmNDZjM2RiOTlmODhhOGVkYTM.-5as\">Now substitute the known quantities into the equation and solve.<\/p>\n<p id=\"x-ck12-9zf\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212129\/7fb79e9ad9bcb40194be760d22e3cdf5.png\" alt=\"P_2=frac{387 text{ kPa} times 425 text{ mL}}{1750 text{ mL}}=94.0 text{ kPa}\" width=\"282\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-nkb\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-Y2JlZjk5YTA0ZGY0NDNkZWZkZGU0YTZlZmU3OWYxOWU.-2qy\">The volume has increased to slightly over 4 times its original value and so the pressure is decreased by about <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212130\/c1689389012aed4922e2871602dab8bc.png\" alt=\"frac{1}{4}{th}\" width=\"24\" height=\"23\" \/> . The pressure is in kPa and the value has three significant figures. Note that any pressure or volume units can be used as long as they are consistent throughout the problem.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-NmZjNTA3ZTFmOTliNDgxOWRmNmIxMWU5NzMzMDIxZjc.-ler\">\n<li>The volume of a gas is inversely proportional to temperature.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-ZjRmZTgxMGFkNmYzZWFhZGViYzllOGZjOWIxZDdhZjQ.-5an\">Do the problems at the link below:<\/p>\n<p id=\"x-ck12-MDVhNjVlMmMyMjViMWU3N2U4OTA0ZDcxZTM5MzJiZWM.-wk0\"><a href=\"http:\/\/www.concord.org\/~ddamelin\/chemsite\/g_gasses\/handouts\/Boyle_Problems.pdf\"> http:\/\/www.concord.org\/~ddamelin\/chemsite\/g_gasses\/handouts\/Boyle_Problems.pdf <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-0em\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-N2U2ZDkxMWYwMjNjZTY4ZGUxNzg1ZDk2ZGVlOGU5Zjg.-xjk\">\n<li>What does \u201cinversely\u201d mean in this law?<\/li>\n<li>Explain Boyle\u2019s law in terms of the kinetic-molecular theory of gases.<\/li>\n<li>Does it matter what units are used?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-ZjQzYTMwY2U4MjgyNTBkZWZlYThlZmM2YjMwNWFlMWY.-bky\">\n<li><strong> Boyle\u2019s law: <\/strong> The volume of a given mass of gas varies inversely with the pressure when the temperature is kept constant.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-Q2hhcmxlcydzIExhdw..\">Charles&#8217;s Law<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-YjAyNzVlYWE1MTQ0YWU4YjRmZWI4Zjk1YjMwZjY1NjU.-akf\">\n<li>State Charles\u2019 Law.<\/li>\n<li>Use this law to perform calculations involving volume-temperature relationships.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-N2Q4MmRkZTA4NjhmZjdlNmFiM2EwMWFkY2MwNDg1NGI.-wju\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212130\/20140811155524496248.jpeg\" alt=\"When bread is baked, the carbon dioxide inside expands\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-YTVjOWMzMzFlOTdkODBkYmNkYWQ4ZGVhMDQ1MTQ1ZjQ.-m5s\"><strong> How do you bake bread? <\/strong><\/p>\n<p id=\"x-ck12-MDU1MmViYjllZWVlNTM2NzkxNWViNmNjNTk1ZGY4NTU.-oy2\">Everybody enjoys the smell and taste of freshly-baked bread. It is light and fluffy as a result of the action of yeast on sugar. The yeast converts the sugar to carbon dioxide, which at high temperatures causes the dough to expand. The end-result is an enjoyable treat, especially when covered with melted butter.<\/p>\n<h3>Charles\u2019s Law<\/h3>\n<div id=\"x-ck12-YThhNjRlYWNjMGVjMjBmNzM3MzBmOGVlZGM5NzNiYjY.-y3m\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-tp3\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MzY5MjA3NS00MS0xOS0y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212132\/20140811155524707039.png\" alt=\"As a gas is heated at constant pressure, its volume increases\" longdesc=\"As%20a%20container%20of%20confined%20gas%20is%20heated%2C%20its%20molecules%20increase%20in%20kinetic%20energy%20and%20push%20the%20movable%20piston%20outward%2C%20resulting%20in%20an%20increase%20in%20volume.\" \/><\/p>\n<p><strong> Figure 14.8 <\/strong><\/p>\n<p id=\"x-ck12-MTUzMmUxNjQzMjZiY2Y4ZGQ3MThmNmE1YjExZmUyMTk.-jfu\">As a container of confined gas is heated, its molecules increase in kinetic energy and push the movable piston outward, resulting in an increase in volume.<\/p>\n<\/div>\n<p id=\"x-ck12-YWI3MzJmZDQzMDM1OTM4YjNjNDFlYjBlYWY1MTI0ODE.-zhd\">French physicist Jacques Charles (1746-1823) studied the effect of temperature on the volume of a gas at constant pressure. <strong> Charles\u2019s law <\/strong> states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stoppage of molecular motion.<\/p>\n<p id=\"x-ck12-Nzc2NzQ2ODk5OGUyZWM3ODM3NGYwYzc5MTAwYjhiZGE.-ma4\">Mathematically, the direct relationship of Charles\u2019s law can be represented by the following equation:<\/p>\n<p id=\"x-ck12-pvf\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212133\/68ec4151a7f8c52092668cc72efd0265.png\" alt=\"frac{V}{T}=k\" width=\"50\" height=\"37\" \/><\/p>\n<p id=\"x-ck12-NGE5NzZlOTVlNWFmMjM2ZmIwODQ1ZTFjOTMxYTFhNjc.-lc0\">As with Boyle\u2019s law, <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212122\/d301c6452e642017007fb7cd755ce08b.png\" alt=\"k\" width=\"9\" height=\"13\" \/> is constant only for a given gas sample. The <strong> Table <\/strong> below shows temperature and volume data for a set amount of gas at a constant pressure. The third column is the constant for this particular data set and is always equal to the volume divided by the Kelvin temperature.<\/p>\n<table id=\"x-ck12-MmJhZjZhNWU4OGVlMDlhMjg3MjhiZjUzZmE1ODE2NjI.-1ne\" class=\"x-ck12-nofloat\">\n<caption>Temperature-Volume Data<\/caption>\n<tbody>\n<tr>\n<td><strong> Temperature (K) <\/strong><\/td>\n<td><strong> Volume (mL) <\/strong><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212133\/de68e314bf5efae1e99b1782323d02a9.png\" alt=\"frac{V}{T}=kleft (frac{mL}{K}right)\" width=\"88\" height=\"24\" \/><\/td>\n<\/tr>\n<tr>\n<td>50<\/td>\n<td>20<\/td>\n<td>0.40<\/td>\n<\/tr>\n<tr>\n<td>100<\/td>\n<td>40<\/td>\n<td>0.40<\/td>\n<\/tr>\n<tr>\n<td>150<\/td>\n<td>60<\/td>\n<td>0.40<\/td>\n<\/tr>\n<tr>\n<td>200<\/td>\n<td>80<\/td>\n<td>0.40<\/td>\n<\/tr>\n<tr>\n<td>300<\/td>\n<td>120<\/td>\n<td>0.40<\/td>\n<\/tr>\n<tr>\n<td>500<\/td>\n<td>200<\/td>\n<td>0.40<\/td>\n<\/tr>\n<tr>\n<td>1000<\/td>\n<td>400<\/td>\n<td>0.40<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"x-ck12-NzYxZTI2Zjc0NTYxYzhmNDQ0NDI2ZDE4ZDEwMzMzY2U.-vmq\">When this data is graphed, the result is a straight line, indicative of a direct relationship, shown in <strong> Figure <\/strong> below .<\/p>\n<div id=\"x-ck12-NGFlYTMwM2RlMDUwNDI1M2Y2MTcyODQzZTBlMzlhY2Y.-vas\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-doy\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MzY5MjExMS0yLTMzLTM.\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212133\/20140811155524837584.png\" alt=\"The volume of a gas increases as the Kelvin temperature increases\" longdesc=\"The%20volume%20of%20a%20gas%20increases%20as%20the%20Kelvin%20temperature%20increases.\" \/><\/p>\n<p><strong> Figure 14.9 <\/strong><\/p>\n<p id=\"x-ck12-M2YyNDJmY2ZkNDQyYmQ0ZDI5ZjQ1ZDU0Y2IwMzJmYmU.-b6p\">The volume of a gas increases as the Kelvin temperature increases.<\/p>\n<\/div>\n<p id=\"x-ck12-MmQxZWEwODdiZTAyMTBhYmY5OWQ0YWZjZjEyNTllYTY.-fn5\">Notice that the line goes exactly toward the origin, meaning that as the absolute temperature of the gas approaches zero, its volume approaches zero. However, when a gas is brought to extremely cold temperatures, its molecules would eventually condense into the liquid state before reaching absolute zero. The temperature at which this change into the liquid state occurs varies for different gases.<\/p>\n<p id=\"x-ck12-NDJlYzE1YjgxZGE2ZTZiYTdiZTU4YTkzNjZiNmZjYjc.-ihc\">Charles\u2019s Law can also be used to compare changing conditions for a gas. Now we use\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/29c883aa1a216bc9157974e94c993965.png\" alt=\"V_1\" width=\"16\" height=\"16\" \/> and\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212135\/5ef2d3527b9eadc8e4f588dbe3795272.png\" alt=\"T_1\" width=\"16\" height=\"16\" \/> to stand for the initial volume and temperature of a gas, while <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> and\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212135\/d64b622498e5207084db58e1dea5b037.png\" alt=\"T_2\" width=\"16\" height=\"15\" \/> stand for the final volume and temperature. The mathematical relationship of Charles\u2019s Law becomes:<\/p>\n<p id=\"x-ck12-2kp\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212135\/6bb0892b51cef8a85ef101eb68a425e5.png\" alt=\"frac{V_1}{T_1}=frac{V_2}{T_2}\" width=\"63\" height=\"40\" \/><\/p>\n<p id=\"x-ck12-YmQ0ZThmMzUwZTFlMTI1ODcxZDEyMGZhY2EyMzBiNTg.-cm8\">This equation can be used to calculate any one of the four quantities if the other three are known. The direct relationship will only hold if the temperatures are expressed in Kelvin. Temperatures in Celsius will not work. Recall the relationship that K\u00a0=\u00a0\u00b0C\u00a0+\u00a0273.<\/p>\n<h4>Sample Problem: Charles\u2019s Law<\/h4>\n<p id=\"x-ck12-NDdkNmU2OGRiMTRmNzMwMmU0MTc5OTM5M2UxMWFkMjc.-inz\">A balloon is filled to a volume of 2.20 L at a temperature of 22\u00b0C. The balloon is then heated to a temperature of 71\u00b0C. Find the new volume of the balloon.<\/p>\n<p id=\"x-ck12-OGU2YWEyYzY0NzAwMDUxZjI4NjFjY2E4MjYyNmNhN2I.-ckj\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-ki5\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-s6x\">\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212135\/2f94b6d4455bd48ff87d466ed526654b.png\" alt=\"V_1=2.20 text{ L}\" width=\"91\" height=\"17\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212136\/b4d715ba982209d5f333b1baf3727991.png\" alt=\"T_1=22^circ text{C}=295 text{ K}\" width=\"151\" height=\"17\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212137\/b4893bb931993217110b5288da684114.png\" alt=\"T_2=71^circ text{C}=344 text{ K}\" width=\"151\" height=\"16\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-6jl\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-9qb\">\n<li><em> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212137\/bbda48a6df75d3defbe58c3b3ce2ebe6.png\" alt=\"V_2= ? text{ L}\" width=\"62\" height=\"16\" \/><\/em><\/li>\n<\/ul>\n<p id=\"x-ck12-NzRlZWZlMTBkM2Q2NmNkYjJiNGJlODZhMjg2ZDhmMGU.-mvb\">Use Charles\u2019s law to solve for the unknown volume <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212137\/0a4b20dd3f9257612d0c459edf0075c2.png\" alt=\"(V_2)\" width=\"30\" height=\"18\" \/> . The temperatures have first been converted to Kelvin.<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-pfi\"><em> Step 2: Solve. <\/em><\/p>\n<p id=\"x-ck12-Yzg1OGU4MDExNzQ5NGQ4OGU4NjY3ZWZlMTFmNDdhNzQ.-ebu\">First, rearrange the equation algebraically to solve for <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> .<\/p>\n<p id=\"x-ck12-ofi\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212138\/8866248b5555b9388de04b70a852ec31.png\" alt=\"V_2=frac{V_1 times T_2}{T_1}\" width=\"101\" height=\"40\" \/><\/p>\n<p id=\"x-ck12-OTNiYzM1Nzk5YWMwYTdmNDZjM2RiOTlmODhhOGVkYTM.-akc\">Now substitute the known quantities into the equation and solve.<\/p>\n<p id=\"x-ck12-2xd\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212138\/8d605d9b16320240278b22c3ad3b9919.png\" alt=\"V_2=frac{2.20 text{ L} times 344 text{ K}}{295 text{ K}}=2.57 text{ L}\" width=\"236\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-rbu\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-YWRjNGQ0MGFjZWFlODYzN2NkMDI2OGYxMDFmNTRhY2Q.-n2j\">The volume increases as the temperature increases. The result has three significant figures.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-ZTMxNDFkOWExNWYwMGY4YTNhZTc4NjMwOTNiMDhkODg.-n2s\">\n<li>Increasing the temperature of a gas at constant pressure will produce and increase in the volume.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-MjJhNzc5ZTg5MmQxOGM2NjRhNjc1Mjk4MGFlODQ0NTQ.-spz\">Perform the calculations at the web site below:<\/p>\n<p id=\"x-ck12-Nzg4MWRiYzg2ZWZjMDM0MGVjNmUwZGIzNjU1MWJjOGI.-kxw\"><a href=\"http:\/\/mmsphyschem.com\/chuckL.pdf\"> http:\/\/mmsphyschem.com\/chuckL.pdf <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-lp0\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-YzQ0OTBlNGYwY2IwYmI0YTgyZmFiOWIyZmI1ZjAyNGI.-qka\">\n<li>Explain Charles\u2019 Law in terms of the kinetic molecular theory.<\/li>\n<li>Why does the temperature need to be in Kelvin?<\/li>\n<li>Does Charles\u2019 law hold when the gas becomes a liquid?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-Njg1MjU3NDcxYzdhN2JkMjg1NTI0OGNkZDRhOTU2NWQ.-6or\">\n<li><strong> Charles\u2019s law: <\/strong> The volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-R2F5LUx1c3NhYydzIExhdw..\">Gay-Lussac&#8217;s Law<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-YzFjMWNmOTQ4Yzg5MGI2NTk5MzlhMDE1YTM5MGVjOGY.-xrq\">\n<li>State Gay-Lussac\u2019s law.<\/li>\n<li>Use this law to perform calculations involving pressure-temperature relationships.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-ZWJjODUzNTZiZDkwOWU5NTI4NjYzODRiZDJhNzdmNTg.-p8q\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212140\/20140811155524966201.jpeg\" alt=\"The temperature of a tank of gas needs to be taken into account when measuring its pressure\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-ODYwYjc2MmNkN2JjNzQ2ZjE1ZjI3ZTczYzI1OGYxZjU.-5ov\"><strong> How much propane is in the tank? <\/strong><\/p>\n<p id=\"x-ck12-NWY4ZjA5MDJmOTU1OGVhZDJjMmVkYjc3ODNiNjVkZTI.-087\">Propane tanks are widely used with barbeque grills. But it\u2019s not fun to find out half-way through your grilling that you\u2019ve run out of gas. You can buy gauges that measure the pressure inside the tank to see how much is left. The gauge measures pressure and will register a higher pressure on a hot day than it will on a cold day. So you need to take the air temperature into account when you decide whether or not to refill the tank before your next cook-out.<\/p>\n<h3>Gay-Lussac\u2019s Law<\/h3>\n<p id=\"x-ck12-Njk5MjJmODk5Y2UxOTE5YmM1NDM2MGMwNGEwMmNkNDg.-zqo\">When the temperature of a sample of gas in a rigid container is increased, the pressure of the gas increases as well. The increase in kinetic energy results in the molecules of gas striking the walls of the container with more force, resulting in a greater pressure. The French chemist Joseph Gay-Lussac (1778-1850) discovered the relationship between the pressure of a gas and its absolute temperature. <strong> Gay-Lussac\u2019s law <\/strong> states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant. Gay-Lussac\u2019s law is very similar to Charles\u2019s law, with the only difference being the type of container. Whereas the container in a Charles\u2019s law experiment is flexible, it is rigid in a Gay-Lussac\u2019s law experiment.<\/p>\n<div id=\"x-ck12-NDU2NjI2ZjZjZjVmYjA2ZjNiOTVmZThhNzQyOGU2NmM.-ryp\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\n<p id=\"x-ck12-oip\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2MzY4ODcwMC03Mi04MS00LjIuNS4y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212142\/20140811155525097528.jpeg\" alt=\"In Gay-Lussac's Law, the pressure of a gas varies directly with the temperature of a gas\" longdesc=\"Joseph%20Louis%20Gay-Lussac.\" \/><\/p>\n<p><strong> Figure 14.10 <\/strong><\/p>\n<p id=\"x-ck12-NzU2YjdkMzFmNTk2NWNlODM4NjViYTM5NWMyMjg4Yjk.-kad\">Joseph Louis Gay-Lussac.<\/p>\n<\/div>\n<p id=\"x-ck12-ZDEwNGNmYjI1MWVmMGE2OGIxYjZlOWM3YTE4YjMyYTI.-r1t\">The mathematical expressions for Gay-Lussac\u2019s law are likewise similar to those of Charles\u2019s law:<\/p>\n<p id=\"x-ck12-YmU1ZDVkMzc1NDJkNzVmOTNhODcwOTQ0NTlmNzY2Nzg.-uxd\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwNjgzMzcwMg..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212142\/5f3e7a33885242475c580acdf25b8f00.png\" alt=\"frac{P}{T}=k quad text{and} quad frac{P_1}{T_1}=frac{P_2}{T_2}\" width=\"182\" height=\"40\" \/><\/p>\n<p id=\"x-ck12-NTc3Mzc1YjczNDNjYjIwMTQxMjg5Y2E4OWI0MWI4ZTg.-ylo\">A graph of pressure vs. temperature also illustrates a direct relationship. As a gas is cooled at constant volume its pressure continually decreases until the gas condenses to a liquid.<\/p>\n<h4>Sample Problem: Gay-Lussac\u2019s Law<\/h4>\n<p id=\"x-ck12-MDhjZjhkNTNjOWYyODZmMWNhZWNkNjg4MzQyYjRjZWU.-tlw\">The gas in an aerosol can is under a pressure of 3.00 atm at a temperature of 25\u00b0C. It is dangerous to dispose of an aerosol can by incineration. What would the pressure in the aerosol can be at a temperature of 845\u00b0C?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-7eh\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-qtz\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-a6j\">\n<li><span class=\"x-ck12-underline\"> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212143\/d70729cd610db66b224f2f9beb8d97e8.png\" alt=\"P_1=3.00 text{atm}\" width=\"112\" height=\"16\" \/><\/span><\/li>\n<li><span class=\"x-ck12-underline\"> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212144\/7ea9fae24f1538e82b6f9bf895af1437.png\" alt=\"T_1=25^circ text{C}=298 text{ K}\" width=\"151\" height=\"17\" \/><\/span><\/li>\n<li><span class=\"x-ck12-underline\"> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212144\/7cd92943119aa0e329d3c6ae9162c438.png\" alt=\"T_2=845^circ text{C}=1118 text{ K}\" width=\"169\" height=\"16\" \/><\/span><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-v9l\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-dwb\">\n<li><em> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212145\/362eb69fcff4f8ce64140e9b9d372da6.png\" alt=\"P_2=? text{atm}\" width=\"83\" height=\"15\" \/><\/em><\/li>\n<\/ul>\n<p id=\"x-ck12-YjY3ZTEwNTY3MzY0ODc4NGMzMDM5ZTg1MjQ1ZmQ3Njg.-btn\">Use Gay-Lussac\u2019s law to solve for the unknown pressure <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212128\/7309a9826c8e30de6d5ca051de922d91.png\" alt=\"(P_2)\" width=\"31\" height=\"18\" \/> . The temperatures have first been converted to Kelvin.<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-fjb\"><em> Step 2: Solve. <\/em><\/p>\n<p id=\"x-ck12-Yzg1OGU4MDExNzQ5NGQ4OGU4NjY3ZWZlMTFmNDdhNzQ.-q4p\">First, rearrange the equation algebraically to solve for <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwNjgzMzcwMw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> .<\/p>\n<p id=\"x-ck12-jaz\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212145\/2573f81dac377b98147ae4290c38664d.png\" alt=\"P_2=frac{P_1 times T_2}{T_1}\" width=\"103\" height=\"40\" \/><\/p>\n<p id=\"x-ck12-OTNiYzM1Nzk5YWMwYTdmNDZjM2RiOTlmODhhOGVkYTM.-b4c\">Now substitute the known quantities into the equation and solve.<\/p>\n<p id=\"x-ck12-mcc\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212146\/6658e7aac6eb46c39a3ae28767d7bf36.png\" alt=\"P_2=frac{3.00 text{ atm} times 1118 text{ K}}{298 text{ K}}=11.3 text{ atm}\" width=\"286\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-neb\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-ZGIxYzgzMmIyNzc0YzY3NmJlZTA5NTkwYTNjYTcwYzY.-pqv\">The pressure increases dramatically due to large increase in temperature.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-MWIwMzllYzdmOTkyMDMzYjUyYTQ5Mzc0MjAxMTA0OWQ.-hgv\">\n<li>Pressure and temperature at constant volume are directly proportional.<\/li>\n<\/ul>\n<h4 id=\"x-ck12-OGMwMDRkN2UzYjhhNWE4NTBhYjI0NTljM2FmYzJmNTc.-wal_4-0rl\">Practice<\/h4>\n<p id=\"x-ck12-YjNkMmJlZTZmNzk4YzU3YTNjY2QzNWQ0NWY1YTM0MzM.-2mw\">Work on the problems found at the web site below:<\/p>\n<p id=\"x-ck12-YmRlYzMxYmYxY2VlNDBiMWI0ZTA4MGY5MWIyYTAzODU.-gku\"><a href=\"https:\/\/web.archive.org\/web\/20160302060512\/http:\/\/www.chemteam.info\/GasLaw\/WS-Gay-Lussac.html\" target=\"_blank\" rel=\"noopener\">Gay Lussac&#8217;s Law<\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-znu\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-ODVmNzk2ZDg5NTExYjYzYmI1Yzk5N2M3NDc4ODUxMWU.-0wn\">\n<li>Explain Gay-Lussac\u2019s Law in terms of the kinetic-molecular theory.<\/li>\n<li>What would a graph of pressure vs. temperature show us?<\/li>\n<li>What is the difference in containers in Charles\u2019 Law and Gay-Lussac\u2019s Law?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-ZGYwMjA5OTQxMTEzM2Q4NzMxNGU2ZmNhNWU2ZjUwMjY.-6xc\">\n<li><strong> Gay-Lussac\u2019s law: <\/strong> The pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-Q29tYmluZWQgR2FzIExhdw..\">Combined Gas Law<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-MTkwN2NhNmM3MDgwY2M4NTQ4MmY2NmM4NjJmMzBhMzA.-nyu\">\n<li>State the combined gas law.<\/li>\n<li>Use the law to calculate parameters in general gas problems.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-OTdiZjMzYjQzMWVmNjJjZmQzMjBjNDBjNzhhODZjMGI.-w5q\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212147\/20140811155525236852.jpeg\" alt=\"A refrigerator uses the combined gas law to dissipate heat\" width=\"200\" \/><\/span><\/p>\n<p id=\"x-ck12-NjE2OTEwMDhmZmNiZTRhZjc3NDFhYTUzNWRkMmFhZTY.-xdz\"><strong> What keeps things cold? <\/strong><\/p>\n<p id=\"x-ck12-MzhkNTk3MjNlZDg5MGFmMzU1YTU1Y2M5MmExZDk3YWE.-bkw\">The modern refrigerator takes advantage of the gas laws to remove heat from a system. Compressed gas in the coils (see above) is allowed to expand. This expansion lowers the temperature of the gas and transfers heat energy from the material in the refrigerator to the gas. As the gas is pumped through the coils, the pressure on the gas compresses it and raises the gas temperature. This heat is then dissipated through the coils into the outside air. As the compressed gas is pumped through the system again, the process repeats itself.<\/p>\n<h3>Combined Gas Law<\/h3>\n<p id=\"x-ck12-YTg4YTM1YjZmNWIyNjcyOTU2NmU1OGM2YWQ1YTlkZjQ.-5wq\">To this point, we have examined the relationships between any two of the variables of <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwNzg1Mjk4Ng..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/dc2a557e4775924079e5de1061889778.png\" alt=\"P\" width=\"14\" height=\"12\" \/> , <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/adb347871c296be723ff3085c0023a28.png\" alt=\"V\" width=\"14\" height=\"12\" \/> , and <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwNzg1Mjk4Nw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/a7284843d100e640c0e00ae83ab66a90.png\" alt=\"T\" width=\"13\" height=\"12\" \/> , while the third variable is held constant. However, situations arise where all three variables change. The <strong> combined gas law <\/strong> expresses the relationship between the pressure, volume, and absolute temperature of a fixed amount of gas. For a combined gas law problem, only the amount of gas is held constant.<\/p>\n<p id=\"x-ck12-YmU1ZDVkMzc1NDJkNzVmOTNhODcwOTQ0NTlmNzY2Nzg.-rzv\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212149\/15e5e85314846399443271ebb3f9bcb0.png\" alt=\"frac{P times V}{T}=k quad text{and} quad frac{P_1 times V_1}{T_1}=frac{P_2 times V_2}{T_2}\" width=\"299\" height=\"40\" \/><\/p>\n<h4>Sample Problem: Combined Gas Law<\/h4>\n<p id=\"x-ck12-OGI0ZTEyNDJlOTdiZWM1OWQ0N2ViMmU4YmQ1N2I4ODg.-r4p\">2.00 L of a gas at 35\u00b0C and 0.833 atm is brought to standard temperature and pressure (STP). What will be the new gas volume?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-8tv\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-tvg\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-n4j\">\n<li><em> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212149\/064576e16ae72453b8d008e85e54c48d.png\" alt=\"P_1=0.833 text{ atm}\" width=\"121\" height=\"16\" \/><\/em><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212150\/604efeb1aaaf796728459e3e8d41ca05.png\" alt=\"V_1=2.00 text{ L}\" width=\"91\" height=\"17\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212150\/999d8e43bf34ed87892a395f377a853c.png\" alt=\"T_1=35^circ text{C}=308 text{ K}\" width=\"151\" height=\"17\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212151\/5d24e643f58e553426043959bc772fc0.png\" alt=\"P_2=1.00 text{ atm}\" width=\"112\" height=\"15\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212151\/bf67d672df5dea053630c457b15ee84b.png\" alt=\"T_2=0^circ text{C}=273 text{ K}\" width=\"142\" height=\"16\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-uwj\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-heb\">\n<li><em> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212152\/51e1944bfa6d5ec4a5ee7088baf64635.png\" alt=\"V_2=? text{ L}\" width=\"62\" height=\"16\" \/><\/em><\/li>\n<\/ul>\n<p id=\"x-ck12-YWFmZDAyZWY3MjlmNTAzMzc5YjVlMjE3MmZlZDg0ZDU.-abl\">Use the combined gas law to solve for the unknown volume <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212137\/0a4b20dd3f9257612d0c459edf0075c2.png\" alt=\"(V_2)\" width=\"30\" height=\"18\" \/> . STP is 273 K and 1 atm. The temperatures have been converted to Kelvin.<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-ex2\"><em> Step 2: Solve <\/em><\/p>\n<p id=\"x-ck12-MDBkZWE2Yzk1YTE0NzIyMmExMGE0NmZmYzU1OWI1YjI.-kjc\">First, rearrange the equation algebraically to solve for <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> .<\/p>\n<p id=\"x-ck12-9jl\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212152\/d2ebee1bf37edce72fd61d9dc57d6902.png\" alt=\"V_2=frac{P_1 times V_1 times T_2}{P_2 times T_1}\" width=\"143\" height=\"40\" \/><\/p>\n<p id=\"x-ck12-OTNiYzM1Nzk5YWMwYTdmNDZjM2RiOTlmODhhOGVkYTM.-zpr\">Now substitute the known quantities into the equation and solve.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwNzg1Mjk4OA..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212152\/f6b309b8131990189c8e66d2820fb70e.png\" alt=\"V_2=frac{0.833 text{ atm} times 2.00 text{ L} times 273 text{ K}}{1.00 text{ atm} times 308 text{ K}}=1.48 text{ L}\" width=\"336\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-fsp\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-M2FlNGFlZjk0OTg4NjIwZDczYjg0NzA3MzVjMjI4ZmU.-g7k\">Both the increase in pressure and the decrease in temperature cause the volume of the gas sample to decrease.\u00a0 Since both changes are relatively small, the volume does not decrease dramatically.<\/p>\n<p id=\"x-ck12-OWVlY2JmZGU4MTcwYmExOTY1NDIwMDEwYzAzZGUxZmE.-rkt\">It may seem challenging to remember all the different gas laws introduced so far. Fortunately, Boyle\u2019s, Charles\u2019s, and Gay-Lussac\u2019s laws can all be easily derived from the combined gas law. For example, consider a situation where a change occurs in the volume and pressure of a gas while the temperature is being held constant. In that case, it can be said that\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212153\/ad0be0d3264b5c6c9cabf30715cc5e0b.png\" alt=\"T_1 = T_2\" width=\"58\" height=\"16\" \/> . Look at the combined gas law and cancel the\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/a7284843d100e640c0e00ae83ab66a90.png\" alt=\"T\" width=\"13\" height=\"12\" \/> variable out from both sides of the equation. What is left over is Boyle\u2019s law:<\/p>\n<p id=\"x-ck12-ZDkwZDdmM2E4ODAyMDBmZTRiOWFiMDUxOGY4MzAwYzg.-wto\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212154\/9db530c3c955711edeb1d9ebd41a60ee.png\" alt=\"P_1 times V_1 = P_2 times V_2\" width=\"140\" height=\"16\" \/> . Likewise, if the pressure is constant, then\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212154\/4d1c4ecefed32beff2939628cb3a3ab9.png\" alt=\"P_1 = P_2\" width=\"61\" height=\"16\" \/> and canceling\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/dc2a557e4775924079e5de1061889778.png\" alt=\"P\" width=\"14\" height=\"12\" \/> out of the equation leaves Charles\u2019s law. If the volume is constant, then\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwNzg1Mjk4OQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212154\/de70d3338a1bb694e07d1b31ad411ad4.png\" alt=\"V_1 = V_2\" width=\"58\" height=\"16\" \/> and canceling\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/adb347871c296be723ff3085c0023a28.png\" alt=\"V\" width=\"14\" height=\"12\" \/> out of the equation leaves Gay-Lussac\u2019s law.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-YTVhNmMyZDEyZDA3ZGJlYWEzMWQwYzEwZmZlNmMzNzI.-imk\">\n<li>The combined gas law shows the relationships among temperature, volume, and pressure.<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212155\/6568f29084a4b11bb32a0ff5d5e40b8f.png\" alt=\"frac{P_1V_1}{T_1} = frac{P_2V_2}{T_2}\" width=\"85\" height=\"25\" \/><\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-ZGIyN2QxMzI1ODI1MWRlNWVlMjE5YTlhNmM5NDJlN2M.-fqy\">Work on the problems at the link below:<\/p>\n<p id=\"x-ck12-YzNhMDI4NGExMjlmY2QzODE1ZGJiMzA4OTEwNDQ5MWE.-o8v\"><a href=\"http:\/\/misterguch.brinkster.net\/WKS001_007_146637.pdf\"> http:\/\/misterguch.brinkster.net\/WKS001_007_146637.pdf <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-b7i\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-NDZmM2ZiYmMzYzljYjVlMGY0OWUxMzBhYzZiNmE4OGY.-dgp\">\n<li>What is the only thing held constant in a combined gas law problem?<\/li>\n<li>If you want to solve for the volume of a gas\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212137\/0a4b20dd3f9257612d0c459edf0075c2.png\" alt=\"(V_2)\" width=\"30\" height=\"18\" \/> and\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/b95fd9f12605d93cbb24fd5fc1d83fbc.png\" alt=\"P_1\" width=\"18\" height=\"16\" \/> is greater than <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/f49ec791ef1ecb63d1c14840f9c2f983.png\" alt=\"P_2\" width=\"18\" height=\"15\" \/> , would you expect\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> to be larger or smaller than <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/29c883aa1a216bc9157974e94c993965.png\" alt=\"V_1\" width=\"16\" height=\"16\" \/> ?<\/li>\n<li>What would be the equation for finding\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/f49ec791ef1ecb63d1c14840f9c2f983.png\" alt=\"P_2\" width=\"18\" height=\"15\" \/> given all the other parameters?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-ZTU5Y2NmZmU0MTZkODQ5Y2M1ZDQ3OGYwNzA2Y2IzMTk.-vsu\">\n<li><strong> combined gas law: <\/strong> Expresses the relationship between the pressure, volume, and absolute temperature of a fixed amount of gas.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-QXZvZ2Fkcm8ncyBMYXc.\">Avogadro&#8217;s Law<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-ZWQ0Y2E1ZjgwYzFmZDQ1MzM4YzdmMmU1MzhlYTBlYTc.-pde\">\n<li>State Avogadro\u2019s Law.<\/li>\n<li>Use this law to perform calculations involving quantities of gases.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-MGM2Zjg3ZGYyNjVmZTk4Yjc4NDBjZTAxMzNiZmFjMWE.-du0\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212155\/20140811155525460045.jpeg\" alt=\"Avogadro's Law tells us that putting gas into a tire increases its pressure\" width=\"200\" \/><\/span><\/p>\n<p id=\"x-ck12-ZjFjODM1NmVlOGVkY2M3MGUwOWRiZTYyZTMwZTE0Zjg.-p6v\"><strong> How much air do you put into a tire? <\/strong><\/p>\n<p id=\"x-ck12-NTZhOWYyNDA0YzNiYzBhMGEyZTI0ZjQwNTdhNWM5MDE.-5qy\">A flat tire is not very useful. It does not cushion the rim of the wheel and creates a very uncomfortable ride. When air is added to the tire, the pressure increases as more molecules of gas are forced into the rigid tire. How much air should be put into a tire depends on the pressure rating for that tire. Too little pressure and the tire will not hold its shape. Too much pressure and the tire could burst.<\/p>\n<h3>Avogadro\u2019s Law<\/h3>\n<p id=\"x-ck12-YTlmZDk4ZTcxNjViYzY2OGExYjkxYWNhZTQxMjA1Nzk.-mjv\">You have learned about Avogadro\u2019s hypothesis: equal volumes of any gas at the same temperature and pressure contain the same number of molecules. It follows that the volume of a gas is directly proportional to the number of moles of gas present in the sample. <strong> Avogadro\u2019s law <\/strong> states that the volume of a gas is directly proportional to the number of moles of gas when the temperature and pressure are held constant. The mathematical expression of Avogadro\u2019s law is<\/p>\n<p id=\"x-ck12-YmU1ZDVkMzc1NDJkNzVmOTNhODcwOTQ0NTlmNzY2Nzg.-ea0\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwODI3MzEzNw..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212156\/4557cd2f339a4ea7c168510243bd9c8a.png\" alt=\"V=k times n quad text{and} quad frac{V_1}{n_1}=frac{V_2}{n_2}\" width=\"211\" height=\"40\" \/><\/p>\n<p id=\"x-ck12-NzVhMWE4YmNlYmRhZDIzNDM0NTI3OWRlYzE3YWU5Njk.-loh\">where\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211002\/c73c4e5aa4eb7c39977bbf98a8cd212a.png\" alt=\"n\" width=\"11\" height=\"8\" \/> is the number of moles of gas and\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwODI3MzEzOA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212122\/d301c6452e642017007fb7cd755ce08b.png\" alt=\"k\" width=\"9\" height=\"13\" \/> is a constant. Avogadro\u2019s law is in evidence whenever you blow up a balloon. The volume of the balloon increases as you add moles of gas to the balloon by blowing it up.<\/p>\n<p id=\"x-ck12-OTBmODhmYWJjNDJmYTEwYmVhNmRmZTM2ZmQyYmQzZTQ.-x6g\">If the container holding the gas is rigid rather than flexible, pressure can be substituted for volume in Avogadro\u2019s law. Adding gas to a rigid container makes the pressure increase.<\/p>\n<h4>Sample Problem: Avogadro\u2019s Law<\/h4>\n<p id=\"x-ck12-YTI1YjgyMmIxMTkyZDFlNWQwYWFmZjdmY2ViZWYzOTA.-2ab\">A balloon has been filled to a volume of 1.90 L with 0.0920 mol of helium gas. If 0.0210 mol of additional helium is added to the balloon while the temperature and pressure are held constant, what is the new volume of the balloon?<\/p>\n<p id=\"x-ck12-OGU2YWEyYzY0NzAwMDUxZjI4NjFjY2E4MjYyNmNhN2I.-ydl\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-iyb\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-dey\">\n<li><em> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212157\/b18c8945f2039f0506b418c4c5664e8d.png\" alt=\"V_1=1.90 text{ L}\" width=\"91\" height=\"17\" \/><\/em><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212157\/b9491744c2ff6024c2a1498e31574164.png\" alt=\"n_1=0.0920 text{ mol}\" width=\"127\" height=\"17\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212158\/f3d75b46fb576905ae99c18e03cde536.png\" alt=\"n_2=0.0920+0.0210=0.1130 text{ mol}\" width=\"272\" height=\"16\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-hot\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-1iu\">\n<li><em> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212152\/51e1944bfa6d5ec4a5ee7088baf64635.png\" alt=\"V_2=? text{ L}\" width=\"62\" height=\"16\" \/><\/em><\/li>\n<\/ul>\n<p id=\"x-ck12-ZTMyNDVhMzcwNTgwYzQ3ZjExY2IwNDhlNGY2OTQ2N2I.-jw1\">Note that the final number of moles has to be calculated by adding the original number of moles to the moles of added helium. Use Avogadro\u2019s law to solve for the final volume.<\/p>\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-fpw\"><em> Step 2: Solve. <\/em><\/p>\n<p id=\"x-ck12-Yzg1OGU4MDExNzQ5NGQ4OGU4NjY3ZWZlMTFmNDdhNzQ.-nxw\">First, rearrange the equation algebraically to solve for <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> .<\/p>\n<p id=\"x-ck12-8pd\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212158\/b1eab62cefeb445a02b827e2c1fcffee.png\" alt=\"V_2=frac{V_1 times n_2}{n_1}\" width=\"101\" height=\"40\" \/><\/p>\n<p id=\"x-ck12-OTNiYzM1Nzk5YWMwYTdmNDZjM2RiOTlmODhhOGVkYTM.-ghy\">Now substitute the known quantities into the equation and solve.<\/p>\n<p id=\"x-ck12-mkt\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212159\/ec3e3cd202cf649f2a46e280d3d59cca.png\" alt=\"V_2=frac{1.90 text{ L} times 0.1130 text{ mol}}{0.0920 text{ mol}}=2.33 text{ L}\" width=\"274\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-wu4\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-NDAxYzUyZmMxYmFlYWQyMTI3MGVmYTc1MWI5N2JlYWQ.-2jc\">Since a relatively small amount of additional helium was added to the balloon, its volume increases slightly.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-ODJkMGZkZWU1YWM0ZjVmNGNiYzFkMTlkNmYyZTA1MDc.-msg\">\n<li>Calculations are shown for relationships between volume and number of moles of a gas.<\/li>\n<\/ul>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-idx\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-NWE1NzA1NzUzYmNiNjE0N2UwNWU5NjgxZmZhMjg4MDY.-daw\">\n<li>What is held constant in the Avogadro\u2019s Law relationship?<\/li>\n<li>What happens if you add gas to a rigid container?<\/li>\n<li>Why does a balloon expand when you add air to it?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-MDhhMmZiMjM4OTI1ZDM2N2I4ZjBhNTI4YjVkMTY1OTU.-tiw\">\n<li><strong> Avogadro\u2019s law: <\/strong> The volume of a gas is directly proportional to the number of moles of gas when the temperature and pressure are held constant.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-SWRlYWwgR2FzIExhdw..\">Ideal Gas Law<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-MGVlOTkwZTY4MjU5NzExYTBmZjQxYmVjYTZiNjNiYjQ.-8xy\">\n<li>Derive the ideal gas law from the combined gas law and Avogadro\u2019s law.<\/li>\n<li>Calculate the value of the ideal gas constant.<\/li>\n<li>Use the ideal gas law to calculate parameters for ideal gases.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-OTRiNjdiZWVlZjZjMGZhOTljMDczNjViNjUyOTRjMWY.-3v1\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212200\/20140811155525625008.jpeg\" alt=\"The ideal gas law can be used to determine the amount of gas present in a system\" width=\"100\" \/><\/span><\/p>\n<p id=\"x-ck12-MmIyZDljNjI4YmRlZWQwYTZhZjc1YTMzYjBlZmQ4YzA.-yxd\"><strong> What chemical reactions require ammonia? <\/strong><\/p>\n<p id=\"x-ck12-MmI1M2Q5YjRlYzE4MGY4MjJiYTZlYWE3OGQ5M2YwZmU.-wvt\">There are a number of chemical reactions that require ammonia. In order to carry out the reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system.<\/p>\n<h3>Ideal Gas Law<\/h3>\n<p id=\"x-ck12-ZTg3NzkxMzViZWJhNDIxNDhiZTcyNjFiMDJjMDdiMTA.-ed4\">The combined gas law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro\u2019s law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these together leaves us with the following equation:<\/p>\n<p id=\"x-ck12-wr3\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwOTEyODc4OQ..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212203\/b18bf265809dd0c3b00092aeae02977d.png\" alt=\"frac{P_1 times V_1}{T_1 times n_1}=frac{P_2 times V_2}{T_2 times n_2}\" width=\"145\" height=\"40\" \/><\/p>\n<p id=\"x-ck12-ZDlmYjBhZTQ3MmRiOWRiYmNjZDFjMTAxYmUwNjBmNzk.-0fr\">As with the other gas laws, we can also say that\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwOTEyODc5MA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212203\/426d96bd9cee0ae6f889fd1cee17c81a.png\" alt=\"frac{left(P times V right)}{left(T times n right)}\" width=\"41\" height=\"29\" \/> is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal.<\/p>\n<p id=\"x-ck12-NTcyNzU3YTgwN2UxMmQ1NjllNzcyMzZkMTQ1OGEyNmU.-5wf\">The <strong> ideal gas law <\/strong> is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> for the constant, the equation becomes:<\/p>\n<p id=\"x-ck12-7qo\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/1a80d20b13965449f86ff0bd84c737d8.png\" alt=\"frac{P times V}{T times n}=R\" width=\"91\" height=\"37\" \/><\/p>\n<p id=\"x-ck12-MGIwODRkZDIxMzlmZWMyYTM1MmI0YzBiYTRhNzQ5MmY.-ht0\">The ideal gas law is conventionally rearranged to look this way, with the multiplication signs omitted:<\/p>\n<p id=\"x-ck12-13m\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/11cf69a276a7ad759359c56f0512e141.png\" alt=\"PV=nRT\" width=\"90\" height=\"12\" \/><\/p>\n<p id=\"x-ck12-NTZmMmI4YzU0M2RjNmY0YzE4MDFiOGExMTlmYmEyNjE.-z5o\">The variable\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> in the equation is called the <strong> ideal gas constant <\/strong> .<\/p>\n<h4>Evaluating the Ideal Gas Constant<\/h4>\n<p id=\"x-ck12-NDU1YzQ4NjBlZjA3ZDE4MzI0ZWU3ODgyNTUwNjc1NDY.-fwa\">The value of <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> , the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. However, pressure is commonly measured in one of three units: kPa, atm, or mmHg. Therefore,\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> can have three different values.<\/p>\n<p id=\"x-ck12-ZTk0N2FlYWJkYmRmZTNmZjM4ZmE5NjgxOTExMTc3N2E.-p0x\">We will demonstrate how\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwOTEyODc5MQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> is calculated when the pressure is measured in kPa. Recall that the volume of 1.00 mol of any gas at STP is measured to be 22.414 L. We can substitute 101.325 kPa for pressure, 22.414 L for volume, and 273.15 K for temperature into the ideal gas equation and solve for <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> .<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212205\/c0037c63a8c77edb80a7778550aee404.png\" alt=\"R=frac{PV}{nT}=frac{101.325 text{kPa} times 22.414 text{ L}}{1.000 text{mol} times 273.15 text{ K}}=8.314 text{kPa} cdot text{L\/K} cdot text{mol}\" width=\"472\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-YmUyYzJhZWZkODU3MTA4NGUxOTcwNGVlNzYyMzI0ODI.-tgm\">This is the value of\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> that is to be used in the ideal gas equation when the pressure is given in kPa. The <strong> Table <\/strong> below shows a summary of this and the other possible values of <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwOTEyODc5Mg..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> . It is important to choose the correct value of\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> to use for a given problem.<\/p>\n<table id=\"x-ck12-ZmVlMThiZTFmNTVkMjg4ZGYzY2UxNGE5NTQyYjllNzU.-br6\" class=\"x-ck12-nofloat\">\n<caption>Values of the Ideal Gas Constant<\/caption>\n<tbody>\n<tr>\n<td><strong> Unit of <\/strong> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/dc2a557e4775924079e5de1061889778.png\" alt=\"P\" width=\"14\" height=\"12\" \/><strong><br \/>\n<\/strong><\/td>\n<td><strong> Unit of <\/strong> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/adb347871c296be723ff3085c0023a28.png\" alt=\"V\" width=\"14\" height=\"12\" \/><\/td>\n<td><strong> Unit of <\/strong> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211002\/c73c4e5aa4eb7c39977bbf98a8cd212a.png\" alt=\"n\" width=\"11\" height=\"8\" \/><strong><br \/>\n<\/strong><\/td>\n<td><strong> Unit of <\/strong> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/a7284843d100e640c0e00ae83ab66a90.png\" alt=\"T\" width=\"13\" height=\"12\" \/><\/td>\n<td><strong> Value and unit of <\/strong> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/><\/td>\n<\/tr>\n<tr>\n<td>kPa<\/td>\n<td>L<\/td>\n<td>mol<\/td>\n<td>K<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212207\/d9e6f5f6e76170eb98b5b1f0eaed65ee.png\" alt=\"8.314 text{J\/K} cdot text{mol}\" width=\"121\" height=\"18\" \/><\/td>\n<\/tr>\n<tr>\n<td>atm<\/td>\n<td>L<\/td>\n<td>mol<\/td>\n<td>K<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212208\/5f341d284dec12741f5809ebf5477684.png\" alt=\"0.08206 text{L} cdot text{atm\/K} cdot text{mol}\" width=\"185\" height=\"18\" \/><\/td>\n<\/tr>\n<tr>\n<td>mmHg<\/td>\n<td>L<\/td>\n<td>mol<\/td>\n<td>K<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212208\/65451fb51aff398ea23a71bd4256f85d.png\" alt=\"62.36 text{ L} cdot text{mmHg\/K} cdot text{mol}\" width=\"188\" height=\"18\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"x-ck12-Mjg1N2M4YzMwZTNiYzhjNjI3Yjg0MWZkODFlZGM1Njg.-xv1\">Notice that the unit for\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> when the pressure is in kPa has been changed to J\/K\u00a0\u2022\u00a0mol. A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule (J).<\/p>\n<h4>Sample Problem: Ideal Gas Law<\/h4>\n<p id=\"x-ck12-MTQyMTY0NDY2ZmRkMjFiMDY2OTNhYWY1YjEzYmMxODQ.-tlt\">What volume is occupied by 3.760 g of oxygen gas at a pressure of 88.4 kPa and a temperature of 19\u00b0C? Assume the oxygen is ideal.<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-nct\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-fhn\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-cy0\">\n<li><em> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212209\/55b6d82cc42bc408a485e0f528f06d6f.png\" alt=\"P = 88.4 text{kPa}\" width=\"106\" height=\"14\" \/><\/em><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212210\/05e91eb9d11d1c8050952111f84da1c7.png\" alt=\"T = 19^circ text{C} = 292 text{ K}\" width=\"146\" height=\"14\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212210\/443c32746f0a9dbd4aac11fda6a45aa2.png\" alt=\"text{mass} O_2 = 3.760 text{ g}\" width=\"145\" height=\"16\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212211\/5c54eeabf3c70df08cb455f8a8907ef7.png\" alt=\"O_2 = 32.00 text{ g\/mol}\" width=\"139\" height=\"18\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212211\/4707dcbdc7b4220c7b11dc57c8d97f93.png\" alt=\"R = 8.314 text{ J\/K} cdot text{mol}\" width=\"159\" height=\"18\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-bkm\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-ysq\">\n<li><em> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212212\/0657dfe2c2c1e6586aa37f082316afef.png\" alt=\"V = ? text{ L}\" width=\"59\" height=\"13\" \/><\/em><\/li>\n<\/ul>\n<p id=\"x-ck12-MGE4NmVlNDJlMGFkMTQ4Y2I4NjRkMTEzY2VmY2JlZGI.-iae\">In order to use the ideal gas law, the number of moles of O <sub> 2 <\/sub> \u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwOTEyODc5Mw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211023\/4168b7a35bc24c982496aef5b556a0a5.png\" alt=\"(n)\" width=\"23\" height=\"18\" \/> must be found from the given mass and the molar mass. Then, use\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212212\/b5a071ec470437e7416a7e999a24b3fc.png\" alt=\"PV = nRT\" width=\"90\" height=\"12\" \/> to solve for the volume of oxygen.<\/p>\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-afs\"><em> Step 2: Solve <\/em> .<\/p>\n<p id=\"x-ck12-sdp\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212213\/a7682991a7f0112a32d04072b051fdaf.png\" alt=\"3.760 text{ g} times frac{1 text{mol} O_2}{32.00 text{ g} O_2}=0.1175 text{mol} O_2\" width=\"299\" height=\"41\" \/><\/p>\n<p id=\"x-ck12-MDZhOTZjMmFlOTE5MGFjMTE0Y2UxYjQ3OTYyN2RkZmY.-pqw\">Rearrange the ideal gas law and solve for <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/adb347871c296be723ff3085c0023a28.png\" alt=\"V\" width=\"14\" height=\"12\" \/> .<\/p>\n<p id=\"x-ck12-wh2\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212214\/a9fa443dfe97998bf81bc7d37b0788f8.png\" alt=\"V=frac{nRT}{P}=frac{0.1175 text{mol} times 8.314 text{ J\/K} cdot text{mol} times 292 text{ K}}{88.4 text{kPa}}=3.23 text{ L } O_2\" width=\"503\" height=\"40\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-vm9\"><em> Step 3: Think about your result <\/em><\/p>\n<p id=\"x-ck12-YWNmNTQwMjBmOGQxNjJhN2NmYjQxZmJkOTk3ZTNlNzg.-bie\">The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume (22.4 L\/mol) since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/a7284843d100e640c0e00ae83ab66a90.png\" alt=\"T\" width=\"13\" height=\"12\" \/> and <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYwOTEyODc5NA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/dc2a557e4775924079e5de1061889778.png\" alt=\"P\" width=\"14\" height=\"12\" \/> . Since a joule (J)\u00a0=\u00a0kPa\u00a0\u2022\u00a0L, the units cancel correctly, leaving a volume in liters.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-ZTk5OTRiY2YyZmNiYTAzODY5MzhjNjI1NThlOWNmZDc.-h8x\">\n<li>The ideal gas constant is calculated.<\/li>\n<li>An example of calculations using the ideal gas law is shown.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-ZGIyN2QxMzI1ODI1MWRlNWVlMjE5YTlhNmM5NDJlN2M.-blq\">Work on the problems at the link below:<\/p>\n<p id=\"x-ck12-NmY2MDNmYTFhOGM4N2Y4MjJiNmU5MjczMTFhYzU0NDk.-qpg\"><a href=\"http:\/\/chemsite.lsrhs.net\/gasses\/handouts\/Ideal_Problems.pdf\"> http:\/\/chemsite.lsrhs.net\/gasses\/handouts\/Ideal_Problems.pdf <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-dmh\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-MDBkYTM5MGMzOWJmOGVmNGZjNGQxNWRkOTA5M2JhMzQ.-654\">\n<li>Which value of\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> will you use if the pressure is given in atm?<\/li>\n<li>You are doing a calculation where the pressure is given in mm Hg. You select 8.314\u00a0J\/K\u00a0\u2022\u00a0mol as your value for <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> . Will you get a correct answer?<\/li>\n<li>How would you check that you have chosen the correct value of\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> for your problem?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-ZTYwZTU0MTExMTZhOGUzNDQ0NzkwNmJlOGRkMTE5N2M.-hkb\">\n<li><strong> ideal gas constant: <\/strong> The variable\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212215\/e096e3ca730e7ebae2c2660b5801dea8.png\" alt=\"R\" width=\"14\" height=\"12\" \/> in the ideal gas law equation.<\/li>\n<li><strong> ideal gas law: <\/strong> A single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-Q2FsY3VsYXRpbmcgdGhlIE1vbGFyIE1hc3Mgb2YgYSBHYXM.\">Calculating the Molar Mass of a Gas<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-Y2E4MjBhYmZjMTllOGEwNTIwNWQ5Nzg2MDFkNDEyMTk.-zy4\">\n<li>Calculate the molar mass of a gas.<\/li>\n<li>Calculate the density of a gas.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-YWExZWY3ZGZiYjViZGY3YWEwOWUwYWYxMmUxYzliOGM.-9uq\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212215\/20140811155525773654.jpeg\" alt=\"Gases with low molar masses such as hydrogen and helium can be used to float balloons\" width=\"200\" \/><\/span><\/p>\n<p id=\"x-ck12-ODZiYWMwYjZjOWViMDUzN2QyMWVmYWViZjk2NjA4MTU.-gio\"><strong> What makes it float? <\/strong><\/p>\n<p id=\"x-ck12-ZTQwNTI2NzQ2YjQ2MmYzYmZkMWVkODg1ZmVkYTJlODM.-s8i\">Helium has long been used in balloons and blimps. Since it is much less dense than air, it will float above the ground. We can buy small balloons filled with helium at stores, but large ones (such as the balloon seen above) are much more expensive and take up a lot more helium.<\/p>\n<h3>Calculating Molar Mass and Density of a Gas<\/h3>\n<p id=\"x-ck12-M2ZjZTlhNGQzNzU5YmIyMmU3ZmU0YmExMDQ3YzYyODg.-yuh\">A chemical reaction, which produces a gas, is performed. The produced gas is then collected and its mass and volume are determined. The molar mass of the unknown gas can be found using the ideal gas law, provided the temperature and pressure of the gas are also known.<\/p>\n<h4>Sample Problem: Molar Mass and the Ideal Gas Law<\/h4>\n<p id=\"x-ck12-MjkyOTkzODk5MGJjMjg5ZmEwZjkwMTY0YjRmMDUyZTQ.-sed\">A certain reaction occurs, producing an oxide of nitrogen as a gas. The gas has a mass of 1.211 g and occupies a volume of 677 mL. The temperature in the laboratory is 23\u00b0C and the air pressure is 0.987 atm. Calculate the molar mass of the gas and deduce its formula. Assume the gas is ideal.<\/p>\n<p id=\"x-ck12-OGU2YWEyYzY0NzAwMDUxZjI4NjFjY2E4MjYyNmNhN2I.-vxd\"><em> Step 1: List the known quantities and plan the problem <\/em> .<\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-em0\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-u3m\">\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212217\/15ee7ca6617ec89c8e9b6af4c30f8368.png\" alt=\"text{mass}=1.211text{ g}\" width=\"118\" height=\"15\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212218\/0938b476f561ff3e3eb8105649e1de84.png\" alt=\"V = 677 text{ ml} = 0.677 text{ L}\" width=\"173\" height=\"14\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212218\/3bab40e282f77f715cc5a084ba006150.png\" alt=\"T= 23^circ text{C}=296 text{ K}\" width=\"146\" height=\"14\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212219\/fcc8aab2b57db7f9d6d576ef7ff81f0f.png\" alt=\"P =0.987 text{atm}\" width=\"116\" height=\"14\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212219\/0338a5ae6c992b4ab8e04603d609fee2.png\" alt=\"R =0.08206 text{ L} cdot text{atm\/K} cdot text{mol}\" width=\"223\" height=\"18\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-rcu\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-3dm\">\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212220\/c528206210b32c4e36374e35fe2432f1.png\" alt=\"n= ? text{mol}\" width=\"73\" height=\"14\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212221\/25947cbe6aff86043fd47cc71379fe36.png\" alt=\"text{molar mass} = ? text{g\/mol}\" width=\"169\" height=\"18\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-OTc0Yzk5MTVlYmQ4YWViMjAxOWYzZWQwZmY2NTMxNzg.-mlj\">First the ideal gas law will be used to solve for the moles of unknown gas <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYxMDQ5NjY2OA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211023\/4168b7a35bc24c982496aef5b556a0a5.png\" alt=\"(n)\" width=\"23\" height=\"18\" \/> . Then the mass of the gas divided by the moles will give the molar mass.<\/p>\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-njd\"><em> Step 2: Solve <\/em> .<\/p>\n<p id=\"x-ck12-3wi\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212221\/c03079f46a2fff832953b85b039e9ee8.png\" alt=\"n=frac{PV}{RT}=frac{0.987 text{atm} times 0.677 text{ L}}{0.08206 text{ L} cdot text{atm\/K} cdot text{mol} times 296 text{ K}}=0.0275 text{mol}\" width=\"457\" height=\"42\" \/><\/p>\n<p id=\"x-ck12-NzdlN2U4YjljYjdkYjdjZThkMDgwN2VlY2NjOTBmMTE.-6om\">Now divide g by mol to get the molar mass.<\/p>\n<p id=\"x-ck12-gau\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212223\/6f092e645d73d411fbd30dffab7aa0ee.png\" alt=\"text{molar mass}=frac{1.211 text{ g}}{0.0275 text{mol}}=44.0 text{g\/mol}\" width=\"309\" height=\"38\" \/><\/p>\n<p id=\"x-ck12-NzlmMDhlMGI3NDhiODcyY2FlNTNjYzJmNTI1Zjc4Mjc.-7sg\">Since N has a molar mass of 14 g\/mol and O has a molar mass of 16 g\/mol, the formula N <sub> 2 <\/sub> O would produce the correct molar mass.<\/p>\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-jqn\"><em> Step 3: Think about your result <\/em><\/p>\n<p id=\"x-ck12-Yjg1MWQ1YzEwMWEyZDI5NTgxMjZkMTY4MjYzYzljMmI.-ghd\">The\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYxMDQ5NjY2OQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212204\/979563bd3dba702df97e7e33b0573496.png\" alt=\"R\" width=\"14\" height=\"12\" \/> value that corresponds to a pressure in atm was chosen for this problem. The calculated molar mass gives a reasonable formula for dinitrogen monoxide.<\/p>\n<h4>Calculating Density of a Gas<\/h4>\n<p id=\"x-ck12-NTJjODQ4YjIzOGRiYTdlMzZmODcxZTU0NTM4NjIyN2M.-xz1\">The ideal gas law can be used to find the density of a gas at conditions that are not standard. For example, we will determine the density of ammonia gas (NH <sub> 3 <\/sub> ) at 0.913 atm and 20\u00b0C, assuming the ammonia is ideal. First, the molar mass of ammonia is calculated to be 17.04 g\/mol. Next, assume exactly 1 mol of ammonia\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212224\/a357460ec50f481e8810da93911a02f7.png\" alt=\"(n = 1)\" width=\"56\" height=\"18\" \/> and calculate the volume that such an amount would occupy at the given temperature and pressure.<\/p>\n<p id=\"x-ck12-7v6\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212224\/bae49338d0e06b594a1d1fb981dcc822.png\" alt=\"V = frac{ nRT}{P}=frac{1.00 text{mol} times 0.08206 text{ L} cdot text{atm\/K}cdot text{mol} times 293 text{ K}}{0.913 text{atm}}=26.3 text{ L}\" width=\"522\" height=\"40\" \/><\/p>\n<p id=\"x-ck12-NDQ1OWIzM2EzYzgyMTc4MjRkMGRhN2M5NzhiNTIxZDA.-dtf\">Now the density can be calculated by dividing the mass of one mole of ammonia by the volume above.<\/p>\n<p id=\"x-ck12-mnz\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212226\/89d5d4933d53b9bc38998e4cccf6822f.png\" alt=\"text{density}=frac{17.04 text{ g}}{26.3 text{ L}}=0.647 text{g\/L}\" width=\"239\" height=\"38\" \/><\/p>\n<p id=\"x-ck12-MmVjYTcxZjNjYTU1NDc3ZTJjYjAwZDcwYTI0ZGQwOTc.-nlm\">As a point of comparison, this density is slightly less than the density of ammonia at STP, which is equal to <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYxMDQ5NjY3Ng..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212227\/8001f341f27809461141f496b63aa091.png\" alt=\"frac{(17.04 text{g\/mol})}{(22.4 text{L\/mol})} = 0.761 text{g\/L}\" width=\"181\" height=\"29\" \/> . It makes sense that the density should be lower compared to that at STP since both the increase in temperature (from 0\u00b0C to 20\u00b0C) and the decrease in pressure (from 1 atm to 0.913 atm) would cause the NH <sub> 3 <\/sub> molecules to spread out a bit further from one another.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-ZmFkMmQzNTEyNjIyZGM0NWE1OGQxMmJiOGEyZTUxZjQ.-ldq\">\n<li>Calculations of molar mass and density of an ideal gas are described.<\/li>\n<\/ul>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-ijl\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-MjhiZWM4OWM1MTIzZTFiNDgwMzBmNGM3ODU4ZGUwMGQ.-b6d\">\n<li>Why do you need the volume, temperature, and pressure of the gas to calculate molar mass?<\/li>\n<li>What assumption about the gas is made in all these calculations?<\/li>\n<li>Why do you need the mass of the gas to calculate the molar mass?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\"><\/div>\n<h1 id=\"x-ck12-R2FzIFN0b2ljaGlvbWV0cnk.\">Gas Stoichiometry<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-YjIwODVmZTdjMmMyZDBjNzMyNDdlZWQzNmU4ZjM3MzU.-t5r\">\n<li>Use the ideal gas law to calculate stoichiometry problems for gases.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-NzhiODljMzE5YzczYzUzMjdjNDU2NzMyNzkyNDFiMDU.-qrz\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212227\/20140811155525942661.jpeg\" alt=\"Knowing the amount of a gas needed is important for many applications\" width=\"400\" \/><\/span><\/p>\n<h4><span class=\"x-ck12-img-inline\"> How is fertilizer produced?<br \/>\n<\/span><\/h4>\n<p id=\"x-ck12-OWI3NjFjMTI3NmNhZWExMzYyMTc3OWNlN2MxYWE3YWE.-fdi\">The Haber cycle reaction of gaseous nitrogen and hydrogen to form ammonia is a critical step in the production of fertilizer from ammonia. It is important to have an excess of the starting materials so that a maximum yield of ammonia can be achieved. By knowing how much ammonia is needed for manufacture of a batch of fertilizer, the proper amounts of nitrogen and hydrogen gases can be incorporated into the process.<\/p>\n<h3>Gas Stoichiometry<\/h3>\n<p id=\"x-ck12-ZGNjYmJhM2NlMTZhOWYyYjc4NjAzNjgzMDczOGI3ODU.-c5l\">You have learned how to use molar volume to solve stoichiometry problems for chemical reactions involving one or more gases at STP. Now, we can use the ideal gas law to expand our treatment of chemical reactions to solve stoichiometry problems for reactions that occur at any temperature and pressure.<\/p>\n<h4>Sample Problem: Gas Stoichiometry and the Ideal Gas Law<\/h4>\n<p id=\"x-ck12-OWJiMTg0MGY5ZWQ3MWVkMGI4ZmY4Y2QxYTNmY2ExZGY.-b5o\">What volume of carbon dioxide is produced by the combustion of 25.21 g of ethanol (C <sub> 2 <\/sub> H <sub> 5 <\/sub> OH)\u00a0at 54\u00b0C and 728 mmHg? Assume the gas is ideal.<\/p>\n<p id=\"x-ck12-ODg1ODBlZjhiMTIwZjJjM2Q5NmJmODc4MDNlZDRhYTA.-ls2\">Before using the ideal gas law, it is necessary to write and balance the chemical equation. Recall that most combustion reactions, the given substance reacts with O <sub> 2 <\/sub> to form CO <sub> 2 <\/sub> and H <sub> 2 <\/sub> O. Here is the balanced equation for the combustion of ethanol.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212229\/da8532c253d207d89961bb44eda89513.png\" alt=\"text{C}_2text{H}_5text{OH}(l)+3text{O}_2(g) rightarrow 2text{CO}_2(g)+3text{H}_2text{O}(l)\" width=\"342\" height=\"18\" \/><\/p>\n<p id=\"x-ck12-ZWRlZGNhY2I2ZjdkODlhOWVlMDkxNTRmNzlmMjUxZDI.-uvx\"><em> Step 1: List the known quantities and solve the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-sbi\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-hup\">\n<li><span class=\"x-ck12-underline\"> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212230\/8895b9609b51970be97e9f8aedb863dc.png\" alt=\"text{mass} text{C}_2text{H}_5text{OH}=25.21 text{ g}\" width=\"192\" height=\"16\" \/><\/span><\/li>\n<li><span class=\"x-ck12-underline\"> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212231\/8c8f86b472bd9d1c10b3d7e3bda54cee.png\" alt=\"text{molar mass} text{C}_2text{H}_5text{OH}=46.08 text{ g\/mol}\" width=\"281\" height=\"18\" \/><\/span><\/li>\n<li><span class=\"x-ck12-underline\"> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212232\/bdd8d0aab73462de83ee60a33c299f81.png\" alt=\"P=728 text{ mmHg}\" width=\"123\" height=\"16\" \/><\/span><\/li>\n<li><span class=\"x-ck12-underline\"> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212233\/816a80e184ef6694c5135dea0d517288.png\" alt=\"T=54^circ text{C}=327 text{ K}\" width=\"146\" height=\"14\" \/><\/span><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-bx5\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-nts\">\n<li><em> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212233\/a8c85ab5d338790007e8d3caf51b2348.png\" alt=\"text{Volume} text{CO}_2=? text{ L}\" width=\"143\" height=\"16\" \/><\/em><\/li>\n<\/ul>\n<p id=\"x-ck12-MDRjYzYyOTJlNjA3MThjOTllNjk0ZjczMjNkMDEzNzc.-6mw\">The number of moles of carbon dioxide gas is first calculated by stoichiometry. Then the ideal gas law is used to calculate the volume of CO <sub> 2 <\/sub> produced.<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-d7z\"><em> Step 2: Solve. <\/em><\/p>\n<p id=\"x-ck12-i9w\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212234\/98be15fbc832bb992b6eaa3261013639.png\" alt=\"25.21 text{ g } text{C}_2text{H}_5text{OH} times frac{1 text{ mol } text{C}_2text{H}_5text{OH}}{46.08 text{ g } text{C}_2text{H}_5text{OH}} times frac{2 text{ mol } text{CO}_2}{1 text{ mol } text{C}_2text{H}_5text{OH}}=1.094 text{ mol } text{C}_2text{H}_5text{OH}\" width=\"603\" height=\"41\" \/><\/p>\n<p id=\"x-ck12-M2ZkOTE2ZDVhM2Y1OTZmYmRjNTQwNGY0Yzc0ODA1ZTg.-ase\">The moles of ethanol\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211023\/4168b7a35bc24c982496aef5b556a0a5.png\" alt=\"(n)\" width=\"23\" height=\"18\" \/> is now substituted into\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212235\/fb2df3ecf87eaa43f2bdf8fd25a59ab0.png\" alt=\"PV=nRT\" width=\"90\" height=\"12\" \/> to solve for the volume.<\/p>\n<p id=\"x-ck12-6lx\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212236\/f803c72bae9ad2aa8a0b1cb827650230.png\" alt=\"V=frac{nRT}{P}=frac{1.094 text{ mol} times 62.36 text{ L} cdot text{mmHg\/K} cdot text{mol} times 327 text{ K}}{728 text{ mmHg}}=30.6 text{ L}\" width=\"535\" height=\"42\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-g2z\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-ZWM3ZjEzYjUxM2ZkMDVlYzg3OTZjMzM1M2IwYWY1ZWQ.-dgx\">The mass of ethanol is slightly more than one half mole, meaning that the mole ratio results in slightly more than one mole of carbon dioxide being produced. Because of the elevated temperature and reduced pressure compared to STP, the resulting volume is larger than 22.4 L.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-Mjg4NjJiNDVjZGFhYzQxNDc4ZjI3MzhkOTdkODA5ZTA.-ixb\">\n<li>The ideal gas law is used to calculate stoichiometry problems for gases.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-MGIxMTE2NjA5NmRjNjU4NzY3YjYxMDkzY2YxYTFjOTA.-iv8\">Solve the problems on the worksheet at this site:<\/p>\n<p id=\"x-ck12-ODc1ZjJlNmYwMDhiNTA5YzRhZDNiOWNiMDEzZTYzMzA.-thm\"><a href=\"http:\/\/misterguch.brinkster.net\/PRA036.pdf\"> http:\/\/misterguch.brinkster.net\/PRA036.pdf <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-ns9\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-NDgwOGNmYWFiN2ZkYjYxZWQ4ODA4OWM0NDYxYzMyYmU.-ubv\">\n<li>Do we need gas conditions to be at STP to calculate stoichiometry problems?<\/li>\n<li>Why do we want to determine the stoichiometry of these reactions?<\/li>\n<li>What assumption are we making about the gases involved?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\"><\/div>\n<h1 id=\"x-ck12-UmVhbCBhbmQgSWRlYWwgR2FzZXM.\">Real and Ideal Gases<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-NTE4NzU5NjA1ZDQzYjBkOGViMTRiNGRjODg5ODQwNmU.-jyp\">\n<li>Define a real gas.<\/li>\n<li>Describe differences between real gases and ideal gases.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-NDk0NjE3M2VlOGFlMDg1ZDgyN2E5YjE1YzIyNGYzODY.-q1i\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212237\/20140811155526027455.png\" alt=\"Intermolecular forces can cause deviations from ideality in gases\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-NjRjZGRhODE5M2NiY2FhNzE3ODY1OGYzM2JmZjM4ODk.-vtw\"><strong> Location, Location, Location <\/strong><\/p>\n<p id=\"x-ck12-NDRhOTA5YjRkYzJjMWE2ZGQ3ZDJjMDBkYmM3OGE1NTc.-1av\">The behavior of a molecule depends a lot on its structure. We can have two compounds with the same number of atoms and yet they act very differently. Ethanol (C <sub> 2 <\/sub> H <sub> 5 <\/sub> OH) is a clear liquid that has a boiling point of about 79\u00b0C. Dimethylether (CH <sub> 3 <\/sub> OCH <sub> 3 <\/sub> ) has the same number of carbons, hydrogens, and oxygens, but boils at a much lower temperature (-25\u00b0C). The difference lies in the amount of intermolecular interaction (strong H-bonds for ethanol, weak van der Waals force for the ether).<\/p>\n<h3>Real and Ideal Gases<\/h3>\n<p id=\"x-ck12-ZDkyZDcwOTI3NmU1NTg1ZGFiNjA0MmY2NjFiMWQwMGI.-upf\">An ideal gas is one that follows the gas laws at all conditions of temperature and pressure. To do so, the gas would need to completely abide by the kinetic-molecular theory. The gas particles would need to occupy zero volume and they would need to exhibit no attractive forces what so ever toward each other. Since neither of those conditions can be true, there is no such thing as an ideal gas. A <strong> real gas <\/strong> is a gas that does not behave according to the assumptions of the kinetic-molecular theory. Fortunately, at the conditions of temperature and pressure that are normally encountered in a laboratory, real gases tend to behave very much like ideal gases.<\/p>\n<p id=\"x-ck12-NDZmN2QyOTExNWJkNDcwZDdmOWE0MDYzOTQyNzA1YWI.-e8n\">Under what conditions then, do gases behave least ideally? When a gas is put under high pressure, its molecules are forced closer together as the empty space between the particles is diminished. A decrease in the empty space means that the assumption that the volume of the particles themselves is negligible is less valid. When a gas is cooled, the decrease in kinetic energy of the particles causes them to slow down. If the particles are moving at slower speeds, the attractive forces between them are more prominent. Another way to view it is that continued cooling the gas will eventually turn it into a liquid and a liquid is certainly not an ideal gas anymore (see liquid nitrogen in the <strong> Figure <\/strong> below ). In summary, a real gas deviates most from an ideal gas at low temperatures and high pressures. Gases are most ideal at high temperature and low pressure.<\/p>\n<div id=\"x-ck12-NTAyNjRkMDljODM3NWJiZDc3NmM2ZmJhNzgyYWVmZGU.-2cf\" class=\"x-ck12-img-thumbnail x-ck12-nofloat\">\n<p id=\"x-ck12-tyd\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2Mzc1NTg3Ni0zNy01Ny0z\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212238\/20140811155526151540.jpeg\" alt=\"Liquid nitrogen is at such low temperatures that it is no longer a gas\" longdesc=\"Nitrogen%20gas%20that%20has%20been%20cooled%20to%2077%20K%20has%20turned%20to%20a%20liquid%20and%20must%20be%20stored%20in%20a%20vacuum%20insulated%20container%20to%20prevent%20it%20from%20rapidly%20vaporizing.\" \/><\/p>\n<p><strong> Figure 14.11 <\/strong><\/p>\n<p id=\"x-ck12-OTk5ZTM3MWRiNDhjM2U5M2Y5MTdjNjgzNGU4YjAyYzM.-cjm\">Nitrogen gas that has been cooled to 77 K has turned to a liquid and must be stored in a vacuum insulated container to prevent it from rapidly vaporizing.<\/p>\n<\/div>\n<p id=\"x-ck12-Mzg3Yjg1ZWVkMjQ3MjJjOWU5YzQ2ZGM1OWNmOWY0Mjc.-x5r\">The <strong> Figure <\/strong> below shows a graph of\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYxMjE2Njg1OA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212239\/f044bae598a4b75a3862b89f6c3a056c.png\" alt=\"frac{PV}{RT}\" width=\"21\" height=\"23\" \/> plotted against pressure for 1 mol of a gas at three different temperatures &#8211; 200 K, 500 K, and 1000 K. An ideal gas would have a value of 1 for that ratio at all temperatures and pressures and the graph would simply be a horizontal line. As can be seen, deviations from an ideal gas occur. As the pressure begins to rise, the attractive forces cause the volume of the gas to be less than expected and the value of\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212239\/f044bae598a4b75a3862b89f6c3a056c.png\" alt=\"frac{PV}{RT}\" width=\"21\" height=\"23\" \/> drops under 1. Continued pressure increase results in the volume of the particles to become significant and the value of\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212239\/f044bae598a4b75a3862b89f6c3a056c.png\" alt=\"frac{PV}{RT}\" width=\"21\" height=\"23\" \/> rises to greater than 1. Notice, that the magnitude of the deviations from ideality is greatest for the gas at 200 K and least for the gas at 1000 K.<\/p>\n<div id=\"x-ck12-OWI3YmI5NWJmNTNmMWI4NjcyNGE1YTM5Yzg0MjQxYzk.-wbi\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-mrc\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2Mzc1NTgwOS02Mi01OS01\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212239\/20140811155526233028.png\" alt=\"Real gases deviate from ideal gases at high pressures and at low temperatures\" longdesc=\"Real%20gases%20deviate%20from%20ideal%20gases%20at%20high%20pressures%20and%20at%20low%20temperatures.\" \/><\/p>\n<p><strong> Figure 14.12 <\/strong><\/p>\n<p id=\"x-ck12-ZDBmNzgyODBjZDEwZmIwNWQ2MDRiOTU2NjM2ZDAzNzY.-o17\">Real gases deviate from ideal gases at high pressures and at low temperatures.<\/p>\n<\/div>\n<p id=\"x-ck12-NDE3Njg1Mjg4MTE2ZTU3OWNkNmQyY2ZiOTBiYWRiNWE.-zdw\">The ideality of a gas also depends on the strength and type of intermolecular attractive forces that exist between the particles. Gases whose attractive forces are weak are more ideal than those with strong attractive forces. At the same temperature and pressure, neon is more ideal than water vapor because neon\u2019s atoms are only attracted by weak dispersion forces, while water vapor\u2019s molecules are attracted by relatively stronger hydrogen bonds. Helium is a more ideal gas than neon because its smaller number of electrons means that helium\u2019s dispersion forces are even weaker than those of neon.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-NWNkMTllNWU1NzUzMzFkMWQ0ZDJlOWIzYjAyMmI1NWE.-7tn\">\n<li>The properties of real gases and their deviations from ideality are described.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-msh\"><em> Questions <\/em><\/p>\n<p id=\"x-ck12-Y2JlMjQ5M2YzMTNmNmRjMzNmZTI0MTMzYzcwM2IzZmY.-ab2\">Use the link below to answer the following questions:<\/p>\n<p id=\"x-ck12-ZGU5NWE1N2U5NGJjYjI3ODFiYmRhOWFlMTJlMmE5MzI.-bvl\"><a href=\"http:\/\/www.adichemistry.com\/physical\/gaseous\/deviation\/van-der-waals-equation.html\"> http:\/\/www.adichemistry.com\/physical\/gaseous\/deviation\/van-der-waals-equation.html <\/a><\/p>\n<ol id=\"x-ck12-ZDk2NGZmMTg4OWVjNzhlZWNhZmNkNzgzZjMxMGQ4Zjk.-vok\">\n<li>What is the compressibility factor for a perfect (ideal) gas?<\/li>\n<li>What does it mean if <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212241\/8befac7dd5d20767941551d93fee0ed7.png\" alt=\"Z&gt;1\" width=\"45\" height=\"13\" \/> ?<\/li>\n<li>What does it mean if <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYxMjE2Njg1OQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212241\/982265d061076cef029721fba31dab33.png\" alt=\"Z&lt;1\" width=\"45\" height=\"13\" \/> ?<\/li>\n<\/ol>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-pxa\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-Y2NmZWU1ODY4ZGE1YjA2YzljOGI0ZTJkNzE5NzEzMDg.-2o0\">\n<li>What becomes more significant as the pressure increases?<\/li>\n<li>Do the attractive forces between gas particles become more prominent at higher or lower temperatures?<\/li>\n<li>Would HCl gas be more or less ideal than helium?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-MmMyMjgyZmVmMmRmZmE3MDYzZTNlZTA4YTUyNDlmNWM.-zfb\">\n<li><strong> real gas: <\/strong> A gas that does not behave according to the assumptions of the kinetic-molecular theory.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-RGFsdG9uJ3MgTGF3IG9mIFBhcnRpYWwgUHJlc3N1cmVz\">Dalton&#8217;s Law of Partial Pressures<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-MDY3MDVkOTBhNDJkNWVhN2JhNzYzMWE1Y2QxYWE5MTE.-bxj\">\n<li>Define partial pressure.<\/li>\n<li>State Dalton\u2019s law of partial pressures.<\/li>\n<li>Use this law to calculate pressures of gas mixtures.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-M2JmNzk2NzlmMmY3NGIwNGRlNWY3NjdlOGY0OGI4YjE.-x3j\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212241\/20140811155526314531.jpeg\" alt=\"Venus has a high partial pressure of carbon dioxide and nitrogen\" width=\"250\" \/><\/span><\/p>\n<h4 id=\"x-ck12-NWVlNTZlMjQyMmMyOGQ2MzIwZDhkZTVlYjU2ZGEwZTY.-4ly_11-ojp\">Is there oxygen available on Venus?<\/h4>\n<p id=\"x-ck12-OWJhOTI4ZDYxMjExYjZlZDk4MjM2ZDVmZjRlNThmZjQ.-g0t\">The atmosphere of Venus is markedly different from that of Earth. The gases in the Venusian atmosphere are 96.5% carbon dioxide and 3% nitrogen. The atmospheric pressure on Venus is roughly 92 times that of Earth, so the amount of nitrogen on Venus would contribute a pressure well over 2700 mm Hg. And there is no oxygen present, so we couldn\u2019t breathe there. Not that we would want to go to Venus \u2013 the surface temperature is usually over 460\u00b0C.<\/p>\n<h3>Dalton\u2019s Law of Partial Pressures<\/h3>\n<p id=\"x-ck12-ZDYzYzQ3MDU4MjUxNDFiZmFhOTUyYmIxYjk4ODFhYTE.-w4c\">Gas pressure results from collisions between gas particles and the inside walls of their container. If more gas is added to a rigid container, the gas pressure increases. The identities of the two gases do not matter. John Dalton, the English chemist who proposed the atomic theory, also studied mixtures of gases. He found that each gas in a mixture exerts a pressure independently of every other gas in the mixture. For example, our atmosphere is composed of about 78% nitrogen and 21% oxygen, with smaller amounts of several other gases making up the rest. Since nitrogen makes up 78% of the gas particles in a given sample of air, it exerts 78% of the pressure. If the overall atmospheric pressure is 1.00 atm, then the pressure of just the nitrogen in the air is 0.78 atm. The pressure of the oxygen in the air is 0.21 atm.<\/p>\n<p id=\"x-ck12-NWNjZTIxYWM1OWIwZTBiNjg2Njk0ZDEzODJiYjQxM2U.-3q5\">The <strong> partial pressure <\/strong> of a gas is the contribution that gas makes to the total pressure when the gas is part of a mixture. The partial pressure of a gas is indicated by a\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYxMjgyNjQ2Mw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212148\/dc2a557e4775924079e5de1061889778.png\" alt=\"P\" width=\"14\" height=\"12\" \/> with a subscript that is the symbol or formula of that gas. The partial pressure of nitrogen is represented by\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYxMjgyNjQ2NA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212243\/28beb3b275887c063f1f720929c422b2.png\" alt=\"P_{N_2}\" width=\"27\" height=\"17\" \/> . <strong> Dalton\u2019s law of partial pressures <\/strong> states that the total pressure of a mixture of gases is equal to the sum of all of the partial pressures of the component gases. Dalton\u2019s law can be expressed with the following equation:<\/p>\n<p id=\"x-ck12-ysv\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212243\/dc796502abd45ae9ca6892fa10b9b1fe.png\" alt=\"P_{text{total}}=P_1+P_2+P_3+ldots\" width=\"205\" height=\"16\" \/><\/p>\n<p id=\"x-ck12-ZjEyMmZiNmI2NjIyMDI3NmEyMzRjNmQ4MmVjZGM1OTI.-6sh\">The <strong> Figure <\/strong> below shows two gases that are in separate, equal-sized containers at the same temperature and pressure. Each exerts a different pressure,\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/b95fd9f12605d93cbb24fd5fc1d83fbc.png\" alt=\"P_1\" width=\"18\" height=\"16\" \/> and <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212125\/f49ec791ef1ecb63d1c14840f9c2f983.png\" alt=\"P_2\" width=\"18\" height=\"15\" \/> , reflective of the number of particles in the container. On the right, the two gases are combined into the same container, with no volume change. The total pressure of the gas mixture is equal to the sum of the individual pressures. If\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212244\/273489be1c6bb8b9fbd5c8559112efa8.png\" alt=\"P_1 = 300 text{ mmHg}\" width=\"128\" height=\"17\" \/> and <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212244\/daa4b3a9457f11fff4ff03fa2aea9830.png\" alt=\"P_2 = 500 text{ mmHg}\" width=\"128\" height=\"16\" \/> , then <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYxMjgyNjQ2NQ..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212245\/512977e45cf92f9bca54108929ca5623.png\" alt=\"P_{text{Total}}=800 text{ mmHg}\" width=\"151\" height=\"17\" \/> .<\/p>\n<div id=\"x-ck12-N2M0MmU0NDIxMjg3MTk5ZTEwYjRjMjBkYTFjYTlkOTU.-usn\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-dxk\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2Mzc1ODExNi0xNS0xMS0xMA..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212246\/20140811155526465849.png\" alt=\"Dalton's law states that the pressure of a gas mixture is equal to the partial pressure of the components\" longdesc=\"Dalton%E2%80%99s%20law%20says%20that%20the%20pressure%20of%20a%20gas%20mixture%20is%20equal%20to%20the%20partial%20pressures%20of%20the%20combining%20gases.\" \/><\/p>\n<p><strong> Figure 14.13 <\/strong><\/p>\n<p id=\"x-ck12-YzE1MjkzNzQ0NWRjYjNhODNmZWY3NjcxM2ZhMTAwZWI.-acd\">Dalton\u2019s law says that the pressure of a gas mixture is equal to the partial pressures of the combining gases.<\/p>\n<\/div>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-MDA2YmNlNThmNjljZDNhYzVkZTA0YjIzMDZjMWNjMzM.-j0h\">\n<li>The total pressure in a system is equal to the sums of the partial pressures of the gases present.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-NDU5MThkNzZhZDRiYmFiYmM1ZGM1ZmRiNjI1Y2QyYzA.-tvi\">Review the concepts at the link below and work the sample problems:<\/p>\n<p id=\"x-ck12-NDgxMjFkMzA0NTNmN2E0MDRlMDA3ZTM0MDY5YTk5ZDQ.-t1g\"><a href=\"http:\/\/www.kentchemistry.com\/links\/GasLaws\/dalton.htm\"> http:\/\/www.kentchemistry.com\/links\/GasLaws\/dalton.htm <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-eeu\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-ZWJjZmYxZjgzNDVlYjYyODljNzcxZTFhMTA3YWJmMzc.-bmk\">\n<li>What is the foundation for Dalton\u2019s law?<\/li>\n<li>Argon makes up about 0.93% of our atmosphere. If the atmospheric pressure is 760 mm Hg, what is the pressure contributed by argon?<\/li>\n<li>On a given day, the water vapor in the air is 2.5%. If the partial pressure of the vapor is 19.4 mm Hg, what is the atmospheric pressure?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-ZjJhNzE4MGMxNTdkZGYwMDdlNGQ0OTcyMDM0ZGIxODI.-sd2\">\n<li><strong> Partial pressure: <\/strong> The contribution that gas makes to the total pressure when the gas is part of a mixture.<\/li>\n<li><strong> Dalton\u2019s law of partial pressures: <\/strong> The total pressure of a mixture of gases is equal to the sum of all of the partial pressures of the component gases.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-TW9sZSBGcmFjdGlvbg..\">Mole Fraction<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-OWFkOTlkY2Q1MzM4ZTllNGUxZDg1Y2VjZmFhYTU3ZDg.-ter\">\n<li>Define mole fraction.<\/li>\n<li>Perform calculations involving mole fractions.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-YTcwYmVhNGMxODY3NjVjMWM0N2Q3ODlhNzcyYWM0MTY.-rve\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212247\/20140811155526612018.jpeg\" alt=\"The changing mole fraction of sulfur dioxide is a mixed blessing\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-ZDU1ZWJkMGNmOTc0MWNiNmUxYzljYjAyYTIwMmNjMjc.-3is\"><strong> The mixed blessing of sulfur dioxide <\/strong><\/p>\n<p id=\"x-ck12-ZWNlYTg2MTlmNjkzNzlkOThhNmIwYTk2ODFkMTA0MGU.-erg\">Sulfur dioxide is a by-product of many processes, both natural and human-made. Massive amounts of this gas are released during volcanic eruptions such as the one seen above on the Big Island (Hawaii). Humans produce sulfur dioxide by burning coal. The gas has a cooling effect when in the atmosphere by reflecting sunlight back away from the earth. However, sulfur dioxide is also a component of smog and acid rain, both of which are harmful to the environment. Many efforts have been made to reduce SO <sub> 2 <\/sub> levels to lower acid rain production. An unforeseen complication: as we lower the concentration of this gas in the atmosphere, we lower its ability to cool and then we have global warming concerns.<\/p>\n<h3>Mole Fraction<\/h3>\n<p id=\"x-ck12-MDQyZjA4ZTJlNjk2NDE4ODk2NjM5NzQ3MDFhYzY4Njk.-yrq\">One way to express relative amounts of substances in a mixture is with the mole fraction. <strong> Mole fraction <\/strong> \u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212249\/0124750beaa395c3119a87d9c19f2789.png\" alt=\"(X)\" width=\"28\" height=\"18\" \/> is the ratio of moles of one substance in a mixture to the total number of moles of all substances. For a mixture of two substances,\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> and <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/00025e2ee78e7b712fbb42f74f2d6cb7.png\" alt=\"B\" width=\"14\" height=\"12\" \/> , the mole fractions of each would be written as follows:<\/p>\n<p id=\"x-ck12-YmU1ZDVkMzc1NDJkNzVmOTNhODcwOTQ0NTlmNzY2Nzg.-zx9\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212249\/3c708d707a41f03b77c141794a18fb87.png\" alt=\"X_A= frac{text{mol} A}{text{mol} A+text{mol} B} quad text{and} quad X_B=frac{text{mol} B}{text{mol} A+text{mol} B}\" width=\"409\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-MzhlNDU0M2Y0MmZlYzA1MDI4NzYwZmYyYWJmOTRiMmI.-xr8\">If a mixture consists of 0.50 mol\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> and 1.00 mol <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/00025e2ee78e7b712fbb42f74f2d6cb7.png\" alt=\"B\" width=\"14\" height=\"12\" \/> , then the mole fraction of\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> would be <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212250\/28245fb4dc2eab5bad3743911a8f1571.png\" alt=\"X_A=frac{0.5}{1.5} = 0.33\" width=\"126\" height=\"24\" \/> .\u00a0 Similarly, the mole fraction of\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/00025e2ee78e7b712fbb42f74f2d6cb7.png\" alt=\"B\" width=\"14\" height=\"12\" \/> would be <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212251\/00d6fe0bf594a44daa1ff7af4a776b56.png\" alt=\"X_B =frac{1.0}{1.5} = 0.67\" width=\"126\" height=\"24\" \/> .<\/p>\n<p id=\"x-ck12-Y2M5M2FiMTlmYWZiZGY5NjAxMzIyMWJkMTE0NDY0ZjU.-m8y\">Mole fraction is a useful quantity for analyzing gas mixtures in conjunction with Dalton\u2019s law of partial pressures. Consider the following situation: A 20.0 liter vessel contains 1.0 mol of hydrogen gas at a pressure of 600 mmHg. Another 20.0 liter vessel contains 3.0 mol of helium at a pressure of 1800 mmHg. These two gases are mixed together in an identical 20.0 liter vessel. Because each will exert its own pressure according to Dalton\u2019s law, we can express the partial pressures as follows:<\/p>\n<p id=\"x-ck12-YmU1ZDVkMzc1NDJkNzVmOTNhODcwOTQ0NTlmNzY2Nzg.-qox\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212251\/09e954a86ef27a64811ada0fe15d0537.png\" alt=\"P_{H_2}=X_{H_2} times P_{text{Total}} quad text{and} quad P_{He}=X_{He} times P_{text{Total}}\" width=\"365\" height=\"17\" \/><\/p>\n<p id=\"x-ck12-NzBjM2NiODE0OTQwYmRmM2YzZjhkNDVkZTE1ZTA0OTA.-ewj\">The partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. For our mixture of hydrogen and helium:<\/p>\n<p id=\"x-ck12-YmU1ZDVkMzc1NDJkNzVmOTNhODcwOTQ0NTlmNzY2Nzg.-jum\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212253\/0c0019ee084d9c55406b47b1d7e2cf9b.png\" alt=\"X_{H_2}=frac{1.0 text{mol}}{1.0 text{mol}+3.0 text{mol}}=0.25 quad text{and} quad X_{He}=frac{3.0 text{mol}}{1.0 text{mol} + 3.0 text{mol}}=0.75\" width=\"572\" height=\"39\" \/><\/p>\n<p id=\"x-ck12-MjQ4ODRlNzAzNTMzYTU3MzIzYTk0OWI3ZDYwNDIwNzg.-dsc\">The total pressure according to Dalton\u2019s law is <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212254\/a35110bf11c36dac7a069303648e986c.png\" alt=\"600 text{ mmHg} + 1800 text{ mmHg} = 2400 text{ mmHg}\" width=\"319\" height=\"16\" \/> . So, each partial pressure will be:<\/p>\n<p id=\"x-ck12-fwz\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212255\/cf654c3d8ff44acf9656dcfbcdc889cd.png\" alt=\"&amp; P_{H_2}=0.25 times 2400 text{ mmHg}=600 text{ mmHg} \\&amp; P_{He}=0.75 times 2400 text{ mmHg}=1800 text{ mmHg}\" width=\"320\" height=\"42\" \/><\/p>\n<p id=\"x-ck12-ODBlMTNhYjlkOWQ0YTE2OGFlMjRjZDg4OWQ4MWEzNWE.-5rp\">The partial pressures of each gas in the mixture don\u2019t change since they were mixed into the same size vessel and the temperature was not changed.<\/p>\n<h4>Sample Problem: Dalton\u2019s Law<\/h4>\n<p id=\"x-ck12-NjY0NmIwNDFhZmM1NjQ2MDkyMTkyYWYwZTBiZDdlMjg.-cop\">A flask contains a mixture of 1.24 moles of hydrogen gas and 2.91 moles of oxygen gas. If the total pressure is 104 kPa, what is the partial pressure of each gas?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-utm\"><em> Step 1: List the known quantities and plan the problem <\/em> .<\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-0ju\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-ZjRlMTgwMTJjM2RhNGYxMzIxZjQxYjMxMmIwODhlY2Q.-j0s\">\n<li>1.24 mol H <sub> 2 <\/sub><\/li>\n<li>2.91 mol O <sub> 2 <\/sub><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212256\/fe4864d1f5f23f08708b43e2ab118bd2.png\" alt=\"P_{text{Total}}=104 text{kPa}\" width=\"129\" height=\"17\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-mux\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-6qx\">\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212257\/128358483d347569abcf233b1977c445.png\" alt=\"P_{H_2}=? text{kPa}\" width=\"93\" height=\"18\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212257\/5ad22f50a3cb357fc8f14ceab55a2d6e.png\" alt=\"P_{O_2}=? text{kPa}\" width=\"92\" height=\"18\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-ZWZjNzRkYTQzYTVhNjBlMTFiMDlhM2I2YzFmYjBkNTQ.-mds\">First, the mole fraction of each gas can be determined. Then, the partial pressure can be calculated by multiplying the mole fraction by the total pressure.<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-hsv\"><em> Step 2: Solve <\/em> .<\/p>\n<p id=\"x-ck12-6ik\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212258\/4a7e9cd4699c81f04274bc8325e98a43.png\" alt=\"&amp; X_{H_2}=frac{1.24 text{mol}}{1.24 text{mol} + 2.91 text{mol}}=0.299 &amp;&amp; X_{O_2}=frac{2.91 text{mol}}{1.24 text{mol} + 2.91 text{mol}}=0.701 \\&amp; P_{H_2}=0.299 times 104 text{ kPa}=31.1 text{ kPa} &amp;&amp; P_{O_2}=0.701 times 104 text{ kPa}=72.9 text{ kPa}\" width=\"574\" height=\"65\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-s3f\"><em> Step 3: Think about your result <\/em> .<\/p>\n<p id=\"x-ck12-YzZhZjZiZGFmMmIyZWVhMjZjODExYWM3NTc0MWMyOTk.-lwh\">The hydrogen is slightly less than one third of the mixture, so it exerts slightly less than one third of the total pressure.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-ZmIwNmY3MjRiNjFjNTYzYjVjMjliOTk0Mzk3ZDM2MmY.-afk\">\n<li>Use of the mole fraction allows calculation to be made for mixtures of gases.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-pc8\"><em> Questions <\/em><\/p>\n<p id=\"x-ck12-YmMzM2NmMmUyYzQ1NmUxNTQ2YTNmYjA2MWZlY2FkODA.-odu\">Watch the video at the link below and answer the following questions:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Mole Fraction and Mole Percent | Given Moles of Gases  | www.whitwellhigh.com\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/7BaX__s-4Ls?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<ol id=\"x-ck12-ZWJhNDUwMTNlZmM3Y2E3NWZjYjJmNDQ4ZGM1NjczMDY.-grk\">\n<li>What is mole percent?<\/li>\n<li>Do the mole fractions add up to 1.00?<\/li>\n<li>What other way could you calculate the mole fraction of oxygen once you have the mole fraction of nitrogen?<\/li>\n<\/ol>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-ebi\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-YmFmMjgyNmNlMDkwZDM5ODBmMjQwMGRhZmJjZjI1MmM.-c02\">\n<li>What is mole fraction?<\/li>\n<li>How do you determine partial pressure of a gas when given the mole fraction and the total pressure?<\/li>\n<li>In a gas mixture containing equal numbers of moles of two gases, what can you say about the partial pressures of each gas?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-MjMzYjRhNDk1ZDFhYWQ2NGNhZDA1MDBhNjZkZDBhYjc.-njd\">\n<li><strong> Mole fraction <\/strong> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212249\/0124750beaa395c3119a87d9c19f2789.png\" alt=\"(X)\" width=\"28\" height=\"18\" \/><strong> : <\/strong> The ratio of moles of one substance in a mixture to the total number of moles of all substances.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-R2FzIENvbGxlY3Rpb24gYnkgV2F0ZXIgRGlzcGxhY2VtZW50\">Gas Collection by Water Displacement<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-NGE2NzM5OTk5MmRhYjQxMDIwMzBjZDk5YjQxMGRjMmM.-hqb\">\n<li>Calculate volumes of dry gases obtained after collecting over water.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-NThiNjgwMzQ2NTQzNjVhYmE3OGZmNzllNzhlN2FkMTU.-jwe\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212300\/20140811155526744059.png\" alt=\"The pressure of gases collected over water can be determined by using the atmospheric pressure in the room\" width=\"400\" \/><\/span><\/p>\n<p id=\"x-ck12-NzBjNDNhMTBkM2VkMzE3MjY1YmVmMmEwNGUyNjU5NmQ.-5mw\"><strong> What is the pressure? <\/strong><\/p>\n<p id=\"x-ck12-MjFjZjViMzQ3M2M3NGRjMjgwYWY3MTRhYTdiMGM2YWE.-3lk\">You need to do a lab experiment where hydrogen gas is generated. In order to calculate the yield of gas, you have to know the pressure inside the tube where the gas is collected. But how can you get a barometer in there? Very simple: you don\u2019t. All you need is the atmospheric pressure in the room. As the gas pushed out the water, it is pushing against the atmosphere, so the pressure inside is equal to the pressure outside.<\/p>\n<h3>Gas Collection by Water Displacement<\/h3>\n<p id=\"x-ck12-NjdiMDQ2N2I1MWExMWRlYTQ0NDBhYmZlNDlhMjQ2ODg.-glt\">Gases that are produced in laboratory experiments are often collected by a technique called <strong> water displacement <\/strong> (see <strong> Figure <\/strong> below ). A bottle is filled with water and placed upside-down in a pan of water. The reaction flask is fitted with rubber tubing which is then fed under the bottle of water. As the gas is produced in the reaction flask, it exits through the rubber tubing and displaces the water in the bottle. When the bottle is full of the gas, it can be sealed with a lid.<\/p>\n<div id=\"x-ck12-MDM5MDYzN2ZhZmJlZTk4ODllZjFhNzA4YzAwNzFhYzk.-eff\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<p id=\"x-ck12-1tt\"><img decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2Mzc1ODQ1Ny0zNy0xNi0y\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212301\/20140811155526874362.png\" alt=\"A gas produced in a chemical reaction can be collected by water displacement.\" longdesc=\"A%20gas%20produced%20in%20a%20chemical%20reaction%20can%20be%20collected%20by%20water%20displacement.\" \/><\/p>\n<p><strong> Figure 14.14 <\/strong><\/p>\n<p id=\"x-ck12-NzVjZDljODZhZDQwOGFkZmIxY2U1NmY3NmNiM2JiNDE.-tqt\">A gas produced in a chemical reaction can be collected by water displacement.<\/p>\n<\/div>\n<p id=\"x-ck12-YTgzZmRiNzE3YjkzYThkMzE3N2QzZWI2NTA4MzYyYzA.-yvf\">Because the gas is collected over water, it is not pure but is mixed with vapor from the evaporation of the water. Dalton\u2019s law can be used to calculate the amount of the desired gas by subtracting the contribution of the water vapor.<\/p>\n<p id=\"x-ck12-gbt\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTQwMDYxMzQzODg4MQ..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212303\/7bf208c74a09188dc83bb4d88013e23c.png\" alt=\"P_{text{Total}} &amp;=P_g+P_{H_2 O} qquad P_g text{ is the pressure of the desired gas}\\P_g &amp;=P_{text{Total}}- P_{H_2 O}\" width=\"462\" height=\"45\" \/><\/p>\n<p id=\"x-ck12-Yjg0MGMyMjBjNjBmMGJmZjM0MTM3ZTdhNGRjODEyOGE.-zrl\">In order to solve a problem, it is necessary to know the vapor pressure of water at the temperature of the reaction (see <strong> Table <\/strong> below). The sample problem illustrates the use of Dalton\u2019s law when a gas is collected over water.<\/p>\n<table>\n<thead>\n<tr>\n<th colspan=\"4\">Vapor Pressure of Water (mmHg) at Selected Temperatures (\u00b0C)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Temperature (\u00b0C)<\/th>\n<th>Vapor Pressure (mmHg)<\/th>\n<th>Temperature (\u00b0C)<\/th>\n<th>Vapor Pressure (mmHg)<\/th>\n<\/tr>\n<tr>\n<td>0<\/td>\n<td>4.58<\/td>\n<td>40<\/td>\n<td>55.32<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>6.54<\/td>\n<td>45<\/td>\n<td>71.88<\/td>\n<\/tr>\n<tr>\n<td>10<\/td>\n<td>9.21<\/td>\n<td>50<\/td>\n<td>92.51<\/td>\n<\/tr>\n<tr>\n<td>15<\/td>\n<td>12.79<\/td>\n<td>55<\/td>\n<td>118.04<\/td>\n<\/tr>\n<tr>\n<td>20<\/td>\n<td>17.54<\/td>\n<td>60<\/td>\n<td>149.38<\/td>\n<\/tr>\n<tr>\n<td>25<\/td>\n<td>23.76<\/td>\n<td>65<\/td>\n<td>187.54<\/td>\n<\/tr>\n<tr>\n<td>30<\/td>\n<td>31.82<\/td>\n<td>70<\/td>\n<td>233.7<\/td>\n<\/tr>\n<tr>\n<td>35<\/td>\n<td>42.18<\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Sample Problem: Gas Collected by Water Displacement<\/h4>\n<p id=\"x-ck12-NDI4YzllZTZmOWEyNzBjY2I2NWQ1ZjQzYjg4ZjI3ODg.-3sz\">A certain experiment generates 2.58 L of hydrogen gas, which is collected over water. The temperature is 20\u00b0C and the atmospheric pressure is 98.60 kPa. Find the volume that the dry hydrogen would occupy at STP.<\/p>\n<p id=\"x-ck12-dkc\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-ykk\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-cy3\">\n<li><span class=\"x-ck12-underline\"> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212304\/006c3d6caa49cb5e202603952dba195f.png\" alt=\"V_{text{Total}} =2.58 text{ L}\" width=\"114\" height=\"17\" \/><\/span><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTQwMDYxMzQzODg4Mw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212304\/dc11e1e2cbee5090672917d93547ea32.png\" alt=\"T=20^ circ text{C}=293 text{ K}\" width=\"146\" height=\"14\" \/><\/li>\n<li><span class=\"x-ck12-underline\"> <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212305\/37bd71ce8cb54b5b5aa31b7d661d6160.png\" alt=\"P_{text{Total}} =98.60 text{ kPa}=739.7 text{ mmHg}\" width=\"266\" height=\"17\" \/><\/span><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-ykd\"><span class=\"x-ck12-underline\"> Unknown <em><br \/>\n<\/em> <\/span><\/p>\n<ul id=\"x-ck12-ifv\">\n<li><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212306\/2f65228bce106ab00ed628005b1f47fa.png\" alt=\"V_{H_2} text{at} STP= ? text{ L}\" width=\"140\" height=\"18\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-Y2VkOTFjYmIxYjRhYmVhZjQzYmQ2YzJhZjI2NzBiNWU.-kgq\">The atmospheric pressure is converted from kPa to mmHg in order to match units with the table. The sum of the pressures of the hydrogen and the water vapor is equal to the atmospheric pressure. The pressure of the hydrogen is found by subtraction. Then, the volume of the gas at STP can be calculated by using the combined gas law.<\/p>\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-4yd\"><em> Step 2: Solve. <\/em><\/p>\n<p id=\"x-ck12-3jx\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212306\/6c113d922b4e57a5765d16193c0ce8c2.png\" alt=\"P_{H_2}=P_{text{Total}}-P_{H_2 O}=739.7 text{ mmHg} -17.54 text{ mmHg}=722.2 text{ mmHg}\" width=\"524\" height=\"17\" \/><\/p>\n<p id=\"x-ck12-MWU2OGE5MzZmNTk5NmMxNmI5YjJhMDIyYzlhOWIzMmM.-dgz\">Now the combined gas law is used, solving for <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTQwMDYxMzQzODg4NA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212126\/fa8b087810f399546cd0b219e38f3e1b.png\" alt=\"V_2\" width=\"16\" height=\"15\" \/> , the volume of hydrogen at STP.<\/p>\n<p id=\"x-ck12-vft\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212307\/38c5ef825b1c1b16462d721800e47177.png\" alt=\"V_2=frac{P_1 times V_1 times T_2}{P_2 times T_1}=frac{722.2 text{ mmHg} times 2.58 text{ L} times 273 text{ K}}{760 text{ mmHg} times 293 text{ K}}=2.28 text{ L } H_2\" width=\"511\" height=\"41\" \/><\/p>\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-pv0\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-MWZhNjY4NGY5OTUwN2RjNWVkNjQ2MDY2NGYzMDE0YTA.-dqo\">If the hydrogen gas were to be collected at STP and without the presence of the water vapor, its volume would be 2.28 L. This is less than the actual collected volume because some of that is water vapor. The conversion using STP is useful for stoichiometry purposes.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-ZjMyMzQ1NjQxODdjODQ0NzhjNWMwOWQ4MmEzM2I0MTI.-8sw\">\n<li>The vapor pressure due to water in a sample can be corrected for in order to get the true value for the pressure of the gas.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-teg\"><em> Questions <\/em><\/p>\n<p id=\"x-ck12-YmMzM2NmMmUyYzQ1NmUxNTQ2YTNmYjA2MWZlY2FkODA.-afu\">Watch the video at the link below and answer the following questions:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Collection of oxygen gas through water displacement lab\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/xmL2Pax4yUQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<ol id=\"x-ck12-MTg0ZjQ4MWJmYmQ5NTU1OTFhOWM5ZDUxZDEzMjMxNDY.-jip\">\n<li>What was the thistle tube used for?<\/li>\n<li>How did the instructor tests for oxygen?<\/li>\n<li>Did you observe any unsafe lab practices in the video?<\/li>\n<li>What would have happened to the splint if carbon dioxide had been collected?<\/li>\n<\/ol>\n<h4>Review<\/h4>\n<p><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-ZWEyYWIwN2U4ZjllZTVlYjcxM2Q3YTIxNGRiMzY2M2I.-ygo\">\n<li>Why is gas collected over water not pure?<\/li>\n<li>Why would we want to correct for water vapor?<\/li>\n<li>A student wants to collect his gas over diethyl ether (vapor pressure of 530 mm Hg at 25\u00b0C). Is this a good idea? Explain your answer.<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-MzNkNDQ2MmYxZjA0ZTg1ZjEyYjdmOTJhZDhmOTRmMTY.-9vg\">\n<li><strong> water displacement: <\/strong> Collection of a gas over water.<\/li>\n<\/ul>\n<\/div>\n<h1 id=\"x-ck12-RGlmZnVzaW9uIGFuZCBFZmZ1c2lvbiBhbmQgR3JhaGFtJ3MgTGF3\">Diffusion and Effusion and Graham&#8217;s Law<\/h1>\n<div class=\"x-ck12-data-objectives\">\n<ul id=\"x-ck12-MDc3OWVlN2RiZDc0OTU3Y2Y1Y2VkZTI2NDc2MDFlZWI.-ub9\">\n<li>Define diffusion and effusion.<\/li>\n<li>State Graham\u2019s law.<\/li>\n<li>Use Graham\u2019s law to perform calculations involving movement of gases.<\/li>\n<\/ul>\n<\/div>\n<p id=\"x-ck12-MmE4Yzc2MzJhZWQyMDE4NTcyMTRjNzAyY2Y3YWI3MjE.-vle\"><span class=\"x-ck12-img-inline\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212310\/20140811155526959330.png\" alt=\"A classic experiment to find the rate of diffusion for gases uses hydrochloric acid and ammonia\" width=\"450\" \/><\/span><\/p>\n<p id=\"x-ck12-ODYwMDBlYTZmMjBiNmZlOWVmMzdiMGYwZmIzODUyYTU.-tvq\"><strong> How do we know how fast a gas moves? <\/strong><\/p>\n<p id=\"x-ck12-Njg5MWRkOWIzOGI2YWM5NDljZjI1OTY3M2ZlOTRkNDY.-ypb\">We usually cannot see gases, so we need ways to detect their movements indirectly. The relative rates of diffusion of ammonia to hydrogen chloride can be observed in a simple experiment. Cotton balls are soaked with solutions of ammonia and hydrogen chloride (hydrochloric acid) and attached to two different rubber stoppers. These are simultaneously plugged into either end of a long glass tube. The vapors of each travel down the tube at different rates. Where the vapors meet, they react to form ammonium chloride (NH <sub> 4 <\/sub> Cl), a white solid that appears in the glass tube as a ring.<\/p>\n<h3>Graham\u2019s Law<\/h3>\n<p id=\"x-ck12-ODA4ZGIxYTRiNjA1MTQ4YzY5OGQ0YWYzOGIwYzNjYjA.-ehk\">When a person opens a bottle of perfume in one corner of a large room, it doesn\u2019t take very long for the scent to spread throughout the entire room. Molecules of the perfume evaporate and the vapor spreads out to fill the entire space. <strong> Diffusion <\/strong> is the tendency of molecules to move from an area of high concentration to an area of low concentration until the concentration is uniform. While gases diffuse rather quickly, liquids diffuse much more slowly. Solids essentially do not diffuse.<\/p>\n<p id=\"x-ck12-OGNhODJkZWQ1YzBiMTQ3MzVkNGI1Mjc2NjQ1OTVlYWU.-2px\">Video of bromine diffusion:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Diffusion of Bromine vapor\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/R_xDe004oTQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p id=\"x-ck12-ZjUyZmQ0ZjJiM2IwYmFhNDY2YTA0OTgzMTk1NzBhMTQ.-d8t\">A related process to diffusion is the effusion. <strong> Effusion <\/strong> is the process of a confined gas escaping through a tiny hole in its container. Effusion can be observed by the fact that a helium-filled balloon will stop floating and sink to the floor after a day or so. This is because the helium gas effuses through tiny pores in the balloon. Both diffusion and effusion are related to the speed at which various gas molecules move. Gases that have a lower molar mass effuse and diffuse at a faster rate than gases that have a higher molar mass.<\/p>\n<p id=\"x-ck12-Mjk5ZDVhMDA5NzFkOWI5ZDMzZDlkMzdlZDg0ZmZiOTk.-lgz\">Scottish chemist Thomas Graham (1805-1869) studied the rates of effusion and diffusion of gases. <strong> Graham\u2019s law <\/strong> states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. Graham\u2019s law can be understood by comparing two gases ( <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYxOTU4ODEwMw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> and <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/00025e2ee78e7b712fbb42f74f2d6cb7.png\" alt=\"B\" width=\"14\" height=\"12\" \/> ) at the same temperature, meaning the gases have the same kinetic energy. The kinetic energy of a moving object is given by the equation <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212312\/50ba5f09928eb42fb31d9b0d2f6e4854.png\" alt=\"KE =frac{1}{2}mv^2\" width=\"96\" height=\"23\" \/> ,<\/p>\n<p id=\"x-ck12-NDJiZjk4ZjQ2MzlhZDY4MjIzYTEwYjc0MjlhZWMyZGE.-ckf\">where\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211009\/de65f173073697541b369ca6047387a8.png\" alt=\"m\" width=\"16\" height=\"8\" \/> is mass and\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYxOTU4ODEwNA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19210944\/ab157ad37ef6efcf3559476aa5ec6133.png\" alt=\"v\" width=\"9\" height=\"8\" \/> is velocity. Setting the kinetic energies of the two gases equal to one another gives:<\/p>\n<p id=\"x-ck12-qnu\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212313\/b3bc5c29051f9d8156106e18b7e5adeb.png\" alt=\"frac{1}{2}m_Av^2_A=frac{1}{2}m_Bv^2_B\" width=\"138\" height=\"37\" \/><\/p>\n<p id=\"x-ck12-YmMzY2MxZmEzNzg3YzNmZmJjZGU1MjRjYmQ3OGVlNzE.-0mk\">The equation can be rearranged to solve for the ratio of the velocity of gas\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> to the velocity of gas\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212313\/cddfbc8003d4628a6a87301133d54c3f.png\" alt=\"Bleft(frac{v_A}{v_B}right)\" width=\"56\" height=\"34\" \/> .<\/p>\n<p id=\"x-ck12-ODAyYmE5MzhjOTc5YWE5MzAyOTFkOTY0Njk4YWMzYmY.-fbq\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2NjYxOTU4ODEwNQ..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212313\/07a84d6027d74eedc76b326f7240482c.png\" alt=\"frac{v^2_A}{v^2_B}=frac{m_B}{m_A} text{ which becomes } frac{v_A}{v_B}=sqrt{frac{m_B}{m_A}}\" width=\"301\" height=\"47\" \/><\/p>\n<p id=\"x-ck12-N2FiOWVlMTNiMGYxNTZiZGFiNTc1NjI0Mjg3OGMyOWM.-wn8\">For the purposes of comparing the rates of effusion or diffusion of two gases at the same temperature, the molar masses of each gas can be used in the equation for <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211009\/de65f173073697541b369ca6047387a8.png\" alt=\"m\" width=\"16\" height=\"8\" \/> .<\/p>\n<h4>Sample Problem: Graham\u2019s Law<\/h4>\n<p id=\"x-ck12-ZDUzYjg0NGQxMzZmMjZkNTJkZDhhN2FmYzlkOTNlYTc.-yrt\">Calculate the ratio of diffusion rates of ammonia gas (NH <sub> 3 <\/sub> ) to hydrogen chloride (HCl) at the same temperature and pressure.<\/p>\n<p id=\"x-ck12-OGU2YWEyYzY0NzAwMDUxZjI4NjFjY2E4MjYyNmNhN2I.-tqe\"><em> Step 1: List the known quantities and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-kdy\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-M2Q3ZjVmZTM3YjFlOTU3OGQ0ZjMxMTk1M2FmOGIzNjE.-qvq\">\n<li>molar mass NH <sub> 3 <\/sub> = 17.04 g\/mol<\/li>\n<li>molar mass HCl = 36.46 g\/mol<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-bqu\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-ZGU5MzNiY2VmMTI1ZWNlY2YxNmE2OTMwNzUwNTE3YjE.-sqs\">\n<li>velcoity ratio\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212314\/1dcf1549ad77a06a51f20268d8f21e59.png\" alt=\"dfrac{v_{text{NH}_3}}{v_{text{HCl}}}\" width=\"36\" height=\"37\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-ZTFiZjBkMGM5ODZhYjBkZWY3YmNjYzRkZjQ4Yjc2MDU.-0zg\">Substitute the molar masses of the gases into Graham\u2019s law and solve for the ratio.<\/p>\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-cnq\"><em> Step 2: Solve. <\/em><\/p>\n<p id=\"x-ck12-tnf\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212315\/dddb559fd0c6a5257db93b5b44578c43.png\" alt=\"frac{v_{NH_3}}{v_{HCl}}=sqrt{frac{36.46 text{ g\/mol}}{17.04 text{ g\/mol}}}=1.46\" width=\"235\" height=\"56\" \/><\/p>\n<p id=\"x-ck12-YmYwNmQwN2NiZGZhZjNmZTRhYjQzODE4NmU1MDIxNWI.-ak8\">The rate of diffusion of ammonia is 1.46 times faster than the rate of diffusion of hydrogen chloride.<\/p>\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-ork\"><em> Step 3: Think about your result <\/em><\/p>\n<p id=\"x-ck12-NTk2ZDY1NDhjMjI1MTA4OTY2MzllM2Q1YzU5MGZjYWE.-bdx\">Since ammonia has a smaller molar mass than hydrogen chloride, the velocity of its molecules is greater and the velocity ratio is larger than 1.<\/p>\n<h4>Summary<\/h4>\n<ul id=\"x-ck12-Mzg3OWM5YWMyYTRiNmE4YmQ1MmYwOWY5ZjE1ZTRjZGU.-7kb\">\n<li>The processes of gas diffusion and effusion are described.<\/li>\n<li>Graham\u2019s law relates the molecular mass of a gas to its rate of diffusion or effusion.<\/li>\n<\/ul>\n<h4>Practice<\/h4>\n<p id=\"x-ck12-Yjc3YzRjOTEyYmFmY2ZkMTdhOGMxM2VhZDRiMzQyMjU.-n25\">Read the material on the link below and do the practice problems:<\/p>\n<p id=\"x-ck12-ZGFjY2IzM2Q1MGUxMTdmNjc1NDkwY2U1YjcxMmYwNzE.-odl\"><a href=\"http:\/\/www.kentchemistry.com\/links\/GasLaws\/GrahamsLaw.htm\"> http:\/\/www.kentchemistry.com\/links\/GasLaws\/GrahamsLaw.htm <\/a><\/p>\n<h4>Review<\/h4>\n<p id=\"x-ck12-NmNlM2JkYzM5ZTU0NDcyNTAzOGUwZTg3Yjc3MWRiMWU.-orq\"><em> Questions <\/em><\/p>\n<ol id=\"x-ck12-ZDU1MWJiY2FmNWQ2Nzg1MmViOTA0ZjRkNTQ5ZWU2ODY.-tfs\">\n<li>Why can you smell food cooking when you are in the next room?<\/li>\n<li>Why does a helium-filled balloon gradually sink?<\/li>\n<li>What does temperature have to do with gas kinetic energies?<\/li>\n<\/ol>\n<div class=\"x-ck12-data-problem-set\"><\/div>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-MzFjY2RjZmMwYTY2YjY5YjM5NTJiNDdmYzZlNWNhMjM.-mnw\">\n<li><strong> diffusion: <\/strong> The tendency of molecules to move from an area of high concentration to an area of low concentration until the concentration is uniform.<\/li>\n<li><strong> effusion: <\/strong> The process of a confined gas escaping through a tiny hole in its container.<\/li>\n<li><strong> Graham\u2019s law: <\/strong> The rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas.<\/li>\n<\/ul>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q836080\">Show References<\/span><\/p>\n<div id=\"q836080\" class=\"hidden-answer\" style=\"display: none\">\n<h2>References<\/h2>\n<ol>\n<li>Image copyright Jandrie Lombard, 2014. <a href=\"http:\/\/www.shutterstock.com\"> http:\/\/www.shutterstock.com <\/a> .<\/li>\n<li>Ian Myles. <a href=\"http:\/\/www.flickr.com\/photos\/imphotography\/3754144111\/\"> http:\/\/www.flickr.com\/photos\/imphotography\/3754144111\/ <\/a> .<\/li>\n<li>Flickr: rick. <a href=\"http:\/\/www.flickr.com\/photos\/spine\/309730216\/\"> http:\/\/www.flickr.com\/photos\/spine\/309730216\/ <\/a> .<\/li>\n<li>Courtesy of US Navy Photographer&#8217;s Mate 2nd Class Damon J. Moritz. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Basketball_game.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Basketball_game.jpg <\/a> .<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung. .<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung. .<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung. .<\/li>\n<li>Courtesy of NOAA Photo Library, NOAA Central Library; OAR\/ERL\/National Severe Storms Laboratory (NSSL). <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Nssl0020.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Nssl0020.jpg <\/a> .<\/li>\n<li>Johann Kerseboom. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Robert_Boyle_0001.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Robert_Boyle_0001.jpg <\/a> .<\/li>\n<li>CK-12 Foundation &#8211; Wade Baxter. CK-12 Foundation .<\/li>\n<li>Courtesy of Petty Officer 3rd Class Charles Oki, US Navy. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Navy_baking_bread.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Navy_baking_bread.jpg <\/a> .<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung. .<\/li>\n<li>CK-12 Foundation &#8211; Wade Baxter. CK-12 Foundation .<\/li>\n<li>Courtesy of Robert Kaufmann, FEMA. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:FEMA_-_22585_-_Photograph_by_Robert_Kaufmann_taken_on_02-27-2006_in_Louisiana.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:FEMA_-_22585_-_Photograph_by_Robert_Kaufmann_taken_on_02-27-2006_in_Louisiana.jpg <\/a> .<\/li>\n<li>. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Joseph_louis_gay-lussac.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Joseph_louis_gay-lussac.jpg <\/a> .<\/li>\n<li>User:Calipper\/It.Wikipedia. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Frigorifero_Ignis_componenti.JPG\"> http:\/\/commons.wikimedia.org\/wiki\/File:Frigorifero_Ignis_componenti.JPG <\/a> .<\/li>\n<li>User:Martijn\/Nl.Wikipedia. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Fietspomp.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Fietspomp.jpg <\/a> .<\/li>\n<li>User:Masur\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Gas_cylinder_ammonia.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Gas_cylinder_ammonia.jpg <\/a> .<\/li>\n<li>User:GeorgHH\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:HighFlyer_Hamburg02.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:HighFlyer_Hamburg02.jpg <\/a> .<\/li>\n<li>User:J\u00fc\/Wikimedia Commons. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Isomere_Ethanol_Dimethylether.png\"> http:\/\/commons.wikimedia.org\/wiki\/File:Isomere_Ethanol_Dimethylether.png <\/a> .<\/li>\n<li>Flickr: andrechinn. <a href=\"http:\/\/www.flickr.com\/photos\/andrec\/2699842079\/\"> http:\/\/www.flickr.com\/photos\/andrec\/2699842079\/ <\/a> .<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung. .<\/li>\n<li>Courtesy of NASA\/JPL. <a href=\"http:\/\/photojournal.jpl.nasa.gov\/catalog\/pia00124\"> http:\/\/photojournal.jpl.nasa.gov\/catalog\/pia00124 <\/a> .<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung. .<\/li>\n<li>Courtesy of J. D. Griggs, US Geological Survey. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Pahoeoe_fountain_sharpen.jpg\"> http:\/\/commons.wikimedia.org\/wiki\/File:Pahoeoe_fountain_sharpen.jpg <\/a> .<\/li>\n<li>Laura Guerin. CK-12 Foundation .<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung. .<\/li>\n<li>CK-12 Foundation &#8211; Christopher Auyeung. .<\/li>\n<\/ol>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1043\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry Concepts Intermediate. <strong>Authored by<\/strong>: Calbreath, Baxter, et al.. <strong>Provided by<\/strong>: CK12.org. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\">http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":1507,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry Concepts Intermediate\",\"author\":\"Calbreath, Baxter, et al.\",\"organization\":\"CK12.org\",\"url\":\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1043","chapter","type-chapter","status-publish","hentry"],"part":2335,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/1043","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/users\/1507"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/1043\/revisions"}],"predecessor-version":[{"id":3694,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/1043\/revisions\/3694"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/parts\/2335"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/1043\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=1043"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=1043"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=1043"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=1043"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}