{"id":2090,"date":"2015-04-22T20:23:02","date_gmt":"2015-04-22T20:23:02","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=2090"},"modified":"2016-08-09T02:01:41","modified_gmt":"2016-08-09T02:01:41","slug":"relating-pressure-volume-amount-and-temperature-the-ideal-gas-law-needs-formulas","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/chapter\/relating-pressure-volume-amount-and-temperature-the-ideal-gas-law-needs-formulas\/","title":{"raw":"Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law","rendered":"Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>LEARNING OBJECTIVES<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Identify the mathematical relationships between the various properties of gases<\/li>\r\n \t<li>Use the ideal gas law, and related gas laws, to compute the values of various gas properties under specified conditions<\/li>\r\n<\/ul>\r\n<\/div>\r\nDuring the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could fly (Figure 1), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, pressure, volume, temperature, and amount of gas. Although their measurements were not precise by today\u2019s standards, they were able to determine the mathematical relationships between pairs of these variables (e.g., pressure and temperature, pressure and volume) that hold for an <em data-effect=\"italics\">ideal<\/em> gas\u2014a hypothetical construct that real gases approximate under certain conditions. Eventually, these individual laws were combined into a single equation\u2014the <em data-effect=\"italics\">ideal gas law<\/em>\u2014that relates gas quantities for gases and is quite accurate for low pressures and moderate temperatures. We will consider the key developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas law.\r\n\r\n[caption id=\"attachment_595\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-595\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205454\/CNX_Chem_09_02_Ballooning-1024x279.jpg\" alt=\"This figure includes three images. Image a is a black and white image of a hydrogen balloon apparently being deflated by a mob of people. In image b, a blue, gold, and red balloon is being held to the ground with ropes while positioned above a platform from which smoke is rising beneath the balloon. In c, an image is shown in grey on a peach-colored background of an inflated balloon with vertical striping in the air. It appears to have a basket attached to its lower side. A large stately building appears in the background.\" width=\"1024\" height=\"279\" \/> Figure 1. In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.[\/caption]\r\n\r\n<section id=\"fs-idm39174352\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Pressure and Temperature: Amontons\u2019s Law<\/h2>\r\n<p id=\"fs-idm131642544\">Imagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (Figure 2) and the pressure increases.<\/p>\r\n\r\n\r\n[caption id=\"attachment_596\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-596\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205456\/CNX_Chem_09_01_Amontons1-1024x489.jpg\" alt=\"This figure includes three similar diagrams. In the first diagram to the left, a rigid spherical container of a gas to which a pressure gauge is attached at the top is placed in a large beaker of water, indicated in light blue, atop a hot plate. The needle on the pressure gauge points to the far left on the gauge. The diagram is labeled \u201clow P\u201d above and \u201chot plate off\u201d below. The second similar diagram also has the rigid spherical container of gas placed in a large beaker from which light blue wavy line segments extend from the top of the liquid in the beaker. The beaker is situated on top of a slightly reddened circular area. The needle on the pressure gauge points straight up, or to the middle on the gauge. The diagram is labeled \u201cmedium P\u201d above and \u201chot plate on medium\u201d below. The third diagram also has the rigid spherical container of gas placed in a large beaker in which bubbles appear near the liquid surface and several wavy light blue line segments extend from the surface out of the beaker. The beaker is situated on top of a bright red circular area. The needle on the pressure gauge points to the far right on the gauge. The diagram is labeled \u201chigh P\u201d above and \u201chot plate on high\u201d below.\" width=\"1024\" height=\"489\" \/> Figure 2. The effect of temperature on gas pressure: When the hot plate is off, the pressure of the gas in the sphere is relatively low. As the gas is heated, the pressure of the gas in the sphere increases.[\/caption]\r\n<p id=\"fs-idm111816736\">This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in Figure 3. We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then <em data-effect=\"italics\">P<\/em> and <em data-effect=\"italics\">T<\/em> are directly proportional (again, when <em data-effect=\"italics\">volume and moles of gas are held constant<\/em>); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor.<\/p>\r\n\r\n\r\n[caption id=\"attachment_597\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-597\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205458\/CNX_Chem_09_02_Amontons2-1024x329.jpg\" alt=\"This figure includes a table and a graph. The table has 3 columns and 7 rows. The first row is a header, which labels the columns \u201cTemperature, degrees C,\u201d \u201cTemperature, K,\u201d and \u201cPressure, k P a.\u201d The first column contains the values from top to bottom negative 150, negative 100, negative 50, 0, 50, and 100. The second column contains the values from top to bottom 173, 223, 273, 323, 373, and 423. The third column contains the values 36.0, 46.4, 56.7, 67.1, 77.5, and 88.0. A graph appears to the right of the table. The horizontal axis is labeled \u201cTemperature ( K ).\u201d with markings and labels provided for multiples of 100 beginning at 0 and ending at 500. The vertical axis is labeled \u201cPressure ( k P a )\u201d with markings and labels provided for multiples of 10, beginning at 0 and ending at 100. Six data points from the table are plotted on the graph with black dots. These dots are connected with a solid black line. A dashed line extends from the data point furthest to the left to the origin. The graph shows a positive linear trend.\" width=\"1024\" height=\"329\" \/> Figure 3. For a constant volume and amount of air, the pressure and temperature are directly proportional, provided the temperature is in kelvin. (Measurements cannot be made at lower temperatures because of the condensation of the gas.) When this line is extrapolated to lower pressures, it reaches a pressure of 0 at \u2013273 \u00b0C, which is 0 on the kelvin scale and the lowest possible temperature, called absolute zero.[\/caption]\r\n<p id=\"fs-idm33029376\">Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the <em data-effect=\"italics\">P<\/em>-<em data-effect=\"italics\">T<\/em> relationship for gases is known as either <strong><span data-type=\"term\">Amontons\u2019s law<\/span><\/strong> or <strong><span data-type=\"term\">Gay-Lussac\u2019s law<\/span><\/strong>. Under either name, it states that <em data-effect=\"italics\">the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant<\/em>. Mathematically, this can be written:<\/p>\r\n\r\n<div id=\"fs-idp58663824\" data-type=\"equation\">[latex]P\\propto T\\text{ or }P=\\text{constant}\\times T\\text{ or }P=k\\times T[\/latex]<\/div>\r\n<p id=\"fs-idp12374848\">where \u221d means \u201cis proportional to,\u201d and <em data-effect=\"italics\">k<\/em> is a proportionality constant that depends on the identity, amount, and volume of the gas.<\/p>\r\n<p id=\"fs-idm100617456\">For a confined, constant volume of gas, the ratio [latex]\\frac{P}{T}[\/latex] is therefore constant (i.e., [latex]\\frac{P}{T}=k[\/latex] ). If the gas is initially in \u201cCondition 1\u201d (with <em data-effect=\"italics\">P<\/em> = <em data-effect=\"italics\">P<\/em><sub>1<\/sub> and <em data-effect=\"italics\">T<\/em> = <em data-effect=\"italics\">T<\/em><sub>1<\/sub>), and then changes to \u201cCondition 2\u201d (with <em data-effect=\"italics\">P<\/em> = <em data-effect=\"italics\">P<\/em><sub>2<\/sub> and <em data-effect=\"italics\">T<\/em> = <em data-effect=\"italics\">T<\/em><sub>2<\/sub>), we have that [latex]\\frac{{P}_{1}}{{T}_{1}}=k[\/latex] and [latex]\\frac{{P}_{2}}{{T}_{2}}=k,[\/latex] which reduces to [latex]\\frac{{P}_{1}}{{T}_{1}}=\\frac{{P}_{2}}{{T}_{2}}.[\/latex] This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called <strong><span data-type=\"term\">absolute zero<\/span><\/strong>). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.)<\/p>\r\n\r\n<div id=\"fs-idm66260928\" class=\"textbox shaded\" data-type=\"example\">\r\n<h3>Example 1<\/h3>\r\n<h4><strong>Predicting Change in Pressure with Temperature<\/strong><\/h4>\r\n<p id=\"fs-idm61839504\">A can of hair spray is used until it is empty except for the propellant, isobutane gas.<\/p>\r\n<p id=\"fs-idp96109888\">(a) On the can is the warning \u201cStore only at temperatures below 120 \u00b0F (48.8 \u00b0C). Do not incinerate.\u201d Why?<\/p>\r\n<p id=\"fs-idm53382240\">(b) The gas in the can is initially at 24 \u00b0C and 360 kPa, and the can has a volume of 350 mL. If the can is left in a car that reaches 50 \u00b0C on a hot day, what is the new pressure in the can?<\/p>\r\n\r\n<h4 id=\"fs-idm152475936\"><span data-type=\"title\">Solution<\/span><\/h4>\r\n(a) The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)\r\n<p id=\"fs-idp91458240\">(b) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons\u2019s\/Gay-Lussac\u2019s law. Taking <em data-effect=\"italics\">P<\/em><sub>1<\/sub> and <em data-effect=\"italics\">T<\/em><sub>1<\/sub> as the initial values, <em data-effect=\"italics\">T<\/em><sub>2<\/sub> as the temperature where the pressure is unknown and <em data-effect=\"italics\">P<\/em><sub>2<\/sub> as the unknown pressure, and converting \u00b0C to K, we have:<\/p>\r\n\r\n<div id=\"fs-idp253098544\" data-type=\"equation\">[latex]\\frac{{P}_{1}}{{T}_{1}}=\\frac{{P}_{2}}{{T}_{2}}\\phantom{\\rule{0.4em}{0ex}}\\text{which means that}\\phantom{\\rule{0.4em}{0ex}}\\frac{360\\text{kPa}}{297\\text{K}}=\\frac{{P}_{2}}{323\\text{K}}[\/latex]<\/div>\r\n<p id=\"fs-idm99533072\">Rearranging and solving gives: [latex]{P}_{2}=\\frac{360\\text{kPa}\\times 323\\cancel{\\text{K}}}{297\\cancel{\\text{K}}}=390\\text{kPa}[\/latex]<\/p>\r\n\r\n<h4 id=\"fs-idm87475200\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nA sample of nitrogen, N<sub>2<\/sub>, occupies 45.0 mL at 27 \u00b0C and 600 torr. What pressure will it have if cooled to \u201373 \u00b0C while the volume remains constant?\r\n<div id=\"fs-idp78288768\" data-type=\"note\">\r\n<div style=\"text-align: right;\" data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer: <\/strong>\u00a0400 torr<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-idp20892224\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Volume and Temperature: Charles\u2019s Law<\/h2>\r\n<p id=\"fs-idm19167872\">If we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let\u2019s say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up.<\/p>\r\n\r\n<div id=\"fs-idp27422192\" class=\"chemistry link-to-learning textbox\" data-type=\"note\">\r\n<p id=\"fs-idp60978480\">This video shows how cooling and heating a gas causes its volume to decrease or increase, respectively.<\/p>\r\nhttps:\/\/youtu.be\/ZgTTUuJZAFs\r\n\r\n<\/div>\r\n<p id=\"fs-idp49208960\">These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Volume-temperature data for a 1-mole sample of methane gas at 1 atm are listed and graphed in Figure 4.<\/p>\r\n\r\n\r\n[caption id=\"attachment_599\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-599\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205500\/CNX_Chem_09_02_Charles2-1024x379.jpg\" alt=\"This figure includes a table and a graph. The table has 3 columns and 6 rows. The first row is a header, which labels the columns \u201cTemperature, degrees C,\u201d \u201cTemperature, K,\u201d and \u201cPressure, k P a.\u201d The first column contains the values from top to bottom negative 100, negative 50, 0, 100, and 200. The second column contains the values from top to bottom 173, 223, 273, 373, and 473. The third column contains the values 14.10, 18.26, 22.40, 30.65, and 38.88. A graph appears to the right of the table. The horizontal axis is labeled \u201cTemperature ( K ).\u201d with markings and labels provided for multiples of 100 beginning at 0 and ending at 300. The vertical axis is labeled \u201cVolume ( L )\u201d with marking and labels provided for multiples of 10, beginning at 0 and ending at 30. Five data points from the table are plotted on the graph with black dots. These dots are connected with a solid black line. The graph shows a positive linear trend.\" width=\"1024\" height=\"379\" \/> Figure 4. The volume and temperature are linearly related for 1 mole of methane gas at a constant pressure of 1 atm. If the temperature is in kelvin, volume and temperature are directly proportional. The line stops at 111 K because methane liquefies at this temperature; when extrapolated, it intersects the graph\u2019s origin, representing a temperature of absolute zero.[\/caption]\r\n<p id=\"fs-idp38621856\">The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles\u2019s law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre C\u00e9sar Charles. <strong><span data-type=\"term\">Charles\u2019s law<\/span><\/strong> states that <em data-effect=\"italics\">the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant<\/em>.<\/p>\r\n<p id=\"fs-idp174747472\">Mathematically, this can be written as:<\/p>\r\n\r\n<div id=\"fs-idm59931152\" data-type=\"equation\">[latex]V\\propto T\\text{or}V=\\text{constant}\\cdot T\\text{or}V=k\\cdot T\\text{or}{V}_{1}\\text{\/}{T}_{1}={V}_{2}\\text{\/}{T}_{2}[\/latex]<\/div>\r\n<p id=\"fs-idp134528\">with <em data-effect=\"italics\">k<\/em> being a proportionality constant that depends on the amount and pressure of the gas.<\/p>\r\n<p id=\"fs-idp8180592\">For a confined, constant pressure gas sample, [latex]\\frac{V}{T}[\/latex] is constant (i.e., the ratio = <em data-effect=\"italics\">k<\/em>), and as seen with the <em data-effect=\"italics\">V<\/em>-<em data-effect=\"italics\">T<\/em> relationship, this leads to another form of Charles\u2019s law: [latex]\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}.[\/latex]<\/p>\r\n\r\n<div id=\"fs-idm45782128\" class=\"textbox shaded\" data-type=\"example\">\r\n<h3><strong>Example 2<\/strong><\/h3>\r\n<h4 id=\"fs-idp8413888\"><strong><span data-type=\"title\">Predicting Change in Volume with Temperature<\/span><\/strong><\/h4>\r\nA sample of carbon dioxide, CO<sub>2<\/sub>, occupies 0.300 L at 10 \u00b0C and 750 torr. What volume will the gas have at 30 \u00b0C and 750 torr?\r\n<h4 id=\"fs-idp207430880\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nBecause we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles\u2019s law. Taking <em data-effect=\"italics\">V<\/em><sub>1<\/sub> and <em data-effect=\"italics\">T<\/em><sub>1<\/sub> as the initial values, <em data-effect=\"italics\">T<\/em><sub>2<\/sub> as the temperature at which the volume is unknown and <em data-effect=\"italics\">V<\/em><sub>2<\/sub> as the unknown volume, and converting \u00b0C into K we have:\r\n<div id=\"fs-idp22541168\" data-type=\"equation\">[latex]\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}\\phantom{\\rule{0.4em}{0ex}}\\text{which means that}\\phantom{\\rule{0.4em}{0ex}}\\frac{0.300\\text{L}}{283\\text{K}}=\\frac{{V}_{2}}{303\\text{K}}[\/latex]<\/div>\r\n<p id=\"fs-idp170393776\">Rearranging and solving gives: [latex]{V}_{2}=\\frac{0.300\\text{L}\\times \\text{303}\\cancel{\\text{K}}}{283\\cancel{\\text{K}}}=0.321\\text{L}[\/latex]<\/p>\r\n<p id=\"fs-idp111256368\">This answer supports our expectation from Charles\u2019s law, namely, that raising the gas temperature (from 283 K to 303 K) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L).<\/p>\r\n\r\n<h4 id=\"fs-idp28974272\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nA sample of oxygen, O<sub>2<\/sub>, occupies 32.2 mL at 30 \u00b0C and 452 torr. What volume will it occupy at \u201370 \u00b0C and the same pressure?\r\n<div id=\"fs-idm7218016\" data-type=\"note\">\r\n<div style=\"text-align: right;\" data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer: <\/strong>\u00a021.6 mL<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm51209248\" class=\"textbox shaded\" data-type=\"example\">\r\n<h3>Example 3<\/h3>\r\n<h4><strong>Measuring Temperature with a Volume Change<\/strong><\/h4>\r\nTemperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of 150.0 cm<sup>3<\/sup> when immersed in a mixture of ice and water (0.00 \u00b0C). When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is 131.7 cm<sup>3<\/sup>. Find the temperature of boiling ammonia on the kelvin and Celsius scales.\r\n<h4 id=\"fs-idm38946752\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nA volume change caused by a temperature change at constant pressure means we should use Charles\u2019s law. Taking <em data-effect=\"italics\">V<\/em><sub>1<\/sub> and <em data-effect=\"italics\">T<\/em><sub>1<\/sub> as the initial values, <em data-effect=\"italics\">T<\/em><sub>2<\/sub> as the temperature at which the volume is unknown and <em data-effect=\"italics\">V<\/em><sub>2<\/sub> as the unknown volume, and converting \u00b0C into K we have:\r\n<div id=\"fs-idp40102160\" data-type=\"equation\">[latex]\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}\\text{ which means that }\\frac{150.0{\\text{cm}}^{3}}{273.15\\text{K}}=\\frac{131.7{\\text{cm}}^{3}}{{T}_{2}}[\/latex]<\/div>\r\n<p id=\"fs-idp6920384\">Rearrangement gives [latex]{T}_{2}=\\frac{131.7{\\cancel{\\text{cm}}}^{3}\\times 273.15\\text{K}}{150.0{\\cancel{\\text{cm}}}^{3}}=239.8\\text{K}[\/latex]<\/p>\r\n<p id=\"fs-idm36267744\">Subtracting 273.15 from 239.8 K, we find that the temperature of the boiling ammonia on the Celsius scale is \u201333.4 \u00b0C.<\/p>\r\n\r\n<h4 id=\"fs-idm26636752\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nWhat is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm?\r\n<div id=\"fs-idm69251328\" data-type=\"note\">\r\n<div style=\"text-align: right;\" data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer: \u00a0<\/strong>635 mL<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-idm198797136\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Volume and Pressure: Boyle\u2019s Law<\/h2>\r\n<p id=\"fs-idm87156432\">If we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say 25 \u00b0C. If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases. This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in Figure 5.<\/p>\r\n\r\n\r\n[caption id=\"attachment_600\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-600\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205502\/CNX_Chem_09_03_BoylesLaw1-1024x911.jpg\" alt=\"This figure contains a diagram and two graphs. The diagram shows a syringe labeled with a scale in m l or c c with multiples of 5 labeled beginning at 5 and ending at 30. The markings halfway between these measurements are also provided. Attached at the top of the syringe is a pressure gauge with a scale marked by fives from 40 on the left to 5 on the right. The gauge needle rests between 10 and 15, slightly closer to 15. The syringe plunger position indicates a volume measurement about halfway between 10 and 15 m l or c c. The first graph is labeled \u201cV ( m L )\u201d on the horizontal axis and \u201cP ( p s i )\u201d on the vertical axis. Points are labeled at 5, 10, 15, 20, and 25 m L with corresponding values of 39.0, 19.5, 13.0, 9.8, and 6.5 p s i. The points are connected with a smooth curve that is declining at a decreasing rate of change. The second graph is labeled \u201cV ( m L )\u201d on the horizontal axis and \u201c1 divided by P ( p s i )\u201d on the vertical axis. The horizontal axis is labeled at multiples of 5, beginning at zero and extending up to 35 m L. The vertical axis is labeled by multiples of 0.02, beginning at 0 and extending up to 0.18. Six points indicated by black dots on this graph are connected with a black line segment showing a positive linear trend.\" width=\"1024\" height=\"911\" \/> Figure 5. When a gas occupies a smaller volume, it exerts a higher pressure; when it occupies a larger volume, it exerts a lower pressure (assuming the amount of gas and the temperature do not change). Since P and V are inversely proportional, a graph of 1\/P vs. V is linear.[\/caption]\r\n<p id=\"fs-idm298625584\">Unlike the <em data-effect=\"italics\">P<\/em>-<em data-effect=\"italics\">T<\/em> and <em data-effect=\"italics\">V<\/em>-<em data-effect=\"italics\">T<\/em> relationships, pressure and volume are not directly proportional to each other. Instead, <em data-effect=\"italics\">P<\/em> and <em data-effect=\"italics\">V<\/em> exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written:<\/p>\r\n\r\n<div data-type=\"equation\">[latex]P\\alpha 1\\text{\/}V\\text{ or }P=k\\cdot 1\\text{\/}V\\text{ or }P\\cdot V=k\\text{ or }{P}_{1}{V}_{1}={P}_{2}{V}_{2}[\/latex]<\/div>\r\n\r\n[caption id=\"attachment_601\" align=\"alignright\" width=\"300\"]<img class=\"wp-image-601\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205504\/CNX_Chem_09_02_Boyleslaw2.jpg\" alt=\"This diagram shows two graphs. In a, a graph is shown with volume on the horizontal axis and pressure on the vertical axis. A curved line is shown on the graph showing a decreasing trend with a decreasing rate of change. In b, a graph is shown with volume on the horizontal axis and one divided by pressure on the vertical axis. A line segment, beginning at the origin of the graph, shows a positive, linear trend.\" width=\"300\" height=\"153\" \/> Figure 6. The relationship between pressure and volume is inversely proportional. (a) The graph of P vs. V is a parabola, whereas (b) the graph of (1\/P) vs. V is linear.[\/caption]\r\n\r\nwith <em data-effect=\"italics\">k<\/em> being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure [latex]\\left(\\frac{1}{P}\\right)[\/latex] versus the volume (<em data-effect=\"italics\">V<\/em>), or the inverse of volume [latex]\\left(\\frac{1}{V}\\right)[\/latex] versus the pressure (<em data-effect=\"italics\">V<\/em>). Graphs with curved lines are difficult to read accurately at low or high values of the variables, and they are more difficult to use in fitting theoretical equations and parameters to experimental data. For those reasons, scientists often try to find a way to \u201clinearize\u201d their data. If we plot <em data-effect=\"italics\">P<\/em> versus <em data-effect=\"italics\">V<\/em>, we obtain a hyperbola (see Figure 6).\r\n<p id=\"fs-idm207692112\">The relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as <strong><span data-type=\"term\">Boyle\u2019s law<\/span><\/strong>: <em data-effect=\"italics\">The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured.<\/em><\/p>\r\n\r\n<div id=\"fs-idp10418864\" class=\"textbox shaded\" data-type=\"example\">\r\n<h3>Example 4<\/h3>\r\n<h4><strong>Volume of a Gas Sample<\/strong><\/h4>\r\n<p id=\"fs-idm205798208\">The sample of gas in Figure 5 has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL, using:<\/p>\r\n<p id=\"fs-idm127753872\">(a) the <em data-effect=\"italics\">P<\/em>-<em data-effect=\"italics\">V<\/em> graph in\u00a0Figure 5<\/p>\r\n<p id=\"fs-idm129344864\">(b) the [latex]\\frac{1}{P}[\/latex] vs. <em data-effect=\"italics\">V<\/em> graph in\u00a0Figure 5<\/p>\r\n<p id=\"fs-idm92820144\">(c) the Boyle\u2019s law equation<\/p>\r\n<p id=\"fs-idm215125536\">Comment on the likely accuracy of each method.<\/p>\r\n\r\n<h4 id=\"fs-idm126059440\"><span data-type=\"title\">Solution<\/span><\/h4>\r\n(a) Estimating from the <em data-effect=\"italics\">P<\/em>-<em data-effect=\"italics\">V<\/em> graph gives a value for <em data-effect=\"italics\">P<\/em> somewhere around 27 psi.\r\n<p id=\"fs-idm112322224\">(b) Estimating from the [latex]\\frac{1}{P}[\/latex] versus <em data-effect=\"italics\">V<\/em> graph give a value of about 26 psi.<\/p>\r\n<p id=\"fs-idm124214336\">(c) From Boyle\u2019s law, we know that the product of pressure and volume (<em data-effect=\"italics\">PV<\/em>) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have <em data-effect=\"italics\">P<\/em><sub>1<\/sub><em data-effect=\"italics\">V<\/em><sub>1<\/sub> = <em data-effect=\"italics\">k<\/em> and <em data-effect=\"italics\">P<\/em><sub>2<\/sub><em data-effect=\"italics\">V<\/em><sub>2<\/sub> = <em data-effect=\"italics\">k<\/em> which means that <em data-effect=\"italics\">P<\/em><sub>1<\/sub><em data-effect=\"italics\">V<\/em><sub>1<\/sub> = <em data-effect=\"italics\">P<\/em><sub>2<\/sub><em data-effect=\"italics\">V<\/em><sub>2<\/sub>.<\/p>\r\n<p id=\"fs-idm207111184\">Using <em data-effect=\"italics\">P<\/em><sub>1<\/sub> and <em data-effect=\"italics\">V<\/em><sub>1<\/sub> as the known values 0.993 atm and 2.40 mL, <em data-effect=\"italics\">P<\/em><sub>2<\/sub> as the pressure at which the volume is unknown, and <em data-effect=\"italics\">V<\/em><sub>2<\/sub> as the unknown volume, we have:<\/p>\r\n\r\n<div id=\"fs-idp70444944\" data-type=\"equation\">[latex]{P}_{1}{V}_{1}={P}_{2}{V}_{2}\\text{or}13.0\\text{psi}\\times 15.0\\text{mL}={P}_{2}\\times 7.5\\text{mL}[\/latex]<\/div>\r\n<p id=\"fs-idm167382656\">Solving:<\/p>\r\n\r\n<div id=\"fs-idp79285632\" data-type=\"equation\">[latex]{V}_{2}=\\frac{13.0\\text{psi}\\times 15.0\\cancel{\\text{mL}}}{7.5\\cancel{\\text{mL}}}=26\\text{mL}[\/latex]<\/div>\r\n<p id=\"fs-idm146893296\">It was more difficult to estimate well from the <em data-effect=\"italics\">P<\/em>-<em data-effect=\"italics\">V<\/em> graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow.<\/p>\r\n\r\n<h4 id=\"fs-idm111326864\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nThe sample of gas in Figure 5 has a volume of 30.0 mL at a pressure of 6.5 psi. Determine the volume of the gas at a pressure of 11.0 mL, using:\r\n<p id=\"fs-idm161358576\">(a) the <em data-effect=\"italics\">P<\/em>-<em data-effect=\"italics\">V<\/em> graph in\u00a0Figure 5<\/p>\r\n<p id=\"fs-idm130543696\">(b) the [latex]\\frac{1}{P}[\/latex] vs. <em data-effect=\"italics\">V<\/em> graph in\u00a0Figure 5<\/p>\r\n<p id=\"fs-idm208892896\">(c) the Boyle\u2019s law equation<\/p>\r\n<p id=\"fs-idm265464784\">Comment on the likely accuracy of each method.<\/p>\r\n\r\n<div id=\"fs-idp7933296\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:\u00a0<\/strong>(a) about 17\u201318 mL; (b) ~18 mL; (c) 17.7 mL; it was more difficult to estimate well from the <em data-effect=\"italics\">P<\/em>-<em data-effect=\"italics\">V<\/em> graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm202428608\" class=\"chemistry everyday-life\" data-type=\"note\">\r\n<div class=\"textbox shaded\" data-type=\"title\">\r\n<h3>Chemistry in Action:\u00a0Breathing and Boyle\u2019s Law<\/h3>\r\nWhat do you do about 20 times per minute for your whole life, without break, and often without even being aware of it? The answer, of course, is respiration, or breathing. How does it work? It turns out that the gas laws apply here. Your lungs take in gas that your body needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your chest cavity and making your lung volume larger. The increase in volume leads to a decrease in pressure (Boyle\u2019s law). This causes air to flow into the lungs (from high pressure to low pressure). When you exhale, the process reverses: Your diaphragm and rib muscles relax, your chest cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle\u2019s law again), and air flows out of the lungs (from high pressure to low pressure). You then breathe in and out again, and again, repeating this Boyle\u2019s law cycle for the rest of your life (Figure 7).\r\n\r\n[caption id=\"attachment_602\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-602\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205505\/CNX_Chem_09_02_BoylesLaw4-1024x769.jpg\" alt=\"This figure contains two diagrams of a cross section of the human head and torso. The first diagram on the left is labeled \u201cInspiration.\u201d It shows curved arrows in gray proceeding through the nasal passages and mouth to the lungs. An arrow points downward from the diaphragm, which is relatively flat, just beneath the lungs. This arrow is labeled \u201cDiaphragm contracts.\u201d At the entrance to the mouth and nasal passages, a label of P subscript lungs equals 1 dash 3 torr lower\u201d is provided. The second, similar diagram, which is labeled \u201cExpiration,\u201d reverses the direction of both arrows. Arrows extend from the lungs out through the nasal passages and mouth. Similarly, an arrow points up to the diaphragm, showing a curved diaphragm and lungs reduced in size from the previous image. This arrow is labeled \u201cDiaphragm relaxes.\u201d At the entrance to the mouth and nasal passages, a label of P subscript lungs equals 1 dash 3 torr higher\u201d is provided.\" width=\"1024\" height=\"769\" \/> Figure 7. Breathing occurs because expanding and contracting lung volume creates small pressure differences between your lungs and your surroundings, causing air to be drawn into and forced out of your lungs.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-idp47884080\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Moles of Gas and Volume: Avogadro\u2019s Law<\/h2>\r\n<p id=\"fs-idm52917120\">The Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed by <strong><span data-type=\"term\">Avogadro\u2019s law<\/span><\/strong>: <em data-effect=\"italics\">For a confined gas, the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both remain constant<\/em>.<\/p>\r\n<p id=\"fs-idp65662512\">In equation form, this is written as:<\/p>\r\n\r\n<div id=\"fs-idm8943936\" data-type=\"equation\">[latex]\\begin{array}{ccccc}V\\propto n&amp; \\text{or}&amp; V=k\\times n&amp; \\text{or}&amp; \\frac{{V}_{1}}{{n}_{1}}=\\frac{{V}_{2}}{{n}_{2}}.\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-idm94717808\">Mathematical relationships can also be determined for the other variable pairs, such as <em data-effect=\"italics\">P<\/em> versus <em data-effect=\"italics\">n<\/em>, and <em data-effect=\"italics\">n<\/em> versus <em data-effect=\"italics\">T<\/em>.<\/p>\r\n\r\n<div id=\"fs-idm50116432\" class=\"chemistry link-to-learning textbox\" data-type=\"note\">\r\n<p id=\"fs-idm91270672\">Visit this interactive <a href=\"http:\/\/phet.colorado.edu\/en\/simulation\/gas-properties\" target=\"_blank\">PhET simulation link to investigate the relationships between pressure, volume, temperature. and amount of gas<\/a>. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws).<\/p>\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-idm68358176\" data-depth=\"1\">\r\n<h2 data-type=\"title\">The Ideal Gas Law<\/h2>\r\n<p id=\"fs-idp161797856\">To this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas:<\/p>\r\n\r\n<ul id=\"fs-idp25403616\" data-bullet-style=\"bullet\">\r\n \t<li>Boyle\u2019s law: <em data-effect=\"italics\">PV<\/em> = constant at constant <em data-effect=\"italics\">T<\/em> and <em data-effect=\"italics\">n<\/em><\/li>\r\n \t<li>Amontons\u2019s law: [latex]\\frac{P}{T}[\/latex] = constant at constant <em data-effect=\"italics\">V<\/em> and <em data-effect=\"italics\">n<\/em><\/li>\r\n \t<li>Charles\u2019s law: [latex]\\frac{V}{T}[\/latex] = constant at constant <em data-effect=\"italics\">P<\/em> and <em data-effect=\"italics\">n<\/em><\/li>\r\n \t<li>Avogadro\u2019s law: [latex]\\frac{V}{n}[\/latex] = constant at constant <em data-effect=\"italics\">P<\/em> and <em data-effect=\"italics\">T<\/em><\/li>\r\n<\/ul>\r\n<p id=\"fs-idp47645536\">Combining these four laws yields the <strong><span data-type=\"term\">ideal gas law<\/span><\/strong>, a relation between the pressure, volume, temperature, and number of moles of a gas:<\/p>\r\n\r\n<div id=\"fs-idp190457648\" data-type=\"equation\">[latex]PV=nRT[\/latex]<\/div>\r\n<p id=\"fs-idp83793536\">where <em data-effect=\"italics\">P<\/em> is the pressure of a gas, <em data-effect=\"italics\">V<\/em> is its volume, <em data-effect=\"italics\">n<\/em> is the number of moles of the gas, <em data-effect=\"italics\">T<\/em> is its temperature on the kelvin scale, and <em data-effect=\"italics\">R<\/em> is a constant called the <strong><span data-type=\"term\">ideal gas constant<\/span><\/strong> or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being 0.08206 L atm mol<sup>\u20131<\/sup> K<sup>\u20131<\/sup> and 8.314 kPa L mol<sup>\u20131<\/sup> K<sup>\u20131<\/sup>.<\/p>\r\n<p id=\"fs-idp100115520\">Gases whose properties of <em data-effect=\"italics\">P<\/em>, <em data-effect=\"italics\">V<\/em>, and <em data-effect=\"italics\">T<\/em> are accurately described by the ideal gas law (or the other gas laws) are said to exhibit <em data-effect=\"italics\">ideal behavior<\/em> or to approximate the traits of an <strong><span data-type=\"term\">ideal gas<\/span><\/strong>. An ideal gas is a hypothetical construct that may be used along with <em data-effect=\"italics\">kinetic molecular theory<\/em> to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the <em data-effect=\"italics\">non-ideal<\/em> behavior observed for many gases at relatively high pressures and low temperatures.<\/p>\r\n<p id=\"fs-idp9263600\">The ideal gas equation contains five terms, the gas constant <em data-effect=\"italics\">R<\/em> and the variable properties <em data-effect=\"italics\">P<\/em>, <em data-effect=\"italics\">V<\/em>, <em data-effect=\"italics\">n<\/em>, and <em data-effect=\"italics\">T<\/em>. Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises.<\/p>\r\n\r\n<div id=\"fs-idm72969744\" class=\"textbox shaded\" data-type=\"example\">\r\n<h3>Example 5<\/h3>\r\n<h4><strong>Using the Ideal Gas Law<\/strong><\/h4>\r\nMethane, CH<sub>4<\/sub>, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of CH<sub>4<\/sub>. What is the volume of this much methane at 25 \u00b0C and 745 torr?\r\n<h4 id=\"fs-idp9667456\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nWe must rearrange <em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em> to solve for <em data-effect=\"italics\">V<\/em>: [latex]V=\\frac{nRT}{P}[\/latex]\r\n<p id=\"fs-idp84134864\">If we choose to use <em data-effect=\"italics\">R<\/em> = 0.08206 L atm mol<sup>\u20131<\/sup> K<sup>\u20131<\/sup>, then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.<\/p>\r\n<p id=\"fs-idm113078592\">Converting into the \u201cright\u201d units:<\/p>\r\n\r\n<div id=\"fs-idp39691024\" data-type=\"equation\">[latex]n=655\\text{g}\\cancel{{\\text{CH}}_{4}}\\times \\frac{1\\text{mol}}{16.043{\\cancel{\\text{g CH}}}_{4}}=40.8\\text{mol}[\/latex]<\/div>\r\n<div id=\"fs-idp101238448\" data-type=\"equation\">[latex]T=25\\text{\\textdegree C}+273=298\\text{K}[\/latex]<\/div>\r\n<div id=\"fs-idp11744096\" data-type=\"equation\">[latex]P=745\\cancel{\\text{torr}}\\times \\frac{1\\text{atm}}{760\\cancel{\\text{torr}}}=0.980\\text{atm}[\/latex]<\/div>\r\n<div id=\"fs-idp248570064\" data-type=\"equation\">[latex]V=\\frac{nRT}{P}=\\frac{\\left(40.8\\cancel{\\text{mol}}\\right)\\left(0.08206\\text{L}\\cancel{{\\text{atm mol}}^{-1}{\\text{K}}^{\\text{-1}}}\\right)\\left(298\\cancel{\\text{K}}\\right)}{0.980\\cancel{\\text{atm}}}=1.02\\times {10}^{3}\\text{L}[\/latex]<\/div>\r\n<p id=\"fs-idm139345632\">It would require 1020 L (269 gal) of gaseous methane at about 1 atm of pressure to replace 1 gal of gasoline. It requires a large container to hold enough methane at 1 atm to replace several gallons of gasoline.<\/p>\r\n\r\n<h4 id=\"fs-idp147095712\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nCalculate the pressure in bar of 2520 moles of hydrogen gas stored at 27 \u00b0C in the 180-L storage tank of a modern hydrogen-powered car.\r\n<div id=\"fs-idp77431312\" data-type=\"note\">\r\n<div style=\"text-align: right;\" data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer: \u00a0<\/strong>350 bar<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idp141407616\">If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: [latex]\\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\\frac{{P}_{2}{V}_{2}}{{T}_{2}}[\/latex] using units of atm, L, and K. Both sets of conditions are equal to the product of <em data-effect=\"italics\">n<\/em> \u00d7 <em data-effect=\"italics\">R<\/em> (where <em data-effect=\"italics\">n<\/em> = the number of moles of the gas and <em data-effect=\"italics\">R<\/em> is the ideal gas law constant).<\/p>\r\n\r\n<div id=\"fs-idp23616944\" class=\"textbox shaded\" data-type=\"example\">\r\n<h3>Example 6<\/h3>\r\n<h4 id=\"fs-idm24051840\"><strong><span data-type=\"title\">Using the Combined Gas Law<\/span><\/strong><\/h4>\r\n[caption id=\"attachment_2085\" align=\"alignright\" width=\"300\"]<img class=\"wp-image-2085\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212034\/CNX_Chem_09_02_Scuba1.jpg\" alt=\"This photograph shows a scuba diver underwater with a tank on his or her back and bubbles ascending from the breathing apparatus.\" width=\"300\" height=\"218\" \/> Figure 8. Scuba divers use compressed air to breathe while underwater. (credit: modification of work by Mark Goodchild)[\/caption]\r\n\r\nWhen filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure 8). If the water temperature is 27 \u00b0C, how many liters of air will such a tank provide to a diver\u2019s lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm?\r\n<p id=\"fs-idm78575136\">Letting <em data-effect=\"italics\">1<\/em> represent the air in the scuba tank and <em data-effect=\"italics\">2<\/em> represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is 37 \u00b0C, we have:<\/p>\r\n\r\n<div id=\"fs-idp224060000\" data-type=\"equation\">[latex]\\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\\frac{{P}_{2}{V}_{2}}{{T}_{2}}\\phantom{\\rule{0.4em}{0ex}}\\rightarrow\\phantom{\\rule{0.4em}{0ex}}\\frac{\\left(153\\text{atm}\\right)\\left(13.2\\text{L}\\right)}{\\left(300\\text{K}\\right)}=\\frac{\\left(3.13\\text{atm}\\right)\\left({V}_{2}\\right)}{\\left(310\\text{K}\\right)}[\/latex]<\/div>\r\n<p id=\"fs-idp72690672\">Solving for <em data-effect=\"italics\">V<\/em><sub>2<\/sub>:<\/p>\r\n\r\n<div id=\"fs-idp102261440\" data-type=\"equation\">[latex]{V}_{2}=\\frac{\\left(153\\cancel{\\text{atm}}\\right)\\left(13.2\\text{L}\\right)\\left(310\\text{K}\\right)}{\\left(300\\text{K}\\right)\\left(3.13\\cancel{\\text{atm}}\\right)}=667\\text{L}[\/latex]<\/div>\r\n<p id=\"fs-idp54320688\">(Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good \u201cballpark\u201d estimate.)<\/p>\r\n\r\n<h4 id=\"fs-idp128482944\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nA sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 \u00b0C and 0.850 atm. Find the volume of this sample at 0 \u00b0C and 1.00 atm.\r\n<div id=\"fs-idp104450512\" data-type=\"note\">\r\n<div style=\"text-align: right;\" data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer: \u00a0<\/strong>0.538 L<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp8804032\" class=\"chemistry everyday-life\" data-type=\"note\">\r\n<div class=\"textbox shaded\" data-type=\"title\">\r\n<h3 data-type=\"title\">The Interdependence between Ocean Depth and Pressure in Scuba Diving<\/h3>\r\n[caption id=\"attachment_2086\" align=\"alignright\" width=\"300\"]<img class=\"wp-image-2086\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212035\/CNX_Chem_09_02_GreatBarri1.jpg\" alt=\"This picture shows colorful underwater corals and anemones in hues of yellow, orange, green, and brown, surrounded by water that appears blue in color.\" width=\"300\" height=\"225\" \/> Figure 9. Scuba divers, whether at the Great Barrier Reef or in the Caribbean, must be aware of buoyancy, pressure equalization, and the amount of time they spend underwater, to avoid the risks associated with pressurized gases in the body. (credit: Kyle Taylor)[\/caption]\r\n\r\nWhether scuba diving at the Great Barrier Reef in Australia (shown in Figure 9) or in the Caribbean, divers must understand how pressure affects a number of issues related to their comfort and safety.\r\n\r\nPressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as \u201catmospheres absolute\u201d or ATA in the diving community:\u00a0Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level.\r\n\r\nAs a diver descends, the increase in pressure causes the body\u2019s air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization.\r\n\r\nBuoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in his BCD expands because of lower pressure according to Boyle\u2019s law (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and she or he begins to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor.\r\n\r\nThe pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast as at the surface.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-idp117090992\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Standard Conditions of Temperature and Pressure<\/h2>\r\n<p id=\"fs-idp92943776\">We have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a <strong><span data-type=\"term\">standard temperature and pressure (STP)<\/span> <\/strong>for reporting properties of gases: 273.15 K and 1 atm (101.325 kPa). At STP, an ideal gas has a volume of about 22.4 L\u2014this is referred to as the <strong><span data-type=\"term\">standard molar volume<\/span><\/strong> (Figure 10).<\/p>\r\n\r\n\r\n[caption id=\"attachment_2087\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-2087\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212037\/CNX_Chem_09_02_HENH3O21-1024x590.jpg\" alt=\"This figure shows three balloons each filled with H e, N H subscript 2, and O subscript 2 respectively. Beneath the first balloon is the label \u201c4 g of He\u201d Beneath the second balloon is the label, \u201c15 g of N H subscript 2.\u201d Beneath the third balloon is the label \u201c32 g of O subscript 2.\u201d Each balloon contains the same number of molecules of their respective gases.\" width=\"1024\" height=\"590\" \/> Figure 10. Since the number of moles in a given volume of gas varies with pressure and temperature changes, chemists use standard temperature and pressure (273.15 K and 1 atm or 101.325 kPa) to report properties of gases.[\/caption]\r\n\r\n<\/section><section id=\"fs-idp68486960\" class=\"summary\" data-depth=\"1\">\r\n<div class=\"bcc-box bcc-success\">\r\n<h2>Key Concepts and Summary<\/h2>\r\n<section>\r\n<div data-type=\"note\">\r\n\r\nThe behavior of gases can be described by several laws based on experimental observations of their properties. The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons\u2019s law). The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles\u2019s law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle\u2019s law). Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro\u2019s law).\r\n<p id=\"fs-idp115685936\">The equations describing these laws are special cases of the ideal gas law, <em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em>, where <em data-effect=\"italics\">P<\/em> is the pressure of the gas, <em data-effect=\"italics\">V<\/em> is its volume, <em data-effect=\"italics\">n<\/em> is the number of moles of the gas, <em data-effect=\"italics\">T<\/em> is its kelvin temperature, and <em data-effect=\"italics\">R<\/em> is the ideal (universal) gas constant.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Key Equations<\/h3>\r\n<ul id=\"fs-idp89091632\" data-bullet-style=\"bullet\">\r\n \t<li><em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em><\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Chemistry End of Chapter Exercises<\/h3>\r\n<section id=\"self-check-questions\">\r\n<ol>\r\n \t<li data-type=\"title\">Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?<\/li>\r\n \t<li data-type=\"title\">Explain how the volume of the bubbles exhausted by a scuba diver (Figure 8) change as they rise to the surface, assuming that they remain intact.<\/li>\r\n \t<li data-type=\"title\">One way to state Boyle\u2019s law is \u201cAll other things being equal, the pressure of a gas is inversely proportional to its volume.\u201d\r\n<ol>\r\n \t<li data-type=\"title\">What is the meaning of the term \u201cinversely proportional?\u201d<\/li>\r\n \t<li data-type=\"title\">What are the \u201cother things\u201d that must be equal?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li data-type=\"title\">An alternate way to state Avogadro\u2019s law is \u201cAll other things being equal, the number of molecules in a gas is directly proportional to the volume of the gas.\u201d\r\n<ol>\r\n \t<li data-type=\"title\">What is the meaning of the term \u201cdirectly proportional?\u201d<\/li>\r\n \t<li data-type=\"title\">What are the \u201cother things\u201d that must be equal?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li data-type=\"title\">How would the graph in Figure 4 change if the number of moles of gas in the sample used to determine the curve were doubled?<\/li>\r\n \t<li data-type=\"title\">How would the graph in Figure 5 change if the number of moles of gas in the sample used to determine the curve were doubled?<\/li>\r\n \t<li data-type=\"title\">In addition to the data found in Figure 5, what other information do we need to find the mass of the sample of air used to determine the graph?<\/li>\r\n \t<li data-type=\"title\">Determine the volume of 1 mol of CH<sub>4<\/sub> gas at 150 K and 1 atm, using Figure 4.<\/li>\r\n \t<li data-type=\"title\">Determine the pressure of the gas in the syringe shown in Figure 5 when its volume is 12.5 mL, using:\r\n<ol>\r\n \t<li data-type=\"title\">the appropriate graph<\/li>\r\n \t<li data-type=\"title\">Boyle\u2019s law<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li data-type=\"title\">A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 \u00b0C. If the can is thrown into a fire (T = 475 \u00b0C), what will be the pressure in the hot can?<\/li>\r\n \t<li data-type=\"title\">What is the temperature of an 11.2-L sample of carbon monoxide, CO, at 744 torr if it occupies 13.3 L at 55 \u00b0C and 744 torr?<\/li>\r\n \t<li data-type=\"title\">A 2.50-L volume of hydrogen measured at \u2013196 \u00b0C is warmed to 100 \u00b0C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.<\/li>\r\n \t<li data-type=\"title\">A balloon inflated with three breaths of air has a volume of 1.7 L. At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?<img class=\"alignright wp-image-2088\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212038\/CNX_Chem_09_02_WeatherBall_img1.jpg\" alt=\"This image shows a white balloon that appears to have an attached white card. The balloon is being held by a person in an outdoor setting.\" width=\"300\" height=\"225\" \/><\/li>\r\n \t<li data-type=\"title\">A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25 \u00b0C at ground level. What is the volume of the balloon under these conditions?<\/li>\r\n \t<li data-type=\"title\">The volume of an automobile air bag was 66.8 L when inflated at 25 \u00b0C with 77.8 g of nitrogen gas. What was the pressure in the bag in kPa?<\/li>\r\n \t<li data-type=\"title\">How many moles of gaseous boron trifluoride, BF<sub>3<\/sub>, are contained in a 4.3410-L bulb at 788.0 K if the pressure is 1.220 atm? How many grams of BF<sub>3<\/sub>?<\/li>\r\n \t<li data-type=\"title\">Iodine, I<sub>2<\/sub>, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a 73.3-mL bulb that contains 0.292 g of I<sub>2<\/sub> vapor at a pressure of 0.462 atm?<\/li>\r\n \t<li data-type=\"title\">How many grams of gas are present in each of the following cases?\r\n<ol>\r\n \t<li data-type=\"title\">0.100 L of CO<sub>2<\/sub> at 307 torr and 26 \u00b0C<\/li>\r\n \t<li data-type=\"title\">8.75 L of C<sub>2<\/sub>H<sub>4<\/sub>, at 378.3 kPa and 483 K<\/li>\r\n \t<li data-type=\"title\">221 mL of Ar at 0.23 torr and \u201354 \u00b0C<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li data-type=\"title\">A high altitude balloon is filled with 1.41 \u00d7 10<sup>4<\/sup> L of hydrogen at a temperature of 21 \u00b0C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is \u201348 \u00b0C and the pressure is 63.1 torr?<\/li>\r\n \t<li data-type=\"title\">A cylinder of medical oxygen has a volume of 35.4 L, and contains O<sub>2<\/sub> at a pressure of 151 atm and a temperature of 25 \u00b0C. What volume of O<sub>2<\/sub> does this correspond to at normal body conditions, that is, 1 atm and 37 \u00b0C?<\/li>\r\n \t<li data-type=\"title\">A large scuba tank (Figure 8) with a volume of 18 L is rated for a pressure of 220 bar. The tank is filled at 20 \u00b0C and contains enough air to supply 1860 L of air to a diver at a pressure of 2.37 atm (a depth of 45 feet). Was the tank filled to capacity at 20 \u00b0C?<\/li>\r\n \t<li data-type=\"title\">A 20.0-L cylinder containing 11.34 kg of butane, C<sub>4<\/sub>H<sub>10<\/sub>, was opened to the atmosphere. Calculate the mass of the gas remaining in the cylinder if it were opened and the gas escaped until the pressure in the cylinder was equal to the atmospheric pressure, 0.983 atm, and a temperature of 27 \u00b0C.<\/li>\r\n \t<li data-type=\"title\">While resting, the average 70-kg human male consumes 14 L of pure O<sub>2<\/sub> per hour at 25 \u00b0C and 100 kPa. How many moles of O<sub>2<\/sub> are consumed by a 70 kg man while resting for 1.0 h?<\/li>\r\n \t<li data-type=\"title\">For a given amount of gas showing ideal behavior, draw labeled graphs of:\r\n<ol>\r\n \t<li data-type=\"title\">the variation of <em data-effect=\"italics\">P<\/em> with <em data-effect=\"italics\">V\u00a0<\/em><\/li>\r\n \t<li data-type=\"title\">the variation of <em data-effect=\"italics\">V<\/em> with <em data-effect=\"italics\">T\u00a0<\/em><\/li>\r\n \t<li data-type=\"title\">the variation of <em data-effect=\"italics\">P<\/em> with <em data-effect=\"italics\">T\u00a0<\/em><\/li>\r\n \t<li data-type=\"title\">the variation of [latex]\\frac{1}{P}[\/latex] with <em data-effect=\"italics\">V<\/em><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li data-type=\"title\">A liter of methane gas, CH<sub>4<\/sub>, at STP contains more atoms of hydrogen than does a liter of pure hydrogen gas, H<sub>2<\/sub>, at STP. Using Avogadro\u2019s law as a starting point, explain why.<\/li>\r\n \t<li data-type=\"title\">The effect of chlorofluorocarbons (such as CCl<sub>2<\/sub>F<sub>2<\/sub>) on the depletion of the ozone layer is well known. The use of substitutes, such as CH<sub>3<\/sub>CH<sub>2<\/sub>F(<em data-effect=\"italics\">g<\/em>), for the chlorofluorocarbons, has largely corrected the problem. Calculate the volume occupied by 10.0 g of each of these compounds at STP:\r\n<ol>\r\n \t<li data-type=\"title\">CCl<sub>2<\/sub>F<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>)<\/li>\r\n \t<li data-type=\"title\">CH<sub>3<\/sub>CH<sub>2<\/sub>F(<em data-effect=\"italics\">g<\/em>)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li data-type=\"title\">As 1 g of the radioactive element radium decays over 1 year, it produces 1.16 \u00d7 10<sup>18<\/sup> alpha particles (helium nuclei). Each alpha particle becomes an atom of helium gas. What is the pressure in pascal of the helium gas produced if it occupies a volume of 125 mL at a temperature of 25 \u00b0C?<\/li>\r\n \t<li data-type=\"title\">A balloon that is 100.21 L at 21 \u00b0C and 0.981 atm is released and just barely clears the top of Mount Crumpet in British Columbia. If the final volume of the balloon is 144.53 L at a temperature of 5.24 \u00b0C, what is the pressure experienced by the balloon as it clears Mount Crumpet?<\/li>\r\n \t<li data-type=\"title\">If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?<\/li>\r\n \t<li data-type=\"title\">If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?<\/li>\r\n<\/ol>\r\n<\/section><\/div>\r\n<div data-type=\"solution\">\r\n<div class=\"bcc-box bcc-info\"><section id=\"self-check-answers\">\r\n<div data-type=\"exercise\">\r\n<h4>Selected Answers<\/h4>\r\n<div id=\"fs-idp16137760\" data-type=\"exercise\">\r\n<p id=\"fs-idm158859536\" data-type=\"solution\">2. As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle\u2019s law.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idm244332560\" data-type=\"solution\">4. (a) The number of particles in the gas increases as the volume increases. This relationship may be written as <em data-effect=\"italics\">n<\/em> = constant \u00d7 <em data-effect=\"italics\">V<\/em>. It is a direct relationship.(b) The temperature and pressure must be kept constant.<\/div>\r\n<div id=\"fs-idm181098208\" data-type=\"solution\">6. The curve would be farther to the right and higher up, but the same basic shape.<\/div>\r\n<div id=\"fs-idm207209024\" data-type=\"solution\">8. The figure shows the change of 1 mol of CH<sub>4<\/sub> gas as a function of temperature. The graph shows that the volume is about 16.3 to 16.5 L.<\/div>\r\n<div id=\"fs-idm211017328\" data-type=\"exercise\">\r\n<div id=\"fs-idm121334432\" data-type=\"solution\">\r\n\r\n10. The first thing to recognize about this problem is that the volume and moles of gas remain constant. Thus, we can use the combined gas law equation in the form:\r\n<p data-type=\"newline\">\u00a0[latex]\\frac{{P}_{2}}{{T}_{2}}=\\frac{{P}_{1}}{{T1}_{}}[\/latex]<\/p>\r\n[latex]{P}_{2}=\\frac{{P}_{1}{T}_{2}}{{T}_{1}}=1344\\text{torr}\\times \\frac{475+273.15}{23+273.15}=3.40\\times {10}^{3}\\text{torr}[\/latex]\r\n<div id=\"fs-idm221278704\" data-type=\"exercise\">\r\n<div id=\"fs-idm213171488\" data-type=\"solution\">\r\n\r\n12. Apply Charles\u2019s law to compute the volume of gas at the higher temperature:\r\n<p data-type=\"newline\">\u00a0<em data-effect=\"italics\">V<\/em><sub>1<\/sub> = 2.50 L<\/p>\r\n<p data-type=\"newline\">\u00a0<em data-effect=\"italics\">T<\/em><sub>1<\/sub> = \u2013193 \u00b0C = 77.15 K<\/p>\r\n<p data-type=\"newline\">\u00a0<em data-effect=\"italics\">V<\/em><sub>2<\/sub> = ?<\/p>\r\n<p data-type=\"newline\">\u00a0<em data-effect=\"italics\">T<\/em><sub>2<\/sub> = 100 \u00b0C = 373.15 K<\/p>\r\n\r\n<div data-type=\"newline\">\r\n<div id=\"fs-idm213171488\" data-type=\"solution\">\r\n\r\n[latex]\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}[\/latex]\r\n<div data-type=\"newline\">\u00a0[latex]{V}_{2}=\\frac{{V}_{1}{T}_{2}}{{T}_{1}}=\\frac{2.50\\text{L}\\times 373.15\\cancel{\\text{K}}}{77.15\\cancel{\\text{K}}}=12.1\\text{L}[\/latex]<\/div>\r\n14. \u00a0<em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em>\r\n\r\n<\/div>\r\n<div id=\"fs-idm24946256\" data-type=\"exercise\">\r\n<div id=\"fs-idm78551216\" data-type=\"solution\">\r\n\r\n[latex]V=\\frac{nRT}{P}=\\frac{8.80\\cancel{\\text{mol}}\\times 0.08206\\text{L}\\cancel{\\text{atm}}{\\cancel{\\text{mol}}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\times 298.15\\cancel{\\text{K}}}{0.992\\cancel{\\text{atm}}}=217\\text{L}[\/latex]\r\n<div id=\"fs-idm130104720\" data-type=\"problem\">16. [latex]n=\\frac{PV}{RT}\\frac{1.220\\cancel{\\text{atm}}\\left(4.3410\\text{L}\\right)}{\\left(0.08206\\text{L}\\cancel{\\text{atm}}\\text{mol}{\\text{-1}}^{}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(788.0\\cancel{\\text{K}}\\right)}=0.08190\\text{mol}=8.190\\times {10}^{\\text{-2}}\\text{mol}[\/latex]<\/div>\r\n<div id=\"fs-idm95261248\" data-type=\"solution\">\r\n<div data-type=\"newline\">\u00a0[latex]n\\times \\text{molar mass}=8.190\\times {10}^{\\text{-2}}\\cancel{\\text{mol}}\\times 67.8052\\text{g}{\\cancel{\\text{mol}}}^{\\text{-1}}=5.553\\text{g}[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-idm171506192\" data-type=\"exercise\">\r\n<div id=\"fs-idm254882416\" data-type=\"solution\">\r\n<div data-type=\"newline\">\r\n\r\n18. In each of these problems, we are given a volume, pressure, and temperature. We can obtain moles from this information using the molar mass, <em data-effect=\"italics\">m<\/em> = <em data-effect=\"italics\">n\u2133<\/em>, where \u2133 is the molar mass:\r\n<p data-type=\"newline\">[latex]P,V,T\\stackrel{n=PV\\text{\/}RT}{\\to }n,\\stackrel{m=n\\left(\\text{molar mass}\\right)}{\\to }\\text{grams}[\/latex]<\/p>\r\n<p data-type=\"newline\">or we can combine these equations to obtain:<\/p>\r\n<p data-type=\"newline\">[latex]\\text{mass}=m=\\frac{PV}{RT}\\times [\/latex]\u2133<\/p>\r\n(a) \u00a0[latex]\\begin{array}{l}307\\cancel{\\text{torr}}\\times \\frac{1\\text{atm}}{760\\cancel{\\text{torr}}}=0.4039\\text{atm}26\\textdegree C=299.1 K\\\\ \\text{Mass}=m=\\frac{0.4039\\cancel{\\text{atm}}\\left(0.100\\cancel{\\text{L}}\\right)}{0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\left(299.1\\cancel{\\text{K}}\\right)}\\times 44.01\\text{g}{\\text{mol}}^{\\text{-1}}=7.24\\times {10}^{\\text{-2}}\\text{g};\\end{array}[\/latex]\r\n\r\n(b)[latex]\\text{Mass}=m=\\frac{378.3\\cancel{\\text{kPa}}\\left(8.75\\cancel{\\text{L}}\\right)}{8.314\\cancel{\\text{L}}\\cancel{\\text{kPa}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\left(483\\cancel{\\text{K}}\\right)}\\times 28.05376\\text{g}{\\text{mol}}^{\\text{-1}}=23.1\\text{g;}[\/latex]\r\n<div id=\"fs-idm33816112\" data-type=\"solution\">(c) \u00a0[latex]\\begin{array}{l}\\\\ 221\\cancel{\\text{mL}}\\times \\frac{1\\text{L}}{1000\\cancel{\\text{mL}}}=0.221\\text{L}-54\\textdegree C+273.15=219.15\\text{K}\\\\ 0.23\\cancel{\\text{torr}}\\times \\frac{1\\text{atm}}{760\\cancel{\\text{torr}}}=3.03\\times {10}^{\\text{-4}}\\text{atm}\\\\ \\text{Mass}=m=\\frac{3.03\\times {10}^{\\text{-4}}\\cancel{\\text{atm}}\\left(0.221\\cancel{\\text{L}}\\right)}{0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\left(219.15\\cancel{\\text{K}}\\right)}\\times 39.978\\text{g}{\\text{mol}}^{\\text{-1}}=1.5\\times {10}^{\\text{-4}}\\text{g}\\end{array}[\/latex]<\/div>\r\n<div id=\"fs-idm87206720\" data-type=\"exercise\">\r\n<div id=\"fs-idm171827136\" data-type=\"solution\">\r\n<div data-type=\"newline\">\r\n<div id=\"fs-idm33816112\" data-type=\"solution\">20. \u00a0[latex]\\frac{{P}_{2}}{{T}_{2}}=\\frac{{P}_{1}}{{T}_{1}}[\/latex]<\/div>\r\n<div id=\"fs-idm87356096\" data-type=\"solution\">\r\n<div data-type=\"newline\">\u00a0<em data-effect=\"italics\">T<\/em><sub>2<\/sub> = 49.5 + 273.15 = 322.65 K<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]{P}_{2}=\\frac{{P}_{1}{T}_{2}}{{T}_{1}}=149.6\\text{atm}\\times \\frac{322.65}{278.15}=173.5\\text{atm}[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-idm247077296\" data-type=\"exercise\">\r\n<div id=\"fs-idm237039312\" data-type=\"solution\">\r\n\r\n22. Calculate the amount of butane in 20.0 L at 0.983 atm and 27\u00b0C. The original amount in the container does not matter. [latex]n=\\frac{PV}{RT}=\\frac{0.983\\cancel{\\text{atm}}\\times 20.0\\cancel{\\text{L}}}{0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\left(300.1\\cancel{\\text{K}}\\right)}=0.798\\text{mol}[\/latex]\u00a0Mass of butane = 0.798 mol \u00d7 58.1234 g\/mol = 46.4 g\r\n\r\n24.\u00a0For a gas exhibiting ideal behavior:\r\n\r\n<img class=\"aligncenter size-large wp-image-5190\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/08\/23214537\/CNX_Chem_09_02_Exercise25_img-1024x868.jpg\" alt=\"Four graphs are shown. In a, Volume is on the horizontal axis and Pressure is on the vertical axis. A downward trend with a decreasing rate of change is shown by a curved line. The label n, P cons is shown on the graph. In b, Temperature is on the horizontal axis and Volume is on the vertical axis. An increasing linear trend is shown by a straight line segment. The label n, P cons is shown on the graph. In c, Temperature is on the horizontal axis and Pressure is on the vertical axis. An increasing linear trend is shown by a straight line segment. The label n, P cons is shown on the graph. In d, Volume is on the horizontal axis and 1 divided by Pressure is on the vertical axis. An increasing linear trend is shown by a straight line segment on the graph. The label n, P cons is shown on the graph.\" width=\"1024\" height=\"868\" \/>\r\n<div data-type=\"newline\">\r\n<div id=\"fs-idp18583872\" data-type=\"solution\">26. (a) Determine the molar mass of CCl<sub>2<\/sub>F<sub>2<\/sub> then calculate the moles of CCl<sub>2<\/sub>F<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) present. Use the ideal gas law <em data-effect=\"italics\">PV = nRT<\/em> to calculate the volume of CCl<sub>2<\/sub>F<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>):<\/div>\r\n<div id=\"fs-idm286430688\" data-type=\"solution\">\r\n<div data-type=\"newline\">\u00a0[latex]\\text{10.0 g}{\\text{CCl}}_{2}{\\text{F}}_{2}\\times \\frac{1\\text{mol}{\\text{CC1}}_{2}{\\text{F}}_{2}}{120.91\\text{g}{\\text{CCl}}_{2}{\\text{F}}_{2}}=0.0827\\text{mol}{\\text{CCl}}_{2}{\\text{F}}_{2}[\/latex]<\/div>\r\n<div data-type=\"newline\">\u00a0<em data-effect=\"italics\">PV = nRT<\/em>, where <em data-effect=\"italics\">n<\/em> = # mol CCl<sub>2<\/sub>F<sub>2<\/sub><\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">[latex]1\\text{atm}\\times V=0.0827\\text{mol}\\times \\frac{0.0821\\text{L atm}}{\\text{mol K}}\\times 273\\text{K}=1.85\\text{L}{\\text{CCl}}_{2}{\\text{F}}_{2};[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n(b) [latex]10.0\\text{g}{\\text{CH}}_{3}{\\text{CH}}_{2}\\text{F}\\times \\frac{1\\text{mol}{\\text{CH}}_{3}{\\text{CH}}_{2}\\text{F}}{48.07{\\text{g CH}}_{3}{\\text{CH}}_{2}\\text{F}}=0.208\\text{mol}{\\text{CH}}_{3}{\\text{CH}}_{2}\\text{F}[\/latex]\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\"><em data-effect=\"italics\">PV = nRT<\/em>, with <em data-effect=\"italics\">n<\/em> = # mol CH<sub>3<\/sub>CH<sub>2<\/sub>F<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">1 atm \u00d7 <em data-effect=\"italics\">V<\/em> = 0.208 mol \u00d7 0.0821 L atm\/mol K \u00d7 273 K = 4.66 L CH<sub>3<\/sub> CH<sub>2<\/sub> F<\/div>\r\n<\/div>\r\n<div id=\"fs-idm26646416\" data-type=\"exercise\">\r\n<div id=\"fs-idm15239616\" data-type=\"solution\">\r\n\r\n28. Identify the variables in the problem and determine that the combined gas law [latex]\\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\\frac{{P}_{2}{V}_{2}}{{T}_{2}}[\/latex] is the necessary equation to use to solve the problem. Then solve for P<sub>2<\/sub>:\r\n<div data-type=\"newline\">\r\n<div id=\"fs-idm23005472\" data-type=\"exercise\">\r\n<div id=\"fs-idm213823792\" data-type=\"solution\">\r\n<p data-type=\"newline\">\u00a0[latex]\\begin{array}{l}\\\\ \\\\ \\frac{0.981\\text{atm}\\times 100.21\\text{L}}{294\\text{K}}=\\frac{{P}_{2}\\times 144.53\\text{L}}{278.24\\text{atm}}\\\\ {P}_{2}=0.644\\text{atm}\\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm136587680\" data-type=\"exercise\">\r\n<div id=\"fs-idm232238480\" data-type=\"solution\">\r\n<p id=\"fs-idm232238224\">30. The pressure decreases by a factor of 3.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section>\r\n<div data-type=\"glossary\">\r\n<div class=\"bcc-box bcc-success\"><section id=\"glossary\">\r\n<h3>Glossary<\/h3>\r\n<div data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">absolute zero\r\n<\/span><\/strong>temperature at which the volume of a gas would be zero according to Charles\u2019s law.\r\n<div id=\"fs-idp77543648\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">Amontons\u2019s law\r\n<\/span><\/strong>(also, Gay-Lussac\u2019s law) pressure of a given number of moles of gas is directly proportional to its kelvin temperature when the volume is held constant\r\n\r\n<\/div>\r\n<div id=\"fs-idm85214288\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">Avogadro\u2019s law\r\n<\/span><\/strong>volume of a gas at constant temperature and pressure is proportional to the number of gas molecules\r\n\r\n<\/div>\r\n<div id=\"fs-idm51295520\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">Boyle\u2019s law\r\n<\/span><\/strong>volume of a given number of moles of gas held at constant temperature is inversely proportional to the pressure under which it is measured\r\n\r\n<\/div>\r\n<div id=\"fs-idm100735488\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">Charles\u2019s law\r\n<\/span><\/strong>volume of a given number of moles of gas is directly proportional to its kelvin temperature when the pressure is held constant\r\n\r\n<\/div>\r\n<div id=\"fs-idm161542192\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">ideal gas\r\n<\/span><\/strong>hypothetical gas whose physical properties are perfectly described by the gas laws\r\n\r\n<\/div>\r\n<div id=\"fs-idm78852976\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">ideal gas constant (<em data-effect=\"italics\">R<\/em>)\r\n<\/span><\/strong>constant derived from the ideal gas equation <em data-effect=\"italics\">R<\/em> = 0.08226 L atm mol<sup>\u20131<\/sup> K<sup>\u20131<\/sup> or 8.314 L kPa mol<sup>\u20131<\/sup> K<sup>\u20131<\/sup>\r\n\r\n<\/div>\r\n<div id=\"fs-idm111878496\" data-type=\"definition\"><span data-type=\"term\"><strong>ideal gas law\r\n<\/strong><\/span>relation between the pressure, volume, amount, and temperature of a gas under conditions derived by combination of the simple gas laws<\/div>\r\n<div id=\"fs-idm158190720\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">standard conditions of temperature and pressure (STP)\r\n<\/span><\/strong>273.15 K (0 \u00b0C) and 1 atm (101.325 kPa)\r\n\r\n<\/div>\r\n<div id=\"fs-idm62647600\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">standard molar volume\r\n<\/span><\/strong>volume of 1 mole of gas at STP, approximately 22.4 L for gases behaving ideally\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>LEARNING OBJECTIVES<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Identify the mathematical relationships between the various properties of gases<\/li>\n<li>Use the ideal gas law, and related gas laws, to compute the values of various gas properties under specified conditions<\/li>\n<\/ul>\n<\/div>\n<p>During the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could fly (Figure 1), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, pressure, volume, temperature, and amount of gas. Although their measurements were not precise by today\u2019s standards, they were able to determine the mathematical relationships between pairs of these variables (e.g., pressure and temperature, pressure and volume) that hold for an <em data-effect=\"italics\">ideal<\/em> gas\u2014a hypothetical construct that real gases approximate under certain conditions. Eventually, these individual laws were combined into a single equation\u2014the <em data-effect=\"italics\">ideal gas law<\/em>\u2014that relates gas quantities for gases and is quite accurate for low pressures and moderate temperatures. We will consider the key developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas law.<\/p>\n<div id=\"attachment_595\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-595\" class=\"size-large wp-image-595\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205454\/CNX_Chem_09_02_Ballooning-1024x279.jpg\" alt=\"This figure includes three images. Image a is a black and white image of a hydrogen balloon apparently being deflated by a mob of people. In image b, a blue, gold, and red balloon is being held to the ground with ropes while positioned above a platform from which smoke is rising beneath the balloon. In c, an image is shown in grey on a peach-colored background of an inflated balloon with vertical striping in the air. It appears to have a basket attached to its lower side. A large stately building appears in the background.\" width=\"1024\" height=\"279\" \/><\/p>\n<p id=\"caption-attachment-595\" class=\"wp-caption-text\">Figure 1. In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.<\/p>\n<\/div>\n<section id=\"fs-idm39174352\" data-depth=\"1\">\n<h2 data-type=\"title\">Pressure and Temperature: Amontons\u2019s Law<\/h2>\n<p id=\"fs-idm131642544\">Imagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (Figure 2) and the pressure increases.<\/p>\n<div id=\"attachment_596\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-596\" class=\"size-large wp-image-596\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205456\/CNX_Chem_09_01_Amontons1-1024x489.jpg\" alt=\"This figure includes three similar diagrams. In the first diagram to the left, a rigid spherical container of a gas to which a pressure gauge is attached at the top is placed in a large beaker of water, indicated in light blue, atop a hot plate. The needle on the pressure gauge points to the far left on the gauge. The diagram is labeled \u201clow P\u201d above and \u201chot plate off\u201d below. The second similar diagram also has the rigid spherical container of gas placed in a large beaker from which light blue wavy line segments extend from the top of the liquid in the beaker. The beaker is situated on top of a slightly reddened circular area. The needle on the pressure gauge points straight up, or to the middle on the gauge. The diagram is labeled \u201cmedium P\u201d above and \u201chot plate on medium\u201d below. The third diagram also has the rigid spherical container of gas placed in a large beaker in which bubbles appear near the liquid surface and several wavy light blue line segments extend from the surface out of the beaker. The beaker is situated on top of a bright red circular area. The needle on the pressure gauge points to the far right on the gauge. The diagram is labeled \u201chigh P\u201d above and \u201chot plate on high\u201d below.\" width=\"1024\" height=\"489\" \/><\/p>\n<p id=\"caption-attachment-596\" class=\"wp-caption-text\">Figure 2. The effect of temperature on gas pressure: When the hot plate is off, the pressure of the gas in the sphere is relatively low. As the gas is heated, the pressure of the gas in the sphere increases.<\/p>\n<\/div>\n<p id=\"fs-idm111816736\">This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in Figure 3. We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then <em data-effect=\"italics\">P<\/em> and <em data-effect=\"italics\">T<\/em> are directly proportional (again, when <em data-effect=\"italics\">volume and moles of gas are held constant<\/em>); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor.<\/p>\n<div id=\"attachment_597\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-597\" class=\"size-large wp-image-597\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205458\/CNX_Chem_09_02_Amontons2-1024x329.jpg\" alt=\"This figure includes a table and a graph. The table has 3 columns and 7 rows. The first row is a header, which labels the columns \u201cTemperature, degrees C,\u201d \u201cTemperature, K,\u201d and \u201cPressure, k P a.\u201d The first column contains the values from top to bottom negative 150, negative 100, negative 50, 0, 50, and 100. The second column contains the values from top to bottom 173, 223, 273, 323, 373, and 423. The third column contains the values 36.0, 46.4, 56.7, 67.1, 77.5, and 88.0. A graph appears to the right of the table. The horizontal axis is labeled \u201cTemperature ( K ).\u201d with markings and labels provided for multiples of 100 beginning at 0 and ending at 500. The vertical axis is labeled \u201cPressure ( k P a )\u201d with markings and labels provided for multiples of 10, beginning at 0 and ending at 100. Six data points from the table are plotted on the graph with black dots. These dots are connected with a solid black line. A dashed line extends from the data point furthest to the left to the origin. The graph shows a positive linear trend.\" width=\"1024\" height=\"329\" \/><\/p>\n<p id=\"caption-attachment-597\" class=\"wp-caption-text\">Figure 3. For a constant volume and amount of air, the pressure and temperature are directly proportional, provided the temperature is in kelvin. (Measurements cannot be made at lower temperatures because of the condensation of the gas.) When this line is extrapolated to lower pressures, it reaches a pressure of 0 at \u2013273 \u00b0C, which is 0 on the kelvin scale and the lowest possible temperature, called absolute zero.<\/p>\n<\/div>\n<p id=\"fs-idm33029376\">Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the <em data-effect=\"italics\">P<\/em>&#8211;<em data-effect=\"italics\">T<\/em> relationship for gases is known as either <strong><span data-type=\"term\">Amontons\u2019s law<\/span><\/strong> or <strong><span data-type=\"term\">Gay-Lussac\u2019s law<\/span><\/strong>. Under either name, it states that <em data-effect=\"italics\">the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant<\/em>. Mathematically, this can be written:<\/p>\n<div id=\"fs-idp58663824\" data-type=\"equation\">[latex]P\\propto T\\text{ or }P=\\text{constant}\\times T\\text{ or }P=k\\times T[\/latex]<\/div>\n<p id=\"fs-idp12374848\">where \u221d means \u201cis proportional to,\u201d and <em data-effect=\"italics\">k<\/em> is a proportionality constant that depends on the identity, amount, and volume of the gas.<\/p>\n<p id=\"fs-idm100617456\">For a confined, constant volume of gas, the ratio [latex]\\frac{P}{T}[\/latex] is therefore constant (i.e., [latex]\\frac{P}{T}=k[\/latex] ). If the gas is initially in \u201cCondition 1\u201d (with <em data-effect=\"italics\">P<\/em> = <em data-effect=\"italics\">P<\/em><sub>1<\/sub> and <em data-effect=\"italics\">T<\/em> = <em data-effect=\"italics\">T<\/em><sub>1<\/sub>), and then changes to \u201cCondition 2\u201d (with <em data-effect=\"italics\">P<\/em> = <em data-effect=\"italics\">P<\/em><sub>2<\/sub> and <em data-effect=\"italics\">T<\/em> = <em data-effect=\"italics\">T<\/em><sub>2<\/sub>), we have that [latex]\\frac{{P}_{1}}{{T}_{1}}=k[\/latex] and [latex]\\frac{{P}_{2}}{{T}_{2}}=k,[\/latex] which reduces to [latex]\\frac{{P}_{1}}{{T}_{1}}=\\frac{{P}_{2}}{{T}_{2}}.[\/latex] This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called <strong><span data-type=\"term\">absolute zero<\/span><\/strong>). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.)<\/p>\n<div id=\"fs-idm66260928\" class=\"textbox shaded\" data-type=\"example\">\n<h3>Example 1<\/h3>\n<h4><strong>Predicting Change in Pressure with Temperature<\/strong><\/h4>\n<p id=\"fs-idm61839504\">A can of hair spray is used until it is empty except for the propellant, isobutane gas.<\/p>\n<p id=\"fs-idp96109888\">(a) On the can is the warning \u201cStore only at temperatures below 120 \u00b0F (48.8 \u00b0C). Do not incinerate.\u201d Why?<\/p>\n<p id=\"fs-idm53382240\">(b) The gas in the can is initially at 24 \u00b0C and 360 kPa, and the can has a volume of 350 mL. If the can is left in a car that reaches 50 \u00b0C on a hot day, what is the new pressure in the can?<\/p>\n<h4 id=\"fs-idm152475936\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>(a) The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)<\/p>\n<p id=\"fs-idp91458240\">(b) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons\u2019s\/Gay-Lussac\u2019s law. Taking <em data-effect=\"italics\">P<\/em><sub>1<\/sub> and <em data-effect=\"italics\">T<\/em><sub>1<\/sub> as the initial values, <em data-effect=\"italics\">T<\/em><sub>2<\/sub> as the temperature where the pressure is unknown and <em data-effect=\"italics\">P<\/em><sub>2<\/sub> as the unknown pressure, and converting \u00b0C to K, we have:<\/p>\n<div id=\"fs-idp253098544\" data-type=\"equation\">[latex]\\frac{{P}_{1}}{{T}_{1}}=\\frac{{P}_{2}}{{T}_{2}}\\phantom{\\rule{0.4em}{0ex}}\\text{which means that}\\phantom{\\rule{0.4em}{0ex}}\\frac{360\\text{kPa}}{297\\text{K}}=\\frac{{P}_{2}}{323\\text{K}}[\/latex]<\/div>\n<p id=\"fs-idm99533072\">Rearranging and solving gives: [latex]{P}_{2}=\\frac{360\\text{kPa}\\times 323\\cancel{\\text{K}}}{297\\cancel{\\text{K}}}=390\\text{kPa}[\/latex]<\/p>\n<h4 id=\"fs-idm87475200\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>A sample of nitrogen, N<sub>2<\/sub>, occupies 45.0 mL at 27 \u00b0C and 600 torr. What pressure will it have if cooled to \u201373 \u00b0C while the volume remains constant?<\/p>\n<div id=\"fs-idp78288768\" data-type=\"note\">\n<div style=\"text-align: right;\" data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer: <\/strong>\u00a0400 torr<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-idp20892224\" data-depth=\"1\">\n<h2 data-type=\"title\">Volume and Temperature: Charles\u2019s Law<\/h2>\n<p id=\"fs-idm19167872\">If we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let\u2019s say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up.<\/p>\n<div id=\"fs-idp27422192\" class=\"chemistry link-to-learning textbox\" data-type=\"note\">\n<p id=\"fs-idp60978480\">This video shows how cooling and heating a gas causes its volume to decrease or increase, respectively.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Liquid Nitrogen Experiments: The Balloon\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ZgTTUuJZAFs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-idp49208960\">These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Volume-temperature data for a 1-mole sample of methane gas at 1 atm are listed and graphed in Figure 4.<\/p>\n<div id=\"attachment_599\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-599\" class=\"size-large wp-image-599\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205500\/CNX_Chem_09_02_Charles2-1024x379.jpg\" alt=\"This figure includes a table and a graph. The table has 3 columns and 6 rows. The first row is a header, which labels the columns \u201cTemperature, degrees C,\u201d \u201cTemperature, K,\u201d and \u201cPressure, k P a.\u201d The first column contains the values from top to bottom negative 100, negative 50, 0, 100, and 200. The second column contains the values from top to bottom 173, 223, 273, 373, and 473. The third column contains the values 14.10, 18.26, 22.40, 30.65, and 38.88. A graph appears to the right of the table. The horizontal axis is labeled \u201cTemperature ( K ).\u201d with markings and labels provided for multiples of 100 beginning at 0 and ending at 300. The vertical axis is labeled \u201cVolume ( L )\u201d with marking and labels provided for multiples of 10, beginning at 0 and ending at 30. Five data points from the table are plotted on the graph with black dots. These dots are connected with a solid black line. The graph shows a positive linear trend.\" width=\"1024\" height=\"379\" \/><\/p>\n<p id=\"caption-attachment-599\" class=\"wp-caption-text\">Figure 4. The volume and temperature are linearly related for 1 mole of methane gas at a constant pressure of 1 atm. If the temperature is in kelvin, volume and temperature are directly proportional. The line stops at 111 K because methane liquefies at this temperature; when extrapolated, it intersects the graph\u2019s origin, representing a temperature of absolute zero.<\/p>\n<\/div>\n<p id=\"fs-idp38621856\">The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles\u2019s law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre C\u00e9sar Charles. <strong><span data-type=\"term\">Charles\u2019s law<\/span><\/strong> states that <em data-effect=\"italics\">the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant<\/em>.<\/p>\n<p id=\"fs-idp174747472\">Mathematically, this can be written as:<\/p>\n<div id=\"fs-idm59931152\" data-type=\"equation\">[latex]V\\propto T\\text{or}V=\\text{constant}\\cdot T\\text{or}V=k\\cdot T\\text{or}{V}_{1}\\text{\/}{T}_{1}={V}_{2}\\text{\/}{T}_{2}[\/latex]<\/div>\n<p id=\"fs-idp134528\">with <em data-effect=\"italics\">k<\/em> being a proportionality constant that depends on the amount and pressure of the gas.<\/p>\n<p id=\"fs-idp8180592\">For a confined, constant pressure gas sample, [latex]\\frac{V}{T}[\/latex] is constant (i.e., the ratio = <em data-effect=\"italics\">k<\/em>), and as seen with the <em data-effect=\"italics\">V<\/em>&#8211;<em data-effect=\"italics\">T<\/em> relationship, this leads to another form of Charles\u2019s law: [latex]\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}.[\/latex]<\/p>\n<div id=\"fs-idm45782128\" class=\"textbox shaded\" data-type=\"example\">\n<h3><strong>Example 2<\/strong><\/h3>\n<h4 id=\"fs-idp8413888\"><strong><span data-type=\"title\">Predicting Change in Volume with Temperature<\/span><\/strong><\/h4>\n<p>A sample of carbon dioxide, CO<sub>2<\/sub>, occupies 0.300 L at 10 \u00b0C and 750 torr. What volume will the gas have at 30 \u00b0C and 750 torr?<\/p>\n<h4 id=\"fs-idp207430880\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>Because we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles\u2019s law. Taking <em data-effect=\"italics\">V<\/em><sub>1<\/sub> and <em data-effect=\"italics\">T<\/em><sub>1<\/sub> as the initial values, <em data-effect=\"italics\">T<\/em><sub>2<\/sub> as the temperature at which the volume is unknown and <em data-effect=\"italics\">V<\/em><sub>2<\/sub> as the unknown volume, and converting \u00b0C into K we have:<\/p>\n<div id=\"fs-idp22541168\" data-type=\"equation\">[latex]\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}\\phantom{\\rule{0.4em}{0ex}}\\text{which means that}\\phantom{\\rule{0.4em}{0ex}}\\frac{0.300\\text{L}}{283\\text{K}}=\\frac{{V}_{2}}{303\\text{K}}[\/latex]<\/div>\n<p id=\"fs-idp170393776\">Rearranging and solving gives: [latex]{V}_{2}=\\frac{0.300\\text{L}\\times \\text{303}\\cancel{\\text{K}}}{283\\cancel{\\text{K}}}=0.321\\text{L}[\/latex]<\/p>\n<p id=\"fs-idp111256368\">This answer supports our expectation from Charles\u2019s law, namely, that raising the gas temperature (from 283 K to 303 K) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L).<\/p>\n<h4 id=\"fs-idp28974272\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>A sample of oxygen, O<sub>2<\/sub>, occupies 32.2 mL at 30 \u00b0C and 452 torr. What volume will it occupy at \u201370 \u00b0C and the same pressure?<\/p>\n<div id=\"fs-idm7218016\" data-type=\"note\">\n<div style=\"text-align: right;\" data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer: <\/strong>\u00a021.6 mL<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm51209248\" class=\"textbox shaded\" data-type=\"example\">\n<h3>Example 3<\/h3>\n<h4><strong>Measuring Temperature with a Volume Change<\/strong><\/h4>\n<p>Temperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of 150.0 cm<sup>3<\/sup> when immersed in a mixture of ice and water (0.00 \u00b0C). When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is 131.7 cm<sup>3<\/sup>. Find the temperature of boiling ammonia on the kelvin and Celsius scales.<\/p>\n<h4 id=\"fs-idm38946752\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>A volume change caused by a temperature change at constant pressure means we should use Charles\u2019s law. Taking <em data-effect=\"italics\">V<\/em><sub>1<\/sub> and <em data-effect=\"italics\">T<\/em><sub>1<\/sub> as the initial values, <em data-effect=\"italics\">T<\/em><sub>2<\/sub> as the temperature at which the volume is unknown and <em data-effect=\"italics\">V<\/em><sub>2<\/sub> as the unknown volume, and converting \u00b0C into K we have:<\/p>\n<div id=\"fs-idp40102160\" data-type=\"equation\">[latex]\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}\\text{ which means that }\\frac{150.0{\\text{cm}}^{3}}{273.15\\text{K}}=\\frac{131.7{\\text{cm}}^{3}}{{T}_{2}}[\/latex]<\/div>\n<p id=\"fs-idp6920384\">Rearrangement gives [latex]{T}_{2}=\\frac{131.7{\\cancel{\\text{cm}}}^{3}\\times 273.15\\text{K}}{150.0{\\cancel{\\text{cm}}}^{3}}=239.8\\text{K}[\/latex]<\/p>\n<p id=\"fs-idm36267744\">Subtracting 273.15 from 239.8 K, we find that the temperature of the boiling ammonia on the Celsius scale is \u201333.4 \u00b0C.<\/p>\n<h4 id=\"fs-idm26636752\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>What is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm?<\/p>\n<div id=\"fs-idm69251328\" data-type=\"note\">\n<div style=\"text-align: right;\" data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer: \u00a0<\/strong>635 mL<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-idm198797136\" data-depth=\"1\">\n<h2 data-type=\"title\">Volume and Pressure: Boyle\u2019s Law<\/h2>\n<p id=\"fs-idm87156432\">If we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say 25 \u00b0C. If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases. This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in Figure 5.<\/p>\n<div id=\"attachment_600\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-600\" class=\"size-large wp-image-600\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205502\/CNX_Chem_09_03_BoylesLaw1-1024x911.jpg\" alt=\"This figure contains a diagram and two graphs. The diagram shows a syringe labeled with a scale in m l or c c with multiples of 5 labeled beginning at 5 and ending at 30. The markings halfway between these measurements are also provided. Attached at the top of the syringe is a pressure gauge with a scale marked by fives from 40 on the left to 5 on the right. The gauge needle rests between 10 and 15, slightly closer to 15. The syringe plunger position indicates a volume measurement about halfway between 10 and 15 m l or c c. The first graph is labeled \u201cV ( m L )\u201d on the horizontal axis and \u201cP ( p s i )\u201d on the vertical axis. Points are labeled at 5, 10, 15, 20, and 25 m L with corresponding values of 39.0, 19.5, 13.0, 9.8, and 6.5 p s i. The points are connected with a smooth curve that is declining at a decreasing rate of change. The second graph is labeled \u201cV ( m L )\u201d on the horizontal axis and \u201c1 divided by P ( p s i )\u201d on the vertical axis. The horizontal axis is labeled at multiples of 5, beginning at zero and extending up to 35 m L. The vertical axis is labeled by multiples of 0.02, beginning at 0 and extending up to 0.18. Six points indicated by black dots on this graph are connected with a black line segment showing a positive linear trend.\" width=\"1024\" height=\"911\" \/><\/p>\n<p id=\"caption-attachment-600\" class=\"wp-caption-text\">Figure 5. When a gas occupies a smaller volume, it exerts a higher pressure; when it occupies a larger volume, it exerts a lower pressure (assuming the amount of gas and the temperature do not change). Since P and V are inversely proportional, a graph of 1\/P vs. V is linear.<\/p>\n<\/div>\n<p id=\"fs-idm298625584\">Unlike the <em data-effect=\"italics\">P<\/em>&#8211;<em data-effect=\"italics\">T<\/em> and <em data-effect=\"italics\">V<\/em>&#8211;<em data-effect=\"italics\">T<\/em> relationships, pressure and volume are not directly proportional to each other. Instead, <em data-effect=\"italics\">P<\/em> and <em data-effect=\"italics\">V<\/em> exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written:<\/p>\n<div data-type=\"equation\">[latex]P\\alpha 1\\text{\/}V\\text{ or }P=k\\cdot 1\\text{\/}V\\text{ or }P\\cdot V=k\\text{ or }{P}_{1}{V}_{1}={P}_{2}{V}_{2}[\/latex]<\/div>\n<div id=\"attachment_601\" style=\"width: 310px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-601\" class=\"wp-image-601\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205504\/CNX_Chem_09_02_Boyleslaw2.jpg\" alt=\"This diagram shows two graphs. In a, a graph is shown with volume on the horizontal axis and pressure on the vertical axis. A curved line is shown on the graph showing a decreasing trend with a decreasing rate of change. In b, a graph is shown with volume on the horizontal axis and one divided by pressure on the vertical axis. A line segment, beginning at the origin of the graph, shows a positive, linear trend.\" width=\"300\" height=\"153\" \/><\/p>\n<p id=\"caption-attachment-601\" class=\"wp-caption-text\">Figure 6. The relationship between pressure and volume is inversely proportional. (a) The graph of P vs. V is a parabola, whereas (b) the graph of (1\/P) vs. V is linear.<\/p>\n<\/div>\n<p>with <em data-effect=\"italics\">k<\/em> being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure [latex]\\left(\\frac{1}{P}\\right)[\/latex] versus the volume (<em data-effect=\"italics\">V<\/em>), or the inverse of volume [latex]\\left(\\frac{1}{V}\\right)[\/latex] versus the pressure (<em data-effect=\"italics\">V<\/em>). Graphs with curved lines are difficult to read accurately at low or high values of the variables, and they are more difficult to use in fitting theoretical equations and parameters to experimental data. For those reasons, scientists often try to find a way to \u201clinearize\u201d their data. If we plot <em data-effect=\"italics\">P<\/em> versus <em data-effect=\"italics\">V<\/em>, we obtain a hyperbola (see Figure 6).<\/p>\n<p id=\"fs-idm207692112\">The relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as <strong><span data-type=\"term\">Boyle\u2019s law<\/span><\/strong>: <em data-effect=\"italics\">The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured.<\/em><\/p>\n<div id=\"fs-idp10418864\" class=\"textbox shaded\" data-type=\"example\">\n<h3>Example 4<\/h3>\n<h4><strong>Volume of a Gas Sample<\/strong><\/h4>\n<p id=\"fs-idm205798208\">The sample of gas in Figure 5 has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL, using:<\/p>\n<p id=\"fs-idm127753872\">(a) the <em data-effect=\"italics\">P<\/em>&#8211;<em data-effect=\"italics\">V<\/em> graph in\u00a0Figure 5<\/p>\n<p id=\"fs-idm129344864\">(b) the [latex]\\frac{1}{P}[\/latex] vs. <em data-effect=\"italics\">V<\/em> graph in\u00a0Figure 5<\/p>\n<p id=\"fs-idm92820144\">(c) the Boyle\u2019s law equation<\/p>\n<p id=\"fs-idm215125536\">Comment on the likely accuracy of each method.<\/p>\n<h4 id=\"fs-idm126059440\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>(a) Estimating from the <em data-effect=\"italics\">P<\/em>&#8211;<em data-effect=\"italics\">V<\/em> graph gives a value for <em data-effect=\"italics\">P<\/em> somewhere around 27 psi.<\/p>\n<p id=\"fs-idm112322224\">(b) Estimating from the [latex]\\frac{1}{P}[\/latex] versus <em data-effect=\"italics\">V<\/em> graph give a value of about 26 psi.<\/p>\n<p id=\"fs-idm124214336\">(c) From Boyle\u2019s law, we know that the product of pressure and volume (<em data-effect=\"italics\">PV<\/em>) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have <em data-effect=\"italics\">P<\/em><sub>1<\/sub><em data-effect=\"italics\">V<\/em><sub>1<\/sub> = <em data-effect=\"italics\">k<\/em> and <em data-effect=\"italics\">P<\/em><sub>2<\/sub><em data-effect=\"italics\">V<\/em><sub>2<\/sub> = <em data-effect=\"italics\">k<\/em> which means that <em data-effect=\"italics\">P<\/em><sub>1<\/sub><em data-effect=\"italics\">V<\/em><sub>1<\/sub> = <em data-effect=\"italics\">P<\/em><sub>2<\/sub><em data-effect=\"italics\">V<\/em><sub>2<\/sub>.<\/p>\n<p id=\"fs-idm207111184\">Using <em data-effect=\"italics\">P<\/em><sub>1<\/sub> and <em data-effect=\"italics\">V<\/em><sub>1<\/sub> as the known values 0.993 atm and 2.40 mL, <em data-effect=\"italics\">P<\/em><sub>2<\/sub> as the pressure at which the volume is unknown, and <em data-effect=\"italics\">V<\/em><sub>2<\/sub> as the unknown volume, we have:<\/p>\n<div id=\"fs-idp70444944\" data-type=\"equation\">[latex]{P}_{1}{V}_{1}={P}_{2}{V}_{2}\\text{or}13.0\\text{psi}\\times 15.0\\text{mL}={P}_{2}\\times 7.5\\text{mL}[\/latex]<\/div>\n<p id=\"fs-idm167382656\">Solving:<\/p>\n<div id=\"fs-idp79285632\" data-type=\"equation\">[latex]{V}_{2}=\\frac{13.0\\text{psi}\\times 15.0\\cancel{\\text{mL}}}{7.5\\cancel{\\text{mL}}}=26\\text{mL}[\/latex]<\/div>\n<p id=\"fs-idm146893296\">It was more difficult to estimate well from the <em data-effect=\"italics\">P<\/em>&#8211;<em data-effect=\"italics\">V<\/em> graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow.<\/p>\n<h4 id=\"fs-idm111326864\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>The sample of gas in Figure 5 has a volume of 30.0 mL at a pressure of 6.5 psi. Determine the volume of the gas at a pressure of 11.0 mL, using:<\/p>\n<p id=\"fs-idm161358576\">(a) the <em data-effect=\"italics\">P<\/em>&#8211;<em data-effect=\"italics\">V<\/em> graph in\u00a0Figure 5<\/p>\n<p id=\"fs-idm130543696\">(b) the [latex]\\frac{1}{P}[\/latex] vs. <em data-effect=\"italics\">V<\/em> graph in\u00a0Figure 5<\/p>\n<p id=\"fs-idm208892896\">(c) the Boyle\u2019s law equation<\/p>\n<p id=\"fs-idm265464784\">Comment on the likely accuracy of each method.<\/p>\n<div id=\"fs-idp7933296\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:\u00a0<\/strong>(a) about 17\u201318 mL; (b) ~18 mL; (c) 17.7 mL; it was more difficult to estimate well from the <em data-effect=\"italics\">P<\/em>&#8211;<em data-effect=\"italics\">V<\/em> graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm202428608\" class=\"chemistry everyday-life\" data-type=\"note\">\n<div class=\"textbox shaded\" data-type=\"title\">\n<h3>Chemistry in Action:\u00a0Breathing and Boyle\u2019s Law<\/h3>\n<p>What do you do about 20 times per minute for your whole life, without break, and often without even being aware of it? The answer, of course, is respiration, or breathing. How does it work? It turns out that the gas laws apply here. Your lungs take in gas that your body needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your chest cavity and making your lung volume larger. The increase in volume leads to a decrease in pressure (Boyle\u2019s law). This causes air to flow into the lungs (from high pressure to low pressure). When you exhale, the process reverses: Your diaphragm and rib muscles relax, your chest cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle\u2019s law again), and air flows out of the lungs (from high pressure to low pressure). You then breathe in and out again, and again, repeating this Boyle\u2019s law cycle for the rest of your life (Figure 7).<\/p>\n<div id=\"attachment_602\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-602\" class=\"size-large wp-image-602\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23205505\/CNX_Chem_09_02_BoylesLaw4-1024x769.jpg\" alt=\"This figure contains two diagrams of a cross section of the human head and torso. The first diagram on the left is labeled \u201cInspiration.\u201d It shows curved arrows in gray proceeding through the nasal passages and mouth to the lungs. An arrow points downward from the diaphragm, which is relatively flat, just beneath the lungs. This arrow is labeled \u201cDiaphragm contracts.\u201d At the entrance to the mouth and nasal passages, a label of P subscript lungs equals 1 dash 3 torr lower\u201d is provided. The second, similar diagram, which is labeled \u201cExpiration,\u201d reverses the direction of both arrows. Arrows extend from the lungs out through the nasal passages and mouth. Similarly, an arrow points up to the diaphragm, showing a curved diaphragm and lungs reduced in size from the previous image. This arrow is labeled \u201cDiaphragm relaxes.\u201d At the entrance to the mouth and nasal passages, a label of P subscript lungs equals 1 dash 3 torr higher\u201d is provided.\" width=\"1024\" height=\"769\" \/><\/p>\n<p id=\"caption-attachment-602\" class=\"wp-caption-text\">Figure 7. Breathing occurs because expanding and contracting lung volume creates small pressure differences between your lungs and your surroundings, causing air to be drawn into and forced out of your lungs.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-idp47884080\" data-depth=\"1\">\n<h2 data-type=\"title\">Moles of Gas and Volume: Avogadro\u2019s Law<\/h2>\n<p id=\"fs-idm52917120\">The Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed by <strong><span data-type=\"term\">Avogadro\u2019s law<\/span><\/strong>: <em data-effect=\"italics\">For a confined gas, the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both remain constant<\/em>.<\/p>\n<p id=\"fs-idp65662512\">In equation form, this is written as:<\/p>\n<div id=\"fs-idm8943936\" data-type=\"equation\">[latex]\\begin{array}{ccccc}V\\propto n& \\text{or}& V=k\\times n& \\text{or}& \\frac{{V}_{1}}{{n}_{1}}=\\frac{{V}_{2}}{{n}_{2}}.\\end{array}[\/latex]<\/div>\n<p id=\"fs-idm94717808\">Mathematical relationships can also be determined for the other variable pairs, such as <em data-effect=\"italics\">P<\/em> versus <em data-effect=\"italics\">n<\/em>, and <em data-effect=\"italics\">n<\/em> versus <em data-effect=\"italics\">T<\/em>.<\/p>\n<div id=\"fs-idm50116432\" class=\"chemistry link-to-learning textbox\" data-type=\"note\">\n<p id=\"fs-idm91270672\">Visit this interactive <a href=\"http:\/\/phet.colorado.edu\/en\/simulation\/gas-properties\" target=\"_blank\">PhET simulation link to investigate the relationships between pressure, volume, temperature. and amount of gas<\/a>. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws).<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idm68358176\" data-depth=\"1\">\n<h2 data-type=\"title\">The Ideal Gas Law<\/h2>\n<p id=\"fs-idp161797856\">To this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas:<\/p>\n<ul id=\"fs-idp25403616\" data-bullet-style=\"bullet\">\n<li>Boyle\u2019s law: <em data-effect=\"italics\">PV<\/em> = constant at constant <em data-effect=\"italics\">T<\/em> and <em data-effect=\"italics\">n<\/em><\/li>\n<li>Amontons\u2019s law: [latex]\\frac{P}{T}[\/latex] = constant at constant <em data-effect=\"italics\">V<\/em> and <em data-effect=\"italics\">n<\/em><\/li>\n<li>Charles\u2019s law: [latex]\\frac{V}{T}[\/latex] = constant at constant <em data-effect=\"italics\">P<\/em> and <em data-effect=\"italics\">n<\/em><\/li>\n<li>Avogadro\u2019s law: [latex]\\frac{V}{n}[\/latex] = constant at constant <em data-effect=\"italics\">P<\/em> and <em data-effect=\"italics\">T<\/em><\/li>\n<\/ul>\n<p id=\"fs-idp47645536\">Combining these four laws yields the <strong><span data-type=\"term\">ideal gas law<\/span><\/strong>, a relation between the pressure, volume, temperature, and number of moles of a gas:<\/p>\n<div id=\"fs-idp190457648\" data-type=\"equation\">[latex]PV=nRT[\/latex]<\/div>\n<p id=\"fs-idp83793536\">where <em data-effect=\"italics\">P<\/em> is the pressure of a gas, <em data-effect=\"italics\">V<\/em> is its volume, <em data-effect=\"italics\">n<\/em> is the number of moles of the gas, <em data-effect=\"italics\">T<\/em> is its temperature on the kelvin scale, and <em data-effect=\"italics\">R<\/em> is a constant called the <strong><span data-type=\"term\">ideal gas constant<\/span><\/strong> or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being 0.08206 L atm mol<sup>\u20131<\/sup> K<sup>\u20131<\/sup> and 8.314 kPa L mol<sup>\u20131<\/sup> K<sup>\u20131<\/sup>.<\/p>\n<p id=\"fs-idp100115520\">Gases whose properties of <em data-effect=\"italics\">P<\/em>, <em data-effect=\"italics\">V<\/em>, and <em data-effect=\"italics\">T<\/em> are accurately described by the ideal gas law (or the other gas laws) are said to exhibit <em data-effect=\"italics\">ideal behavior<\/em> or to approximate the traits of an <strong><span data-type=\"term\">ideal gas<\/span><\/strong>. An ideal gas is a hypothetical construct that may be used along with <em data-effect=\"italics\">kinetic molecular theory<\/em> to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the <em data-effect=\"italics\">non-ideal<\/em> behavior observed for many gases at relatively high pressures and low temperatures.<\/p>\n<p id=\"fs-idp9263600\">The ideal gas equation contains five terms, the gas constant <em data-effect=\"italics\">R<\/em> and the variable properties <em data-effect=\"italics\">P<\/em>, <em data-effect=\"italics\">V<\/em>, <em data-effect=\"italics\">n<\/em>, and <em data-effect=\"italics\">T<\/em>. Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises.<\/p>\n<div id=\"fs-idm72969744\" class=\"textbox shaded\" data-type=\"example\">\n<h3>Example 5<\/h3>\n<h4><strong>Using the Ideal Gas Law<\/strong><\/h4>\n<p>Methane, CH<sub>4<\/sub>, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of CH<sub>4<\/sub>. What is the volume of this much methane at 25 \u00b0C and 745 torr?<\/p>\n<h4 id=\"fs-idp9667456\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>We must rearrange <em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em> to solve for <em data-effect=\"italics\">V<\/em>: [latex]V=\\frac{nRT}{P}[\/latex]<\/p>\n<p id=\"fs-idp84134864\">If we choose to use <em data-effect=\"italics\">R<\/em> = 0.08206 L atm mol<sup>\u20131<\/sup> K<sup>\u20131<\/sup>, then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.<\/p>\n<p id=\"fs-idm113078592\">Converting into the \u201cright\u201d units:<\/p>\n<div id=\"fs-idp39691024\" data-type=\"equation\">[latex]n=655\\text{g}\\cancel{{\\text{CH}}_{4}}\\times \\frac{1\\text{mol}}{16.043{\\cancel{\\text{g CH}}}_{4}}=40.8\\text{mol}[\/latex]<\/div>\n<div id=\"fs-idp101238448\" data-type=\"equation\">[latex]T=25\\text{\\textdegree C}+273=298\\text{K}[\/latex]<\/div>\n<div id=\"fs-idp11744096\" data-type=\"equation\">[latex]P=745\\cancel{\\text{torr}}\\times \\frac{1\\text{atm}}{760\\cancel{\\text{torr}}}=0.980\\text{atm}[\/latex]<\/div>\n<div id=\"fs-idp248570064\" data-type=\"equation\">[latex]V=\\frac{nRT}{P}=\\frac{\\left(40.8\\cancel{\\text{mol}}\\right)\\left(0.08206\\text{L}\\cancel{{\\text{atm mol}}^{-1}{\\text{K}}^{\\text{-1}}}\\right)\\left(298\\cancel{\\text{K}}\\right)}{0.980\\cancel{\\text{atm}}}=1.02\\times {10}^{3}\\text{L}[\/latex]<\/div>\n<p id=\"fs-idm139345632\">It would require 1020 L (269 gal) of gaseous methane at about 1 atm of pressure to replace 1 gal of gasoline. It requires a large container to hold enough methane at 1 atm to replace several gallons of gasoline.<\/p>\n<h4 id=\"fs-idp147095712\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>Calculate the pressure in bar of 2520 moles of hydrogen gas stored at 27 \u00b0C in the 180-L storage tank of a modern hydrogen-powered car.<\/p>\n<div id=\"fs-idp77431312\" data-type=\"note\">\n<div style=\"text-align: right;\" data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer: \u00a0<\/strong>350 bar<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-idp141407616\">If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: [latex]\\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\\frac{{P}_{2}{V}_{2}}{{T}_{2}}[\/latex] using units of atm, L, and K. Both sets of conditions are equal to the product of <em data-effect=\"italics\">n<\/em> \u00d7 <em data-effect=\"italics\">R<\/em> (where <em data-effect=\"italics\">n<\/em> = the number of moles of the gas and <em data-effect=\"italics\">R<\/em> is the ideal gas law constant).<\/p>\n<div id=\"fs-idp23616944\" class=\"textbox shaded\" data-type=\"example\">\n<h3>Example 6<\/h3>\n<h4 id=\"fs-idm24051840\"><strong><span data-type=\"title\">Using the Combined Gas Law<\/span><\/strong><\/h4>\n<div id=\"attachment_2085\" style=\"width: 310px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2085\" class=\"wp-image-2085\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212034\/CNX_Chem_09_02_Scuba1.jpg\" alt=\"This photograph shows a scuba diver underwater with a tank on his or her back and bubbles ascending from the breathing apparatus.\" width=\"300\" height=\"218\" \/><\/p>\n<p id=\"caption-attachment-2085\" class=\"wp-caption-text\">Figure 8. Scuba divers use compressed air to breathe while underwater. (credit: modification of work by Mark Goodchild)<\/p>\n<\/div>\n<p>When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure 8). If the water temperature is 27 \u00b0C, how many liters of air will such a tank provide to a diver\u2019s lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm?<\/p>\n<p id=\"fs-idm78575136\">Letting <em data-effect=\"italics\">1<\/em> represent the air in the scuba tank and <em data-effect=\"italics\">2<\/em> represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is 37 \u00b0C, we have:<\/p>\n<div id=\"fs-idp224060000\" data-type=\"equation\">[latex]\\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\\frac{{P}_{2}{V}_{2}}{{T}_{2}}\\phantom{\\rule{0.4em}{0ex}}\\rightarrow\\phantom{\\rule{0.4em}{0ex}}\\frac{\\left(153\\text{atm}\\right)\\left(13.2\\text{L}\\right)}{\\left(300\\text{K}\\right)}=\\frac{\\left(3.13\\text{atm}\\right)\\left({V}_{2}\\right)}{\\left(310\\text{K}\\right)}[\/latex]<\/div>\n<p id=\"fs-idp72690672\">Solving for <em data-effect=\"italics\">V<\/em><sub>2<\/sub>:<\/p>\n<div id=\"fs-idp102261440\" data-type=\"equation\">[latex]{V}_{2}=\\frac{\\left(153\\cancel{\\text{atm}}\\right)\\left(13.2\\text{L}\\right)\\left(310\\text{K}\\right)}{\\left(300\\text{K}\\right)\\left(3.13\\cancel{\\text{atm}}\\right)}=667\\text{L}[\/latex]<\/div>\n<p id=\"fs-idp54320688\">(Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good \u201cballpark\u201d estimate.)<\/p>\n<h4 id=\"fs-idp128482944\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 \u00b0C and 0.850 atm. Find the volume of this sample at 0 \u00b0C and 1.00 atm.<\/p>\n<div id=\"fs-idp104450512\" data-type=\"note\">\n<div style=\"text-align: right;\" data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer: \u00a0<\/strong>0.538 L<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idp8804032\" class=\"chemistry everyday-life\" data-type=\"note\">\n<div class=\"textbox shaded\" data-type=\"title\">\n<h3 data-type=\"title\">The Interdependence between Ocean Depth and Pressure in Scuba Diving<\/h3>\n<div id=\"attachment_2086\" style=\"width: 310px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2086\" class=\"wp-image-2086\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212035\/CNX_Chem_09_02_GreatBarri1.jpg\" alt=\"This picture shows colorful underwater corals and anemones in hues of yellow, orange, green, and brown, surrounded by water that appears blue in color.\" width=\"300\" height=\"225\" \/><\/p>\n<p id=\"caption-attachment-2086\" class=\"wp-caption-text\">Figure 9. Scuba divers, whether at the Great Barrier Reef or in the Caribbean, must be aware of buoyancy, pressure equalization, and the amount of time they spend underwater, to avoid the risks associated with pressurized gases in the body. (credit: Kyle Taylor)<\/p>\n<\/div>\n<p>Whether scuba diving at the Great Barrier Reef in Australia (shown in Figure 9) or in the Caribbean, divers must understand how pressure affects a number of issues related to their comfort and safety.<\/p>\n<p>Pressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as \u201catmospheres absolute\u201d or ATA in the diving community:\u00a0Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level.<\/p>\n<p>As a diver descends, the increase in pressure causes the body\u2019s air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization.<\/p>\n<p>Buoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in his BCD expands because of lower pressure according to Boyle\u2019s law (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and she or he begins to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor.<\/p>\n<p>The pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast as at the surface.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-idp117090992\" data-depth=\"1\">\n<h2 data-type=\"title\">Standard Conditions of Temperature and Pressure<\/h2>\n<p id=\"fs-idp92943776\">We have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a <strong><span data-type=\"term\">standard temperature and pressure (STP)<\/span> <\/strong>for reporting properties of gases: 273.15 K and 1 atm (101.325 kPa). At STP, an ideal gas has a volume of about 22.4 L\u2014this is referred to as the <strong><span data-type=\"term\">standard molar volume<\/span><\/strong> (Figure 10).<\/p>\n<div id=\"attachment_2087\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2087\" class=\"size-large wp-image-2087\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212037\/CNX_Chem_09_02_HENH3O21-1024x590.jpg\" alt=\"This figure shows three balloons each filled with H e, N H subscript 2, and O subscript 2 respectively. Beneath the first balloon is the label \u201c4 g of He\u201d Beneath the second balloon is the label, \u201c15 g of N H subscript 2.\u201d Beneath the third balloon is the label \u201c32 g of O subscript 2.\u201d Each balloon contains the same number of molecules of their respective gases.\" width=\"1024\" height=\"590\" \/><\/p>\n<p id=\"caption-attachment-2087\" class=\"wp-caption-text\">Figure 10. Since the number of moles in a given volume of gas varies with pressure and temperature changes, chemists use standard temperature and pressure (273.15 K and 1 atm or 101.325 kPa) to report properties of gases.<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idp68486960\" class=\"summary\" data-depth=\"1\">\n<div class=\"bcc-box bcc-success\">\n<h2>Key Concepts and Summary<\/h2>\n<section>\n<div data-type=\"note\">\n<p>The behavior of gases can be described by several laws based on experimental observations of their properties. The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons\u2019s law). The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles\u2019s law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle\u2019s law). Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro\u2019s law).<\/p>\n<p id=\"fs-idp115685936\">The equations describing these laws are special cases of the ideal gas law, <em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em>, where <em data-effect=\"italics\">P<\/em> is the pressure of the gas, <em data-effect=\"italics\">V<\/em> is its volume, <em data-effect=\"italics\">n<\/em> is the number of moles of the gas, <em data-effect=\"italics\">T<\/em> is its kelvin temperature, and <em data-effect=\"italics\">R<\/em> is the ideal (universal) gas constant.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Key Equations<\/h3>\n<ul id=\"fs-idp89091632\" data-bullet-style=\"bullet\">\n<li><em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em><\/li>\n<\/ul>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Chemistry End of Chapter Exercises<\/h3>\n<section id=\"self-check-questions\">\n<ol>\n<li data-type=\"title\">Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?<\/li>\n<li data-type=\"title\">Explain how the volume of the bubbles exhausted by a scuba diver (Figure 8) change as they rise to the surface, assuming that they remain intact.<\/li>\n<li data-type=\"title\">One way to state Boyle\u2019s law is \u201cAll other things being equal, the pressure of a gas is inversely proportional to its volume.\u201d\n<ol>\n<li data-type=\"title\">What is the meaning of the term \u201cinversely proportional?\u201d<\/li>\n<li data-type=\"title\">What are the \u201cother things\u201d that must be equal?<\/li>\n<\/ol>\n<\/li>\n<li data-type=\"title\">An alternate way to state Avogadro\u2019s law is \u201cAll other things being equal, the number of molecules in a gas is directly proportional to the volume of the gas.\u201d\n<ol>\n<li data-type=\"title\">What is the meaning of the term \u201cdirectly proportional?\u201d<\/li>\n<li data-type=\"title\">What are the \u201cother things\u201d that must be equal?<\/li>\n<\/ol>\n<\/li>\n<li data-type=\"title\">How would the graph in Figure 4 change if the number of moles of gas in the sample used to determine the curve were doubled?<\/li>\n<li data-type=\"title\">How would the graph in Figure 5 change if the number of moles of gas in the sample used to determine the curve were doubled?<\/li>\n<li data-type=\"title\">In addition to the data found in Figure 5, what other information do we need to find the mass of the sample of air used to determine the graph?<\/li>\n<li data-type=\"title\">Determine the volume of 1 mol of CH<sub>4<\/sub> gas at 150 K and 1 atm, using Figure 4.<\/li>\n<li data-type=\"title\">Determine the pressure of the gas in the syringe shown in Figure 5 when its volume is 12.5 mL, using:\n<ol>\n<li data-type=\"title\">the appropriate graph<\/li>\n<li data-type=\"title\">Boyle\u2019s law<\/li>\n<\/ol>\n<\/li>\n<li data-type=\"title\">A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 \u00b0C. If the can is thrown into a fire (T = 475 \u00b0C), what will be the pressure in the hot can?<\/li>\n<li data-type=\"title\">What is the temperature of an 11.2-L sample of carbon monoxide, CO, at 744 torr if it occupies 13.3 L at 55 \u00b0C and 744 torr?<\/li>\n<li data-type=\"title\">A 2.50-L volume of hydrogen measured at \u2013196 \u00b0C is warmed to 100 \u00b0C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.<\/li>\n<li data-type=\"title\">A balloon inflated with three breaths of air has a volume of 1.7 L. At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?<img loading=\"lazy\" decoding=\"async\" class=\"alignright wp-image-2088\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212038\/CNX_Chem_09_02_WeatherBall_img1.jpg\" alt=\"This image shows a white balloon that appears to have an attached white card. The balloon is being held by a person in an outdoor setting.\" width=\"300\" height=\"225\" \/><\/li>\n<li data-type=\"title\">A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25 \u00b0C at ground level. What is the volume of the balloon under these conditions?<\/li>\n<li data-type=\"title\">The volume of an automobile air bag was 66.8 L when inflated at 25 \u00b0C with 77.8 g of nitrogen gas. What was the pressure in the bag in kPa?<\/li>\n<li data-type=\"title\">How many moles of gaseous boron trifluoride, BF<sub>3<\/sub>, are contained in a 4.3410-L bulb at 788.0 K if the pressure is 1.220 atm? How many grams of BF<sub>3<\/sub>?<\/li>\n<li data-type=\"title\">Iodine, I<sub>2<\/sub>, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a 73.3-mL bulb that contains 0.292 g of I<sub>2<\/sub> vapor at a pressure of 0.462 atm?<\/li>\n<li data-type=\"title\">How many grams of gas are present in each of the following cases?\n<ol>\n<li data-type=\"title\">0.100 L of CO<sub>2<\/sub> at 307 torr and 26 \u00b0C<\/li>\n<li data-type=\"title\">8.75 L of C<sub>2<\/sub>H<sub>4<\/sub>, at 378.3 kPa and 483 K<\/li>\n<li data-type=\"title\">221 mL of Ar at 0.23 torr and \u201354 \u00b0C<\/li>\n<\/ol>\n<\/li>\n<li data-type=\"title\">A high altitude balloon is filled with 1.41 \u00d7 10<sup>4<\/sup> L of hydrogen at a temperature of 21 \u00b0C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is \u201348 \u00b0C and the pressure is 63.1 torr?<\/li>\n<li data-type=\"title\">A cylinder of medical oxygen has a volume of 35.4 L, and contains O<sub>2<\/sub> at a pressure of 151 atm and a temperature of 25 \u00b0C. What volume of O<sub>2<\/sub> does this correspond to at normal body conditions, that is, 1 atm and 37 \u00b0C?<\/li>\n<li data-type=\"title\">A large scuba tank (Figure 8) with a volume of 18 L is rated for a pressure of 220 bar. The tank is filled at 20 \u00b0C and contains enough air to supply 1860 L of air to a diver at a pressure of 2.37 atm (a depth of 45 feet). Was the tank filled to capacity at 20 \u00b0C?<\/li>\n<li data-type=\"title\">A 20.0-L cylinder containing 11.34 kg of butane, C<sub>4<\/sub>H<sub>10<\/sub>, was opened to the atmosphere. Calculate the mass of the gas remaining in the cylinder if it were opened and the gas escaped until the pressure in the cylinder was equal to the atmospheric pressure, 0.983 atm, and a temperature of 27 \u00b0C.<\/li>\n<li data-type=\"title\">While resting, the average 70-kg human male consumes 14 L of pure O<sub>2<\/sub> per hour at 25 \u00b0C and 100 kPa. How many moles of O<sub>2<\/sub> are consumed by a 70 kg man while resting for 1.0 h?<\/li>\n<li data-type=\"title\">For a given amount of gas showing ideal behavior, draw labeled graphs of:\n<ol>\n<li data-type=\"title\">the variation of <em data-effect=\"italics\">P<\/em> with <em data-effect=\"italics\">V\u00a0<\/em><\/li>\n<li data-type=\"title\">the variation of <em data-effect=\"italics\">V<\/em> with <em data-effect=\"italics\">T\u00a0<\/em><\/li>\n<li data-type=\"title\">the variation of <em data-effect=\"italics\">P<\/em> with <em data-effect=\"italics\">T\u00a0<\/em><\/li>\n<li data-type=\"title\">the variation of [latex]\\frac{1}{P}[\/latex] with <em data-effect=\"italics\">V<\/em><\/li>\n<\/ol>\n<\/li>\n<li data-type=\"title\">A liter of methane gas, CH<sub>4<\/sub>, at STP contains more atoms of hydrogen than does a liter of pure hydrogen gas, H<sub>2<\/sub>, at STP. Using Avogadro\u2019s law as a starting point, explain why.<\/li>\n<li data-type=\"title\">The effect of chlorofluorocarbons (such as CCl<sub>2<\/sub>F<sub>2<\/sub>) on the depletion of the ozone layer is well known. The use of substitutes, such as CH<sub>3<\/sub>CH<sub>2<\/sub>F(<em data-effect=\"italics\">g<\/em>), for the chlorofluorocarbons, has largely corrected the problem. Calculate the volume occupied by 10.0 g of each of these compounds at STP:\n<ol>\n<li data-type=\"title\">CCl<sub>2<\/sub>F<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>)<\/li>\n<li data-type=\"title\">CH<sub>3<\/sub>CH<sub>2<\/sub>F(<em data-effect=\"italics\">g<\/em>)<\/li>\n<\/ol>\n<\/li>\n<li data-type=\"title\">As 1 g of the radioactive element radium decays over 1 year, it produces 1.16 \u00d7 10<sup>18<\/sup> alpha particles (helium nuclei). Each alpha particle becomes an atom of helium gas. What is the pressure in pascal of the helium gas produced if it occupies a volume of 125 mL at a temperature of 25 \u00b0C?<\/li>\n<li data-type=\"title\">A balloon that is 100.21 L at 21 \u00b0C and 0.981 atm is released and just barely clears the top of Mount Crumpet in British Columbia. If the final volume of the balloon is 144.53 L at a temperature of 5.24 \u00b0C, what is the pressure experienced by the balloon as it clears Mount Crumpet?<\/li>\n<li data-type=\"title\">If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?<\/li>\n<li data-type=\"title\">If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?<\/li>\n<\/ol>\n<\/section>\n<\/div>\n<div data-type=\"solution\">\n<div class=\"bcc-box bcc-info\">\n<section id=\"self-check-answers\">\n<div data-type=\"exercise\">\n<h4>Selected Answers<\/h4>\n<div id=\"fs-idp16137760\" data-type=\"exercise\">\n<p id=\"fs-idm158859536\" data-type=\"solution\">2. As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle\u2019s law.<\/p>\n<\/div>\n<div id=\"fs-idm244332560\" data-type=\"solution\">4. (a) The number of particles in the gas increases as the volume increases. This relationship may be written as <em data-effect=\"italics\">n<\/em> = constant \u00d7 <em data-effect=\"italics\">V<\/em>. It is a direct relationship.(b) The temperature and pressure must be kept constant.<\/div>\n<div id=\"fs-idm181098208\" data-type=\"solution\">6. The curve would be farther to the right and higher up, but the same basic shape.<\/div>\n<div id=\"fs-idm207209024\" data-type=\"solution\">8. The figure shows the change of 1 mol of CH<sub>4<\/sub> gas as a function of temperature. The graph shows that the volume is about 16.3 to 16.5 L.<\/div>\n<div id=\"fs-idm211017328\" data-type=\"exercise\">\n<div id=\"fs-idm121334432\" data-type=\"solution\">\n<p>10. The first thing to recognize about this problem is that the volume and moles of gas remain constant. Thus, we can use the combined gas law equation in the form:<\/p>\n<p data-type=\"newline\">\u00a0[latex]\\frac{{P}_{2}}{{T}_{2}}=\\frac{{P}_{1}}{{T1}_{}}[\/latex]<\/p>\n<p>[latex]{P}_{2}=\\frac{{P}_{1}{T}_{2}}{{T}_{1}}=1344\\text{torr}\\times \\frac{475+273.15}{23+273.15}=3.40\\times {10}^{3}\\text{torr}[\/latex]<\/p>\n<div id=\"fs-idm221278704\" data-type=\"exercise\">\n<div id=\"fs-idm213171488\" data-type=\"solution\">\n<p>12. Apply Charles\u2019s law to compute the volume of gas at the higher temperature:<\/p>\n<p data-type=\"newline\">\u00a0<em data-effect=\"italics\">V<\/em><sub>1<\/sub> = 2.50 L<\/p>\n<p data-type=\"newline\">\u00a0<em data-effect=\"italics\">T<\/em><sub>1<\/sub> = \u2013193 \u00b0C = 77.15 K<\/p>\n<p data-type=\"newline\">\u00a0<em data-effect=\"italics\">V<\/em><sub>2<\/sub> = ?<\/p>\n<p data-type=\"newline\">\u00a0<em data-effect=\"italics\">T<\/em><sub>2<\/sub> = 100 \u00b0C = 373.15 K<\/p>\n<div data-type=\"newline\">\n<div id=\"fs-idm213171488\" data-type=\"solution\">\n<p>[latex]\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}[\/latex]<\/p>\n<div data-type=\"newline\">\u00a0[latex]{V}_{2}=\\frac{{V}_{1}{T}_{2}}{{T}_{1}}=\\frac{2.50\\text{L}\\times 373.15\\cancel{\\text{K}}}{77.15\\cancel{\\text{K}}}=12.1\\text{L}[\/latex]<\/div>\n<p>14. \u00a0<em data-effect=\"italics\">PV<\/em> = <em data-effect=\"italics\">nRT<\/em><\/p>\n<\/div>\n<div id=\"fs-idm24946256\" data-type=\"exercise\">\n<div id=\"fs-idm78551216\" data-type=\"solution\">\n<p>[latex]V=\\frac{nRT}{P}=\\frac{8.80\\cancel{\\text{mol}}\\times 0.08206\\text{L}\\cancel{\\text{atm}}{\\cancel{\\text{mol}}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\times 298.15\\cancel{\\text{K}}}{0.992\\cancel{\\text{atm}}}=217\\text{L}[\/latex]<\/p>\n<div id=\"fs-idm130104720\" data-type=\"problem\">16. [latex]n=\\frac{PV}{RT}\\frac{1.220\\cancel{\\text{atm}}\\left(4.3410\\text{L}\\right)}{\\left(0.08206\\text{L}\\cancel{\\text{atm}}\\text{mol}{\\text{-1}}^{}\\cancel{{\\text{K}}^{\\text{-1}}}\\right)\\left(788.0\\cancel{\\text{K}}\\right)}=0.08190\\text{mol}=8.190\\times {10}^{\\text{-2}}\\text{mol}[\/latex]<\/div>\n<div id=\"fs-idm95261248\" data-type=\"solution\">\n<div data-type=\"newline\">\u00a0[latex]n\\times \\text{molar mass}=8.190\\times {10}^{\\text{-2}}\\cancel{\\text{mol}}\\times 67.8052\\text{g}{\\cancel{\\text{mol}}}^{\\text{-1}}=5.553\\text{g}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-idm171506192\" data-type=\"exercise\">\n<div id=\"fs-idm254882416\" data-type=\"solution\">\n<div data-type=\"newline\">\n<p>18. In each of these problems, we are given a volume, pressure, and temperature. We can obtain moles from this information using the molar mass, <em data-effect=\"italics\">m<\/em> = <em data-effect=\"italics\">n\u2133<\/em>, where \u2133 is the molar mass:<\/p>\n<p data-type=\"newline\">[latex]P,V,T\\stackrel{n=PV\\text{\/}RT}{\\to }n,\\stackrel{m=n\\left(\\text{molar mass}\\right)}{\\to }\\text{grams}[\/latex]<\/p>\n<p data-type=\"newline\">or we can combine these equations to obtain:<\/p>\n<p data-type=\"newline\">[latex]\\text{mass}=m=\\frac{PV}{RT}\\times[\/latex]\u2133<\/p>\n<p>(a) \u00a0[latex]\\begin{array}{l}307\\cancel{\\text{torr}}\\times \\frac{1\\text{atm}}{760\\cancel{\\text{torr}}}=0.4039\\text{atm}26\\textdegree C=299.1 K\\\\ \\text{Mass}=m=\\frac{0.4039\\cancel{\\text{atm}}\\left(0.100\\cancel{\\text{L}}\\right)}{0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\left(299.1\\cancel{\\text{K}}\\right)}\\times 44.01\\text{g}{\\text{mol}}^{\\text{-1}}=7.24\\times {10}^{\\text{-2}}\\text{g};\\end{array}[\/latex]<\/p>\n<p>(b)[latex]\\text{Mass}=m=\\frac{378.3\\cancel{\\text{kPa}}\\left(8.75\\cancel{\\text{L}}\\right)}{8.314\\cancel{\\text{L}}\\cancel{\\text{kPa}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\left(483\\cancel{\\text{K}}\\right)}\\times 28.05376\\text{g}{\\text{mol}}^{\\text{-1}}=23.1\\text{g;}[\/latex]<\/p>\n<div id=\"fs-idm33816112\" data-type=\"solution\">(c) \u00a0[latex]\\begin{array}{l}\\\\ 221\\cancel{\\text{mL}}\\times \\frac{1\\text{L}}{1000\\cancel{\\text{mL}}}=0.221\\text{L}-54\\textdegree C+273.15=219.15\\text{K}\\\\ 0.23\\cancel{\\text{torr}}\\times \\frac{1\\text{atm}}{760\\cancel{\\text{torr}}}=3.03\\times {10}^{\\text{-4}}\\text{atm}\\\\ \\text{Mass}=m=\\frac{3.03\\times {10}^{\\text{-4}}\\cancel{\\text{atm}}\\left(0.221\\cancel{\\text{L}}\\right)}{0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\left(219.15\\cancel{\\text{K}}\\right)}\\times 39.978\\text{g}{\\text{mol}}^{\\text{-1}}=1.5\\times {10}^{\\text{-4}}\\text{g}\\end{array}[\/latex]<\/div>\n<div id=\"fs-idm87206720\" data-type=\"exercise\">\n<div id=\"fs-idm171827136\" data-type=\"solution\">\n<div data-type=\"newline\">\n<div id=\"fs-idm33816112\" data-type=\"solution\">20. \u00a0[latex]\\frac{{P}_{2}}{{T}_{2}}=\\frac{{P}_{1}}{{T}_{1}}[\/latex]<\/div>\n<div id=\"fs-idm87356096\" data-type=\"solution\">\n<div data-type=\"newline\">\u00a0<em data-effect=\"italics\">T<\/em><sub>2<\/sub> = 49.5 + 273.15 = 322.65 K<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]{P}_{2}=\\frac{{P}_{1}{T}_{2}}{{T}_{1}}=149.6\\text{atm}\\times \\frac{322.65}{278.15}=173.5\\text{atm}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-idm247077296\" data-type=\"exercise\">\n<div id=\"fs-idm237039312\" data-type=\"solution\">\n<p>22. Calculate the amount of butane in 20.0 L at 0.983 atm and 27\u00b0C. The original amount in the container does not matter. [latex]n=\\frac{PV}{RT}=\\frac{0.983\\cancel{\\text{atm}}\\times 20.0\\cancel{\\text{L}}}{0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\left(300.1\\cancel{\\text{K}}\\right)}=0.798\\text{mol}[\/latex]\u00a0Mass of butane = 0.798 mol \u00d7 58.1234 g\/mol = 46.4 g<\/p>\n<p>24.\u00a0For a gas exhibiting ideal behavior:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-large wp-image-5190\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/08\/23214537\/CNX_Chem_09_02_Exercise25_img-1024x868.jpg\" alt=\"Four graphs are shown. In a, Volume is on the horizontal axis and Pressure is on the vertical axis. A downward trend with a decreasing rate of change is shown by a curved line. The label n, P cons is shown on the graph. In b, Temperature is on the horizontal axis and Volume is on the vertical axis. An increasing linear trend is shown by a straight line segment. The label n, P cons is shown on the graph. In c, Temperature is on the horizontal axis and Pressure is on the vertical axis. An increasing linear trend is shown by a straight line segment. The label n, P cons is shown on the graph. In d, Volume is on the horizontal axis and 1 divided by Pressure is on the vertical axis. An increasing linear trend is shown by a straight line segment on the graph. The label n, P cons is shown on the graph.\" width=\"1024\" height=\"868\" \/><\/p>\n<div data-type=\"newline\">\n<div id=\"fs-idp18583872\" data-type=\"solution\">26. (a) Determine the molar mass of CCl<sub>2<\/sub>F<sub>2<\/sub> then calculate the moles of CCl<sub>2<\/sub>F<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) present. Use the ideal gas law <em data-effect=\"italics\">PV = nRT<\/em> to calculate the volume of CCl<sub>2<\/sub>F<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>):<\/div>\n<div id=\"fs-idm286430688\" data-type=\"solution\">\n<div data-type=\"newline\">\u00a0[latex]\\text{10.0 g}{\\text{CCl}}_{2}{\\text{F}}_{2}\\times \\frac{1\\text{mol}{\\text{CC1}}_{2}{\\text{F}}_{2}}{120.91\\text{g}{\\text{CCl}}_{2}{\\text{F}}_{2}}=0.0827\\text{mol}{\\text{CCl}}_{2}{\\text{F}}_{2}[\/latex]<\/div>\n<div data-type=\"newline\">\u00a0<em data-effect=\"italics\">PV = nRT<\/em>, where <em data-effect=\"italics\">n<\/em> = # mol CCl<sub>2<\/sub>F<sub>2<\/sub><\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">[latex]1\\text{atm}\\times V=0.0827\\text{mol}\\times \\frac{0.0821\\text{L atm}}{\\text{mol K}}\\times 273\\text{K}=1.85\\text{L}{\\text{CCl}}_{2}{\\text{F}}_{2};[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<p>(b) [latex]10.0\\text{g}{\\text{CH}}_{3}{\\text{CH}}_{2}\\text{F}\\times \\frac{1\\text{mol}{\\text{CH}}_{3}{\\text{CH}}_{2}\\text{F}}{48.07{\\text{g CH}}_{3}{\\text{CH}}_{2}\\text{F}}=0.208\\text{mol}{\\text{CH}}_{3}{\\text{CH}}_{2}\\text{F}[\/latex]<\/p>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\"><em data-effect=\"italics\">PV = nRT<\/em>, with <em data-effect=\"italics\">n<\/em> = # mol CH<sub>3<\/sub>CH<sub>2<\/sub>F<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">1 atm \u00d7 <em data-effect=\"italics\">V<\/em> = 0.208 mol \u00d7 0.0821 L atm\/mol K \u00d7 273 K = 4.66 L CH<sub>3<\/sub> CH<sub>2<\/sub> F<\/div>\n<\/div>\n<div id=\"fs-idm26646416\" data-type=\"exercise\">\n<div id=\"fs-idm15239616\" data-type=\"solution\">\n<p>28. Identify the variables in the problem and determine that the combined gas law [latex]\\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\\frac{{P}_{2}{V}_{2}}{{T}_{2}}[\/latex] is the necessary equation to use to solve the problem. Then solve for P<sub>2<\/sub>:<\/p>\n<div data-type=\"newline\">\n<div id=\"fs-idm23005472\" data-type=\"exercise\">\n<div id=\"fs-idm213823792\" data-type=\"solution\">\n<p data-type=\"newline\">\u00a0[latex]\\begin{array}{l}\\\\ \\\\ \\frac{0.981\\text{atm}\\times 100.21\\text{L}}{294\\text{K}}=\\frac{{P}_{2}\\times 144.53\\text{L}}{278.24\\text{atm}}\\\\ {P}_{2}=0.644\\text{atm}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm136587680\" data-type=\"exercise\">\n<div id=\"fs-idm232238480\" data-type=\"solution\">\n<p id=\"fs-idm232238224\">30. The pressure decreases by a factor of 3.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<div data-type=\"glossary\">\n<div class=\"bcc-box bcc-success\">\n<section id=\"glossary\">\n<h3>Glossary<\/h3>\n<div data-type=\"definition\">\n<p><strong><span data-type=\"term\">absolute zero<br \/>\n<\/span><\/strong>temperature at which the volume of a gas would be zero according to Charles\u2019s law.<\/p>\n<div id=\"fs-idp77543648\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">Amontons\u2019s law<br \/>\n<\/span><\/strong>(also, Gay-Lussac\u2019s law) pressure of a given number of moles of gas is directly proportional to its kelvin temperature when the volume is held constant<\/p>\n<\/div>\n<div id=\"fs-idm85214288\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">Avogadro\u2019s law<br \/>\n<\/span><\/strong>volume of a gas at constant temperature and pressure is proportional to the number of gas molecules<\/p>\n<\/div>\n<div id=\"fs-idm51295520\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">Boyle\u2019s law<br \/>\n<\/span><\/strong>volume of a given number of moles of gas held at constant temperature is inversely proportional to the pressure under which it is measured<\/p>\n<\/div>\n<div id=\"fs-idm100735488\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">Charles\u2019s law<br \/>\n<\/span><\/strong>volume of a given number of moles of gas is directly proportional to its kelvin temperature when the pressure is held constant<\/p>\n<\/div>\n<div id=\"fs-idm161542192\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">ideal gas<br \/>\n<\/span><\/strong>hypothetical gas whose physical properties are perfectly described by the gas laws<\/p>\n<\/div>\n<div id=\"fs-idm78852976\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">ideal gas constant (<em data-effect=\"italics\">R<\/em>)<br \/>\n<\/span><\/strong>constant derived from the ideal gas equation <em data-effect=\"italics\">R<\/em> = 0.08226 L atm mol<sup>\u20131<\/sup> K<sup>\u20131<\/sup> or 8.314 L kPa mol<sup>\u20131<\/sup> K<sup>\u20131<\/sup><\/p>\n<\/div>\n<div id=\"fs-idm111878496\" data-type=\"definition\"><span data-type=\"term\"><strong>ideal gas law<br \/>\n<\/strong><\/span>relation between the pressure, volume, amount, and temperature of a gas under conditions derived by combination of the simple gas laws<\/div>\n<div id=\"fs-idm158190720\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">standard conditions of temperature and pressure (STP)<br \/>\n<\/span><\/strong>273.15 K (0 \u00b0C) and 1 atm (101.325 kPa)<\/p>\n<\/div>\n<div id=\"fs-idm62647600\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">standard molar volume<br \/>\n<\/span><\/strong>volume of 1 mole of gas at STP, approximately 22.4 L for gases behaving ideally<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\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-2090\">\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. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Liquid Nitrogen Experiments: The Balloon. <strong>Authored by<\/strong>: Jefferson Lab. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ZgTTUuJZAFs\">https:\/\/youtu.be\/ZgTTUuJZAFs<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/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":17,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"Liquid Nitrogen Experiments: The Balloon\",\"author\":\"Jefferson Lab\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/ZgTTUuJZAFs\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at 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