{"id":1436,"date":"2018-08-14T03:18:50","date_gmt":"2018-08-14T03:18:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/?post_type=chapter&#038;p=1436"},"modified":"2018-08-15T04:34:25","modified_gmt":"2018-08-15T04:34:25","slug":"the-simple-gas-laws","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/the-simple-gas-laws\/","title":{"raw":"8.2 The Simple Gas Laws","rendered":"8.2 The Simple Gas Laws"},"content":{"raw":"<div class=\"textbox learning-objectives\">\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 combined 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>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>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 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<h2>Pressure and Temperature: Gay-Lussac\u2019s Law<\/h2>\r\nImagine 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.\r\n\r\n[caption id=\"attachment_596\" align=\"aligncenter\" width=\"1024\"]<img 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\r\nThis 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>P<\/em> and <em>T<\/em> are directly proportional (again, when <em>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.\r\n\r\n[caption id=\"attachment_597\" align=\"aligncenter\" width=\"1024\"]<img 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\r\nGuillaume 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>P<\/em>-<em>T<\/em> relationship for gases is known as either <strong>Gay-Lussac\u2019s law <\/strong>or<strong> Amontons's Law<\/strong>. Under either name, it states that <em>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:\r\n<p style=\"text-align: center\">[latex]\\large P\\propto T[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\large P=\\text{constant}\\times T[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\large P=k\\times T[\/latex]<\/p>\r\nwhere \u221d means \u201cis proportional to,\u201d and <em>k<\/em> is a proportionality constant that depends on the identity, amount, and volume of the gas.\r\n\r\nFor a confined, constant volume of gas, the ratio [latex]\\large\\frac{P}{T}[\/latex] is therefore constant (i.e., [latex]\\large\\frac{P}{T}=k[\/latex] ). If the gas is initially in \u201cCondition 1\u201d (with <em>P<\/em> = <em>P<\/em><sub>1<\/sub> and <em>T<\/em> = <em>T<\/em><sub>1<\/sub>), and then changes to \u201cCondition 2\u201d (with <em>P<\/em> = <em>P<\/em><sub>2<\/sub> and <em>T<\/em> = <em>T<\/em><sub>2<\/sub>), we have that [latex]\\large\\frac{{P}_{1}}{{T}_{1}}=k[\/latex] and [latex]\\frac{{P}_{2}}{{T}_{2}}=k,[\/latex] which reduces to:\r\n<p style=\"text-align: center\">[latex]\\large\\frac{{P}_{1}}{{T}_{1}}=\\frac{{P}_{2}}{{T}_{2}}[\/latex]<\/p>\r\nThis 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>absolute zero<\/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.)\r\n<div class=\"textbox examples\">\r\n<h3>Example 1: Predicting Change in Pressure with Temperature<\/h3>\r\nA can of hair spray is used until it is empty except for the propellant, isobutane gas.\r\n<ol>\r\n \t<li>On the can is the warning \u201cStore only at temperatures below 120 \u00b0F (48.8 \u00b0C). Do not incinerate.\u201d Why?<\/li>\r\n \t<li>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?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"10179\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"10179\"]\r\n<ol>\r\n \t<li>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.)<\/li>\r\n \t<li>We are looking for a pressure change due to a temperature change at constant volume, so we will use Gay-Lussac\u2019s law. Taking <em>P<\/em><sub>1<\/sub> and <em>T<\/em><sub>1<\/sub> as the initial values, <em>T<\/em><sub>2<\/sub> as the temperature where the pressure is unknown and <em>P<\/em><sub>2<\/sub> as the unknown pressure, and converting \u00b0C to K, we have:\r\n[latex]\\large\\frac{{P}_{1}}{{T}_{1}}=\\frac{{P}_{2}}{{T}_{2}}\\text{ which means that}\\frac{360\\text{ kPa}}{297\\text{ K}}=\\frac{{P}_{2}}{323\\text{ K}}[\/latex]\r\nRearranging and solving gives: [latex]\\large{P}_{2}=\\frac{360\\text{ kPa}\\times 323\\cancel{\\text{K}}}{297\\cancel{\\text{ K}}}=390\\text{ kPa}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/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\r\n[reveal-answer q=\"463971\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"463971\"]400 torr[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Volume and Temperature: Charles\u2019s Law<\/h2>\r\nIf 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.\r\n<div class=\"textbox\">\r\n\r\nThis video shows how cooling and heating a gas causes its volume to decrease or increase, respectively.\r\n\r\nhttps:\/\/youtu.be\/ZgTTUuJZAFs\r\n\r\n<\/div>\r\nThese 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.\r\n\r\n[caption id=\"attachment_599\" align=\"aligncenter\" width=\"1024\"]<img 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\r\nThe 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>Charles\u2019s law<\/strong> states that <em>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>.\r\n\r\nMathematically, this can be written as:\r\n<p style=\"text-align: center\">[latex]\\large V\\propto T[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\large V=\\text{constant}\\cdot T[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\large V=k\\cdot T[\/latex]<\/p>\r\nwith <em>k<\/em> being a proportionality constant that depends on the amount and pressure of the gas.\r\n\r\nFor a confined, constant pressure gas sample, [latex]\\large\\frac{V}{T}[\/latex] is constant (i.e., the ratio = <em>k<\/em>), and as seen with the <em>V<\/em>-<em>T<\/em> relationship, this leads to another form of Charles\u2019s law:\r\n<p style=\"text-align: center\">[latex]\\large\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}[\/latex]<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3><strong>Example 2: Predicting Change in Volume with Temperature<\/strong><\/h3>\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\r\n[reveal-answer q=\"302946\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"302946\"]\r\n\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>V<\/em><sub>1<\/sub> and <em>T<\/em><sub>1<\/sub> as the initial values, <em>T<\/em><sub>2<\/sub> as the temperature at which the volume is unknown and <em>V<\/em><sub>2<\/sub> as the unknown volume, and converting \u00b0C into K we have:\r\n<p style=\"text-align: center\">[latex]\\large\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}\\text{, which means that }\\frac{0.300\\text{ L}}{283\\text{ K}}=\\frac{{V}_{2}}{303\\text{ K}}[\/latex]<\/p>\r\nRearranging and solving gives: [latex]\\large{V}_{2}=\\frac{0.300\\text{L}\\times \\text{303}\\cancel{\\text{ K}}}{283\\cancel{\\text{K}}}=0.321\\text{ L}[\/latex]\r\n\r\nThis 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).\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/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\r\n[reveal-answer q=\"6826\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"6826\"]21.6 mL[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 3: Measuring Temperature with a Volume Change<\/h3>\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\r\n[reveal-answer q=\"476397\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"476397\"]\r\n\r\nA volume change caused by a temperature change at constant pressure means we should use Charles\u2019s law. Taking <em>V<\/em><sub>1<\/sub> and <em>T<\/em><sub>1<\/sub> as the initial values, <em>T<\/em><sub>2<\/sub> as the temperature at which the volume is unknown and <em>V<\/em><sub>2<\/sub> as the unknown volume, and converting \u00b0C into K we have:\r\n<p style=\"text-align: center\">[latex]\\large\\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]<\/p>\r\nRearrangement gives [latex]\\large{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]\r\n\r\nSubtracting 273.15 from 239.8 K, we find that the temperature of the boiling ammonia on the Celsius scale is \u201333.4 \u00b0C.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/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\r\n[reveal-answer q=\"990169\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"990169\"]635 mL[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Volume and Pressure: Boyle\u2019s Law<\/h2>\r\nIf 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.\r\n\r\n[caption id=\"attachment_600\" align=\"aligncenter\" width=\"548\"]<img 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=\"548\" height=\"488\" \/> 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\r\nUnlike the <em>P<\/em>-<em>T<\/em> and <em>V<\/em>-<em>T<\/em> relationships, pressure and volume are not directly proportional to each other. Instead, <em>P<\/em> and <em>V<\/em> exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written:\r\n<p style=\"text-align: center\">[latex]\\large P\\propto 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]<\/p>\r\n\r\n\r\n[caption id=\"attachment_601\" align=\"alignright\" width=\"400\"]<img 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=\"400\" height=\"204\" \/> 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>k<\/em> being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure [latex]\\large\\left(\\frac{1}{P}\\right)[\/latex] versus the volume (<em>V<\/em>), or the inverse of volume [latex]\\large\\left(\\frac{1}{V}\\right)[\/latex] versus the pressure (<em>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>P<\/em> versus <em>V<\/em>, we obtain a hyperbola (see Figure 6).\r\n\r\nThe 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>Boyle\u2019s law<\/strong>: <em>The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured.<\/em>\r\n<div class=\"textbox examples\">\r\n<h3>Example 4: Volume of a Gas Sample<\/h3>\r\nThe 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:\r\n<ol>\r\n \t<li>the <em>P<\/em>-<em>V<\/em> graph in Figure 5<\/li>\r\n \t<li>the [latex]\\large\\frac{1}{P}[\/latex] vs. <em>V<\/em> graph in Figure 5<\/li>\r\n \t<li>the Boyle\u2019s law equation<\/li>\r\n<\/ol>\r\nComment on the likely accuracy of each method.\r\n\r\n[reveal-answer q=\"787877\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"787877\"]\r\n<ol>\r\n \t<li>Estimating from the <em>P<\/em>-<em>V<\/em> graph gives a value for <em>P<\/em> somewhere around 27 psi.<\/li>\r\n \t<li>Estimating from the [latex]\\frac{1}{P}[\/latex] versus <em>V<\/em> graph give a value of about 26 psi.<\/li>\r\n \t<li>From Boyle\u2019s law, we know that the product of pressure and volume (<em>PV<\/em>) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have <em>P<\/em><sub>1<\/sub><em>V<\/em><sub>1<\/sub> = <em>k<\/em> and <em>P<\/em><sub>2<\/sub><em>V<\/em><sub>2<\/sub> = <em>k<\/em> which means that <em>P<\/em><sub>1<\/sub><em>V<\/em><sub>1<\/sub> = <em>P<\/em><sub>2<\/sub><em>V<\/em><sub>2<\/sub>.<\/li>\r\n<\/ol>\r\nUsing <em>P<\/em><sub>1<\/sub> and <em>V<\/em><sub>1<\/sub> as the known values 0.993 atm and 2.40 mL, <em>P<\/em><sub>2<\/sub> as the pressure at which the volume is unknown, and <em>V<\/em><sub>2<\/sub> as the unknown volume, we have:\r\n<p style=\"text-align: center\">[latex]\\large{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]<\/p>\r\nSolving:\r\n<p style=\"text-align: center\">[latex]\\large{V}_{2}=\\frac{13.0\\text{ psi}\\times 15.0\\cancel{\\text{mL}}}{7.5\\cancel{\\text{mL}}}=26\\text{ psi}[\/latex]<\/p>\r\nIt was more difficult to estimate well from the <em>P<\/em>-<em>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.\r\n\r\n[\/hidden-answer]\r\n<h4><strong>Check Your Learning<\/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<ol>\r\n \t<li>the <em>P<\/em>-<em>V<\/em> graph in Figure 5<\/li>\r\n \t<li>the [latex]\\frac{1}{P}[\/latex] vs. <em>V<\/em> graph in Figure 5<\/li>\r\n \t<li>the Boyle\u2019s law equation<\/li>\r\n<\/ol>\r\nComment on the likely accuracy of each method.\r\n\r\n[reveal-answer q=\"773812\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"773812\"]\r\n<ol>\r\n \t<li>about 17\u201318 mL<\/li>\r\n \t<li>~18 mL<\/li>\r\n \t<li>17.7 mL<\/li>\r\n<\/ol>\r\nIt was more difficult to estimate well from the <em>P<\/em>-<em>V<\/em> graph, so (1) is likely more inaccurate than (2); the calculation will be as accurate as the equation and measurements allow.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Chemistry in Action: Breathing 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=\"578\"]<img 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=\"578\" height=\"434\" \/> 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<h2>Moles of Gas and Volume: Avogadro\u2019s Law<\/h2>\r\nThe 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>Avogadro\u2019s law<\/strong>: <em>For a confined gas, the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both remain constant<\/em>.\r\n\r\nIn equation form, this is written as:\r\n<p style=\"text-align: center\">[latex]\\large\\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]<\/p>\r\nMathematical relationships can also be determined for the other variable pairs, such as <em>P<\/em> versus <em>n<\/em>, and <em>n<\/em> versus <em>T<\/em>.\r\n<div class=\"textbox\">Visit this interactive <a href=\"http:\/\/phet.colorado.edu\/en\/simulation\/gas-properties\" target=\"_blank\" rel=\"noopener noreferrer\">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).<\/div>\r\n<div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 5: Measuring Volume with Change in moles<\/h3>\r\nA 2.45 L volume of gas contains 4.5 moles of gas. How many moles of gas are there in 3.87 L if the gas is at constant pressure and temperature?\r\n\r\n[reveal-answer q=\"476388\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"476388\"]\r\n<p id=\"ball-ch06_s04_p07\" class=\"para\">We can set up Avogadro\u2019s law as follows:<\/p>\r\n<p style=\"text-align: center\">[latex]\\large\\frac{{2.45\\text{ L}}}{{4.5\\text{ mol}}}=\\frac{{3.87\\text{ L}}}{{n}_{2}}[\/latex]<\/p>\r\n<p id=\"ball-ch06_s04_p08\" class=\"para\" style=\"text-align: left\">We algebraically rearrange to solve for <em class=\"emphasis\">n<\/em><sub class=\"subscript\">2<\/sub>:<\/p>\r\n<p style=\"text-align: center\">[latex]\\large\\frac{{(3.87\\cancel{\\text{ L})}}{(4.5\\text{ mol})}}{{2.45\\cancel{\\text{ L})}}}={n}_{2}[\/latex]<\/p>\r\n<p id=\"ball-ch06_s04_p09\" class=\"para\" style=\"text-align: left\">The L units cancel, so we solve for <em class=\"emphasis\">n<\/em><sub class=\"subscript\">2<\/sub>:<\/p>\r\n<p style=\"text-align: left\"><span class=\"informalequation\"><span class=\"mathphrase\"><em class=\"emphasis\">n<\/em><sub class=\"subscript\">2<\/sub> = 7.1 moles\r\n<\/span><\/span><\/p>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nA 12.8 L volume of gas contains 3.00 moles of gas. At constant temperature and pressure, what will be the volume of gas if 5.22 moles of gas is added?\r\n\r\n[reveal-answer q=\"990133\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"990133\"]34.1 L[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Combined Gas Law<\/h2>\r\n<p id=\"ball-ch06_s04_p13\" class=\"para editable block\">One thing we notice about all the gas laws is that, collectively, volume and pressure are always in the numerator, and temperature is always in the denominator. This suggests that we can propose a gas law that combines pressure, volume, and temperature. This gas law is known as the <span class=\"margin_term\"><a class=\"glossterm\">combined gas law<\/a><\/span>, and its mathematical form is<\/p>\r\n<p style=\"text-align: center\">[latex]\\large\\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\\frac{{P}_{2}{V}_{2}}{{T}_{2}}\\text{ with n constant}[\/latex]<\/p>\r\n<p id=\"ball-ch06_s04_p14\" class=\"para editable block\">This allows us to follow changes in all three major properties of a gas. Again, the usual warnings apply about how to solve for an unknown algebraically (isolate it on one side of the equation in the numerator), units (they must be the same for the two similar variables of each type), and units of temperature (must be in kelvins).<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 6: Combined gas law<\/h3>\r\nA sample of gas at an initial volume of 8.33 L, an initial pressure of 1.82 atm, and an initial temperature of 286 K simultaneously changes its temperature to 355 K and its volume to 5.72 L. What is the final pressure of the gas?\r\n\r\n[reveal-answer q=\"476387\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"476387\"]\r\n<p id=\"ball-ch06_s04_p16\" class=\"para\">We can use the combined gas law directly; all the units are consistent with each other, and the temperatures are given in Kelvin. Substituting,<\/p>\r\n<p style=\"text-align: center\">[latex]\\large\\frac{{(1.82\\text{ atm})}{(8.33\\text{ L})}}{{286\\text{ K}}}=\\frac{{P}_{2}{(5.72\\text{ L})}}{{355\\text{ K}}}[\/latex]<\/p>\r\n<p id=\"ball-ch06_s04_p17\" class=\"para\">We rearrange this to isolate the <em class=\"emphasis\">P<\/em><sub class=\"subscript\">2<\/sub> variable all by itself. When we do so, certain units cancel:<\/p>\r\n<p style=\"text-align: center\">[latex]\\large\\frac{{(1.82\\text{ atm})}{(8.33\\cancel{\\text{ L}})}{(355\\cancel{\\text{ K}})}}{{(286\\cancel{\\text{ K}})(5.72\\cancel{\\text{ L})}}}={P}_{2}[\/latex]<\/p>\r\n<p id=\"ball-ch06_s04_p18\" class=\"para\">Multiplying and dividing all the numbers, we get<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\"><em class=\"emphasis\">P<\/em><sub class=\"subscript\">2<\/sub> = 3.29 atm<\/span><\/span><\/p>\r\n<p id=\"ball-ch06_s04_p19\" class=\"para\">Ultimately, the pressure increased, which would have been difficult to predict because two properties of the gas were changing.<\/p>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nA sample of gas at an initial volume of 46.7 mL, an initial pressure of 662 torr, and an initial temperature of 266 K simultaneously changes its temperature to 371 K and its pressure to 40 torr. What is the final volume of the gas?\r\n\r\n[reveal-answer q=\"990144\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"990144\"]105 mL[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"ball-ch06_s04_p22\" class=\"para editable block\">As with other gas laws, if you need to determine the value of a variable in the denominator of the combined gas law, you can either cross-multiply all the terms or just take the reciprocal of the combined gas law. Remember, the variable you are solving for must be in the numerator and all by itself on one side of the equation.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\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 (Gay-Lussac's law). The volume of a given amount of 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<h4>Key Equations<\/h4>\r\n<ul>\r\n \t<li>Gay-Lussac's Law: [latex]\\large\\frac{{P}_{1}}{{T}_{1}}=\\frac{{P}_{2}}{{T}_{2}}[\/latex]<\/li>\r\n \t<li>Charles's Law: [latex]\\large\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}[\/latex]<\/li>\r\n \t<li>Boyle's Law: [latex]\\large{P}_{1}{V}_{1}={P}_{2}{V}_{2}[\/latex]<\/li>\r\n \t<li>Combined Gas Law: [latex]\\large\\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\\frac{{P}_{2}{V}_{2}}{{T}_{2}}\\text{ with n constant}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li>Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?<\/li>\r\n \t<li>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>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 style=\"list-style-type: lower-alpha\">\r\n \t<li>What is the meaning of the term \u201cinversely proportional?\u201d<\/li>\r\n \t<li>What are the \u201cother things\u201d that must be equal?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>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 style=\"list-style-type: lower-alpha\">\r\n \t<li>What is the meaning of the term \u201cdirectly proportional?\u201d<\/li>\r\n \t<li>What are the \u201cother things\u201d that must be equal?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>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>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>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>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>Determine the pressure of the gas in the syringe shown in Figure 5 when its volume is 12.5 mL, using:\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>the appropriate graph<\/li>\r\n \t<li>Boyle\u2019s law<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>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>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>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>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?<\/li>\r\n \t<li>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>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>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>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>How many grams of gas are present in each of the following cases?\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>0.100 L of CO<sub>2<\/sub> at 307 torr and 26 \u00b0C<\/li>\r\n \t<li>8.75 L of C<sub>2<\/sub>H<sub>4<\/sub>, at 378.3 kPa and 483 K<\/li>\r\n \t<li>221 mL of Ar at 0.23 torr and \u201354 \u00b0C<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>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>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>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>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>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>For a given amount of gas showing ideal behavior, draw labeled graphs of:\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>the variation of <em>P<\/em> with <em>V <\/em><\/li>\r\n \t<li>the variation of <em>V<\/em> with <em>T <\/em><\/li>\r\n \t<li>the variation of <em>P<\/em> with <em>T <\/em><\/li>\r\n \t<li>the variation of [latex]\\frac{1}{P}[\/latex] with <em>V<\/em><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>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>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>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 style=\"list-style-type: lower-alpha\">\r\n \t<li>CCl<sub>2<\/sub>F<sub>2<\/sub>(<em>g<\/em>)<\/li>\r\n \t<li>CH<sub>3<\/sub>CH<sub>2<\/sub>F(<em>g<\/em>)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>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>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>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>If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?<\/li>\r\n \t<li id=\"ball-ch06_s04_qs01_qd01_qa11\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch06_s04_qs01_p19\" class=\"para\">A gas starts at the conditions 78.9 mL, 3.008 atm, and 56\u00b0C. Its conditions change to 35.6 mL and 2.55 atm. What is its final temperature?<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"ball-ch06_s04_qs01_qd01_qa12\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch06_s04_qs01_p21\" class=\"para\">The initial conditions of a sample of gas are 319 K, 3.087 L, and 591 torr. What is its final pressure if volume is changed to 2.222 L and temperature is changed to 299 K?<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"ball-ch06_s04_qs01_qd01_qa13\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch06_s04_qs01_p23\" class=\"para\">A gas starts with initial pressure of 7.11 atm, initial temperature of 66\u00b0C, and initial volume of 90.7 mL. If its conditions change to 33\u00b0C and 14.33 atm, what is its final volume?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n[reveal-answer q=\"669834\"]Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"669834\"]\r\n\r\n2. As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle\u2019s law.\r\n\r\n4. The answers are as follows:\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>The number of particles in the gas increases as the volume increases. This relationship may be written as <em>n<\/em> = constant \u00d7 <em>V<\/em>. It is a direct relationship.<\/li>\r\n \t<li>The temperature and pressure must be kept constant.<\/li>\r\n<\/ol>\r\n6. The curve would be farther to the right and higher up, but the same basic shape.\r\n\r\n8. 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.\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 style=\"text-align: center\">[latex]\\large\\frac{{P}_{2}}{{T}_{2}}=\\frac{{P}_{1}}{{T1}_{}}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\large{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>\r\n12. Apply Charles\u2019s law to compute the volume of gas at the higher temperature:\r\n<ul>\r\n \t<li><em>V<\/em><sub>1<\/sub> = 2.50 L<\/li>\r\n \t<li><em>T<\/em><sub>1<\/sub> = \u2013193 \u00b0C = 77.15 K<\/li>\r\n \t<li><em>V<\/em><sub>2<\/sub> = ?<\/li>\r\n \t<li><em>T<\/em><sub>2<\/sub> = 100 \u00b0C = 373.15 K<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center\">[latex]\\large\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\large{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]<\/p>\r\n14. <em>PV<\/em> = <em>nRT<\/em>\r\n<p style=\"text-align: center\">[latex]\\large V=\\frac{nRT}{P}=\\frac{8.80\\cancel{\\text{mol}}\\times 0.08206\\text{ L}\\cancel{\\text{atm}}{\\cancel{\\text{mol}}}^{{-1}}\\cancel{{\\text{K}}^{{-1}}}\\times 298.15\\cancel{\\text{K}}}{0.992\\cancel{\\text{atm}}}=217\\text{ L}[\/latex]<\/p>\r\n16. [latex]\\large 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}{{-1}}^{}\\cancel{{\\text{K}}^{{-1}}}\\right)\\left(788.0\\cancel{\\text{K}}\\right)}=0.08190\\text{mol}=8.190\\times {10}^{{-2}}\\text{mol}[\/latex]\r\n\r\n[latex]\\large n\\times \\text{molar mass}=8.190\\times {10}^{{-2}}\\cancel{\\text{mol}}\\times 67.8052\\text{g}{\\cancel{\\text{mol}}}^{{-1}}=5.553\\text{g}[\/latex]\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>m<\/em> = <em>n\u2133<\/em>, where \u2133 is the molar mass:\r\n<p style=\"text-align: center\">[latex]\\large P,V,T\\,\\,\\,{\\xrightarrow{n=PV\\text{\/}RT}}\\,\\,\\,n,\\,\\,\\,{\\xrightarrow{m=n\\left(\\text{molar mass}\\right)}}\\,\\,\\,\\text{grams}[\/latex]<\/p>\r\nor we can combine these equations to obtain:\r\n<p style=\"text-align: center\">[latex]\\large\\text{mass}=m=\\frac{PV}{RT}\\times [\/latex]\u2133<\/p>\r\n\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>[latex]\\large\\begin{array}{l}\\\\307\\cancel{\\text{torr}}\\times \\frac{1\\text{atm}}{760\\cancel{\\text{torr}}}=0.4039\\text{ atm }25^\\circ{\\text{ C}}=299.1 \\text{ 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}}^{{-1}}\\cancel{{\\text{K}}^{{-1}}}\\left(299.1\\cancel{\\text{K}}\\right)}\\times 44.01\\text{g}{\\text{mol}}^{{-1}}=7.24\\times {10}^{{-2}}\\text{g}\\end{array}[\/latex]<\/li>\r\n \t<li>[latex]\\large\\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}}^{{-1}}\\cancel{{\\text{K}}^{{-1}}}\\left(483\\cancel{\\text{K}}\\right)}\\times 28.05376\\text{ g}{\\text{mol}}^{{-1}}=23.1\\text{g}[\/latex]<\/li>\r\n \t<li>[latex]\\large\\begin{array}{l}\\\\ \\\\ 221\\cancel{\\text{mL}}\\times \\frac{1\\text{L}}{1000\\cancel{\\text{mL}}}=0.221\\text{L}-54^{\\circ}\\text{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}^{{-4}}\\text{atm}\\\\ \\text{Mass}=m=\\frac{3.03\\times {10}^{{-4}}\\cancel{\\text{atm}}\\left(0.221\\cancel{\\text{L}}\\right)}{0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{{-1}}\\cancel{{\\text{K}}^{{-1}}}\\left(219.15\\cancel{\\text{K}}\\right)}\\times 39.978\\text{ g}{\\text{mol}}^{{-1}}=1.5\\times {10}^{{-4}}\\text{g}\\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n20. [latex]\\large\\frac{{P}_{2}}{{T}_{2}}=\\frac{{P}_{1}}{{T}_{1}}[\/latex]\r\n\r\n<em>T<\/em><sub>2<\/sub> = 49.5 + 273.15 = 322.65 K\r\n\r\n[latex]\\large{P}_{2}=\\frac{{P}_{1}{T}_{2}}{{T}_{1}}=149.6\\text{atm}\\times \\frac{322.65}{278.15}=173.5\\text{ atm}[\/latex]\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]\\large 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}}^{{-1}}\\cancel{{\\text{K}}^{{-1}}}\\left(300.1\\cancel{\\text{K}}\\right)}=0.798\\text{mol}[\/latex] Mass of butane = 0.798 mol \u00d7 58.1234 g\/mol = 46.4 g\r\n\r\n24. For a gas exhibiting ideal behavior: <img class=\"aligncenter\" 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\" width=\"518\" height=\"439\" \/>\r\n\r\n26. The volume is as follows:\r\n<ol style=\"list-style-type: lower-alpha\">\r\n \t<li>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>g<\/em>) present. Use the ideal gas law <em>PV = nRT<\/em> to calculate the volume of CCl<sub>2<\/sub>F<sub>2<\/sub>(<em>g<\/em>):\r\n[latex]\\large\\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]\r\n<em>PV = nRT<\/em>, where <em>n<\/em> = # mol CCl<sub>2<\/sub>F<sub>2\r\n<\/sub>[latex]\\large1\\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]<\/li>\r\n \t<li>[latex]\\large10.0\\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<em>PV = nRT<\/em>, with <em>n<\/em> = # mol CH<sub>3<\/sub>CH<sub>2<\/sub>F\r\n1 atm \u00d7 <em>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<\/li>\r\n<\/ol>\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<p style=\"text-align: center\">[latex]\\large\\begin{array}{rcl}{}\\frac{0.981\\text{ atm}\\times 100.21\\text{ L}}{294\\text{ K}}&amp;=&amp;\\frac{{P}_{2}\\times 144.53\\text{ L}}{278.24\\text{ atm}}\\\\ {P}_{2}&amp;=&amp;0.644\\text{ atm}\\end{array}[\/latex]<\/p>\r\n30. The pressure decreases by a factor of 3.\r\n\r\n31. 126 K, or \u2212147\u00b0C\r\n\r\n33. 40.6 mL\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>absolute zero: <\/strong>temperature at which the volume of a gas would be zero according to Charles\u2019s law.\r\n\r\n<strong>Avogadro\u2019s law: <\/strong>volume of a gas at constant temperature and pressure is proportional to the number of gas molecules\r\n\r\n<strong>Boyle\u2019s law: <\/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<strong>Charles\u2019s law: <\/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<strong>Gay-Lussac\u2019s law: <\/strong>(also, <strong>Amontons\u2019s law<\/strong>) 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&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\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 combined 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>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>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\" 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<h2>Pressure and Temperature: Gay-Lussac\u2019s Law<\/h2>\n<p>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\" 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>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>P<\/em> and <em>T<\/em> are directly proportional (again, when <em>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\" 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>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>P<\/em>&#8211;<em>T<\/em> relationship for gases is known as either <strong>Gay-Lussac\u2019s law <\/strong>or<strong> Amontons&#8217;s Law<\/strong>. Under either name, it states that <em>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<p style=\"text-align: center\">[latex]\\large P\\propto T[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\large P=\\text{constant}\\times T[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\large P=k\\times T[\/latex]<\/p>\n<p>where \u221d means \u201cis proportional to,\u201d and <em>k<\/em> is a proportionality constant that depends on the identity, amount, and volume of the gas.<\/p>\n<p>For a confined, constant volume of gas, the ratio [latex]\\large\\frac{P}{T}[\/latex] is therefore constant (i.e., [latex]\\large\\frac{P}{T}=k[\/latex] ). If the gas is initially in \u201cCondition 1\u201d (with <em>P<\/em> = <em>P<\/em><sub>1<\/sub> and <em>T<\/em> = <em>T<\/em><sub>1<\/sub>), and then changes to \u201cCondition 2\u201d (with <em>P<\/em> = <em>P<\/em><sub>2<\/sub> and <em>T<\/em> = <em>T<\/em><sub>2<\/sub>), we have that [latex]\\large\\frac{{P}_{1}}{{T}_{1}}=k[\/latex] and [latex]\\frac{{P}_{2}}{{T}_{2}}=k,[\/latex] which reduces to:<\/p>\n<p style=\"text-align: center\">[latex]\\large\\frac{{P}_{1}}{{T}_{1}}=\\frac{{P}_{2}}{{T}_{2}}[\/latex]<\/p>\n<p>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>absolute zero<\/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 class=\"textbox examples\">\n<h3>Example 1: Predicting Change in Pressure with Temperature<\/h3>\n<p>A can of hair spray is used until it is empty except for the propellant, isobutane gas.<\/p>\n<ol>\n<li>On the can is the warning \u201cStore only at temperatures below 120 \u00b0F (48.8 \u00b0C). Do not incinerate.\u201d Why?<\/li>\n<li>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?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q10179\">Show Answer<\/span><\/p>\n<div id=\"q10179\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>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.)<\/li>\n<li>We are looking for a pressure change due to a temperature change at constant volume, so we will use Gay-Lussac\u2019s law. Taking <em>P<\/em><sub>1<\/sub> and <em>T<\/em><sub>1<\/sub> as the initial values, <em>T<\/em><sub>2<\/sub> as the temperature where the pressure is unknown and <em>P<\/em><sub>2<\/sub> as the unknown pressure, and converting \u00b0C to K, we have:<br \/>\n[latex]\\large\\frac{{P}_{1}}{{T}_{1}}=\\frac{{P}_{2}}{{T}_{2}}\\text{ which means that}\\frac{360\\text{ kPa}}{297\\text{ K}}=\\frac{{P}_{2}}{323\\text{ K}}[\/latex]<br \/>\nRearranging and solving gives: [latex]\\large{P}_{2}=\\frac{360\\text{ kPa}\\times 323\\cancel{\\text{K}}}{297\\cancel{\\text{ K}}}=390\\text{ kPa}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q463971\">Show Answer<\/span><\/p>\n<div id=\"q463971\" class=\"hidden-answer\" style=\"display: none\">400 torr<\/div>\n<\/div>\n<\/div>\n<h2>Volume and Temperature: Charles\u2019s Law<\/h2>\n<p>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 class=\"textbox\">\n<p>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>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\" 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>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>Charles\u2019s law<\/strong> states that <em>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>Mathematically, this can be written as:<\/p>\n<p style=\"text-align: center\">[latex]\\large V\\propto T[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\large V=\\text{constant}\\cdot T[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\large V=k\\cdot T[\/latex]<\/p>\n<p>with <em>k<\/em> being a proportionality constant that depends on the amount and pressure of the gas.<\/p>\n<p>For a confined, constant pressure gas sample, [latex]\\large\\frac{V}{T}[\/latex] is constant (i.e., the ratio = <em>k<\/em>), and as seen with the <em>V<\/em>&#8211;<em>T<\/em> relationship, this leads to another form of Charles\u2019s law:<\/p>\n<p style=\"text-align: center\">[latex]\\large\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3><strong>Example 2: Predicting Change in Volume with Temperature<\/strong><\/h3>\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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q302946\">Show Answer<\/span><\/p>\n<div id=\"q302946\" class=\"hidden-answer\" style=\"display: none\">\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>V<\/em><sub>1<\/sub> and <em>T<\/em><sub>1<\/sub> as the initial values, <em>T<\/em><sub>2<\/sub> as the temperature at which the volume is unknown and <em>V<\/em><sub>2<\/sub> as the unknown volume, and converting \u00b0C into K we have:<\/p>\n<p style=\"text-align: center\">[latex]\\large\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}\\text{, which means that }\\frac{0.300\\text{ L}}{283\\text{ K}}=\\frac{{V}_{2}}{303\\text{ K}}[\/latex]<\/p>\n<p>Rearranging and solving gives: [latex]\\large{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>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<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q6826\">Show Answer<\/span><\/p>\n<div id=\"q6826\" class=\"hidden-answer\" style=\"display: none\">21.6 mL<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 3: Measuring Temperature with a Volume Change<\/h3>\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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q476397\">Show Answer<\/span><\/p>\n<div id=\"q476397\" class=\"hidden-answer\" style=\"display: none\">\n<p>A volume change caused by a temperature change at constant pressure means we should use Charles\u2019s law. Taking <em>V<\/em><sub>1<\/sub> and <em>T<\/em><sub>1<\/sub> as the initial values, <em>T<\/em><sub>2<\/sub> as the temperature at which the volume is unknown and <em>V<\/em><sub>2<\/sub> as the unknown volume, and converting \u00b0C into K we have:<\/p>\n<p style=\"text-align: center\">[latex]\\large\\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]<\/p>\n<p>Rearrangement gives [latex]\\large{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>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<\/div>\n<\/div>\n<h4>Check Your Learning<\/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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q990169\">Show Answer<\/span><\/p>\n<div id=\"q990169\" class=\"hidden-answer\" style=\"display: none\">635 mL<\/div>\n<\/div>\n<\/div>\n<h2>Volume and Pressure: Boyle\u2019s Law<\/h2>\n<p>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: 558px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-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=\"548\" height=\"488\" \/><\/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>Unlike the <em>P<\/em>&#8211;<em>T<\/em> and <em>V<\/em>&#8211;<em>T<\/em> relationships, pressure and volume are not directly proportional to each other. Instead, <em>P<\/em> and <em>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<p style=\"text-align: center\">[latex]\\large P\\propto 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]<\/p>\n<div id=\"attachment_601\" style=\"width: 410px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-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=\"400\" height=\"204\" \/><\/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>k<\/em> being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure [latex]\\large\\left(\\frac{1}{P}\\right)[\/latex] versus the volume (<em>V<\/em>), or the inverse of volume [latex]\\large\\left(\\frac{1}{V}\\right)[\/latex] versus the pressure (<em>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>P<\/em> versus <em>V<\/em>, we obtain a hyperbola (see Figure 6).<\/p>\n<p>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>Boyle\u2019s law<\/strong>: <em>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 class=\"textbox examples\">\n<h3>Example 4: Volume of a Gas Sample<\/h3>\n<p>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<ol>\n<li>the <em>P<\/em>&#8211;<em>V<\/em> graph in Figure 5<\/li>\n<li>the [latex]\\large\\frac{1}{P}[\/latex] vs. <em>V<\/em> graph in Figure 5<\/li>\n<li>the Boyle\u2019s law equation<\/li>\n<\/ol>\n<p>Comment on the likely accuracy of each method.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q787877\">Show Answer<\/span><\/p>\n<div id=\"q787877\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Estimating from the <em>P<\/em>&#8211;<em>V<\/em> graph gives a value for <em>P<\/em> somewhere around 27 psi.<\/li>\n<li>Estimating from the [latex]\\frac{1}{P}[\/latex] versus <em>V<\/em> graph give a value of about 26 psi.<\/li>\n<li>From Boyle\u2019s law, we know that the product of pressure and volume (<em>PV<\/em>) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have <em>P<\/em><sub>1<\/sub><em>V<\/em><sub>1<\/sub> = <em>k<\/em> and <em>P<\/em><sub>2<\/sub><em>V<\/em><sub>2<\/sub> = <em>k<\/em> which means that <em>P<\/em><sub>1<\/sub><em>V<\/em><sub>1<\/sub> = <em>P<\/em><sub>2<\/sub><em>V<\/em><sub>2<\/sub>.<\/li>\n<\/ol>\n<p>Using <em>P<\/em><sub>1<\/sub> and <em>V<\/em><sub>1<\/sub> as the known values 0.993 atm and 2.40 mL, <em>P<\/em><sub>2<\/sub> as the pressure at which the volume is unknown, and <em>V<\/em><sub>2<\/sub> as the unknown volume, we have:<\/p>\n<p style=\"text-align: center\">[latex]\\large{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]<\/p>\n<p>Solving:<\/p>\n<p style=\"text-align: center\">[latex]\\large{V}_{2}=\\frac{13.0\\text{ psi}\\times 15.0\\cancel{\\text{mL}}}{7.5\\cancel{\\text{mL}}}=26\\text{ psi}[\/latex]<\/p>\n<p>It was more difficult to estimate well from the <em>P<\/em>&#8211;<em>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<\/div>\n<\/div>\n<h4><strong>Check Your Learning<\/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<ol>\n<li>the <em>P<\/em>&#8211;<em>V<\/em> graph in Figure 5<\/li>\n<li>the [latex]\\frac{1}{P}[\/latex] vs. <em>V<\/em> graph in Figure 5<\/li>\n<li>the Boyle\u2019s law equation<\/li>\n<\/ol>\n<p>Comment on the likely accuracy of each method.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q773812\">Show Answer<\/span><\/p>\n<div id=\"q773812\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>about 17\u201318 mL<\/li>\n<li>~18 mL<\/li>\n<li>17.7 mL<\/li>\n<\/ol>\n<p>It was more difficult to estimate well from the <em>P<\/em>&#8211;<em>V<\/em> graph, so (1) is likely more inaccurate than (2); the calculation will be as accurate as the equation and measurements allow.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Chemistry in Action: Breathing 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: 588px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-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=\"578\" height=\"434\" \/><\/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<h2>Moles of Gas and Volume: Avogadro\u2019s Law<\/h2>\n<p>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>Avogadro\u2019s law<\/strong>: <em>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>In equation form, this is written as:<\/p>\n<p style=\"text-align: center\">[latex]\\large\\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]<\/p>\n<p>Mathematical relationships can also be determined for the other variable pairs, such as <em>P<\/em> versus <em>n<\/em>, and <em>n<\/em> versus <em>T<\/em>.<\/p>\n<div class=\"textbox\">Visit this interactive <a href=\"http:\/\/phet.colorado.edu\/en\/simulation\/gas-properties\" target=\"_blank\" rel=\"noopener noreferrer\">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).<\/div>\n<div>\n<div class=\"textbox examples\">\n<h3>Example 5: Measuring Volume with Change in moles<\/h3>\n<p>A 2.45 L volume of gas contains 4.5 moles of gas. How many moles of gas are there in 3.87 L if the gas is at constant pressure and temperature?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q476388\">Show Answer<\/span><\/p>\n<div id=\"q476388\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"ball-ch06_s04_p07\" class=\"para\">We can set up Avogadro\u2019s law as follows:<\/p>\n<p style=\"text-align: center\">[latex]\\large\\frac{{2.45\\text{ L}}}{{4.5\\text{ mol}}}=\\frac{{3.87\\text{ L}}}{{n}_{2}}[\/latex]<\/p>\n<p id=\"ball-ch06_s04_p08\" class=\"para\" style=\"text-align: left\">We algebraically rearrange to solve for <em class=\"emphasis\">n<\/em><sub class=\"subscript\">2<\/sub>:<\/p>\n<p style=\"text-align: center\">[latex]\\large\\frac{{(3.87\\cancel{\\text{ L})}}{(4.5\\text{ mol})}}{{2.45\\cancel{\\text{ L})}}}={n}_{2}[\/latex]<\/p>\n<p id=\"ball-ch06_s04_p09\" class=\"para\" style=\"text-align: left\">The L units cancel, so we solve for <em class=\"emphasis\">n<\/em><sub class=\"subscript\">2<\/sub>:<\/p>\n<p style=\"text-align: left\"><span class=\"informalequation\"><span class=\"mathphrase\"><em class=\"emphasis\">n<\/em><sub class=\"subscript\">2<\/sub> = 7.1 moles<br \/>\n<\/span><\/span><\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>A 12.8 L volume of gas contains 3.00 moles of gas. At constant temperature and pressure, what will be the volume of gas if 5.22 moles of gas is added?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q990133\">Show Answer<\/span><\/p>\n<div id=\"q990133\" class=\"hidden-answer\" style=\"display: none\">34.1 L<\/div>\n<\/div>\n<\/div>\n<h2>The Combined Gas Law<\/h2>\n<p id=\"ball-ch06_s04_p13\" class=\"para editable block\">One thing we notice about all the gas laws is that, collectively, volume and pressure are always in the numerator, and temperature is always in the denominator. This suggests that we can propose a gas law that combines pressure, volume, and temperature. This gas law is known as the <span class=\"margin_term\"><a class=\"glossterm\">combined gas law<\/a><\/span>, and its mathematical form is<\/p>\n<p style=\"text-align: center\">[latex]\\large\\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\\frac{{P}_{2}{V}_{2}}{{T}_{2}}\\text{ with n constant}[\/latex]<\/p>\n<p id=\"ball-ch06_s04_p14\" class=\"para editable block\">This allows us to follow changes in all three major properties of a gas. Again, the usual warnings apply about how to solve for an unknown algebraically (isolate it on one side of the equation in the numerator), units (they must be the same for the two similar variables of each type), and units of temperature (must be in kelvins).<\/p>\n<div class=\"textbox examples\">\n<h3>Example 6: Combined gas law<\/h3>\n<p>A sample of gas at an initial volume of 8.33 L, an initial pressure of 1.82 atm, and an initial temperature of 286 K simultaneously changes its temperature to 355 K and its volume to 5.72 L. What is the final pressure of the gas?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q476387\">Show Answer<\/span><\/p>\n<div id=\"q476387\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"ball-ch06_s04_p16\" class=\"para\">We can use the combined gas law directly; all the units are consistent with each other, and the temperatures are given in Kelvin. Substituting,<\/p>\n<p style=\"text-align: center\">[latex]\\large\\frac{{(1.82\\text{ atm})}{(8.33\\text{ L})}}{{286\\text{ K}}}=\\frac{{P}_{2}{(5.72\\text{ L})}}{{355\\text{ K}}}[\/latex]<\/p>\n<p id=\"ball-ch06_s04_p17\" class=\"para\">We rearrange this to isolate the <em class=\"emphasis\">P<\/em><sub class=\"subscript\">2<\/sub> variable all by itself. When we do so, certain units cancel:<\/p>\n<p style=\"text-align: center\">[latex]\\large\\frac{{(1.82\\text{ atm})}{(8.33\\cancel{\\text{ L}})}{(355\\cancel{\\text{ K}})}}{{(286\\cancel{\\text{ K}})(5.72\\cancel{\\text{ L})}}}={P}_{2}[\/latex]<\/p>\n<p id=\"ball-ch06_s04_p18\" class=\"para\">Multiplying and dividing all the numbers, we get<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\"><em class=\"emphasis\">P<\/em><sub class=\"subscript\">2<\/sub> = 3.29 atm<\/span><\/span><\/p>\n<p id=\"ball-ch06_s04_p19\" class=\"para\">Ultimately, the pressure increased, which would have been difficult to predict because two properties of the gas were changing.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>A sample of gas at an initial volume of 46.7 mL, an initial pressure of 662 torr, and an initial temperature of 266 K simultaneously changes its temperature to 371 K and its pressure to 40 torr. What is the final volume of the gas?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q990144\">Show Answer<\/span><\/p>\n<div id=\"q990144\" class=\"hidden-answer\" style=\"display: none\">105 mL<\/div>\n<\/div>\n<\/div>\n<p id=\"ball-ch06_s04_p22\" class=\"para editable block\">As with other gas laws, if you need to determine the value of a variable in the denominator of the combined gas law, you can either cross-multiply all the terms or just take the reciprocal of the combined gas law. Remember, the variable you are solving for must be in the numerator and all by itself on one side of the equation.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\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 (Gay-Lussac&#8217;s law). The volume of a given amount of 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<h4>Key Equations<\/h4>\n<ul>\n<li>Gay-Lussac&#8217;s Law: [latex]\\large\\frac{{P}_{1}}{{T}_{1}}=\\frac{{P}_{2}}{{T}_{2}}[\/latex]<\/li>\n<li>Charles&#8217;s Law: [latex]\\large\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}[\/latex]<\/li>\n<li>Boyle&#8217;s Law: [latex]\\large{P}_{1}{V}_{1}={P}_{2}{V}_{2}[\/latex]<\/li>\n<li>Combined Gas Law: [latex]\\large\\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\\frac{{P}_{2}{V}_{2}}{{T}_{2}}\\text{ with n constant}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li>Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?<\/li>\n<li>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>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 style=\"list-style-type: lower-alpha\">\n<li>What is the meaning of the term \u201cinversely proportional?\u201d<\/li>\n<li>What are the \u201cother things\u201d that must be equal?<\/li>\n<\/ol>\n<\/li>\n<li>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 style=\"list-style-type: lower-alpha\">\n<li>What is the meaning of the term \u201cdirectly proportional?\u201d<\/li>\n<li>What are the \u201cother things\u201d that must be equal?<\/li>\n<\/ol>\n<\/li>\n<li>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>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>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>Determine the volume of 1 mol of CH<sub>4<\/sub> gas at 150 K and 1 atm, using Figure 4.<\/li>\n<li>Determine the pressure of the gas in the syringe shown in Figure 5 when its volume is 12.5 mL, using:\n<ol style=\"list-style-type: lower-alpha\">\n<li>the appropriate graph<\/li>\n<li>Boyle\u2019s law<\/li>\n<\/ol>\n<\/li>\n<li>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>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>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>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?<\/li>\n<li>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>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>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>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>How many grams of gas are present in each of the following cases?\n<ol style=\"list-style-type: lower-alpha\">\n<li>0.100 L of CO<sub>2<\/sub> at 307 torr and 26 \u00b0C<\/li>\n<li>8.75 L of C<sub>2<\/sub>H<sub>4<\/sub>, at 378.3 kPa and 483 K<\/li>\n<li>221 mL of Ar at 0.23 torr and \u201354 \u00b0C<\/li>\n<\/ol>\n<\/li>\n<li>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>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>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>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>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>For a given amount of gas showing ideal behavior, draw labeled graphs of:\n<ol style=\"list-style-type: lower-alpha\">\n<li>the variation of <em>P<\/em> with <em>V <\/em><\/li>\n<li>the variation of <em>V<\/em> with <em>T <\/em><\/li>\n<li>the variation of <em>P<\/em> with <em>T <\/em><\/li>\n<li>the variation of [latex]\\frac{1}{P}[\/latex] with <em>V<\/em><\/li>\n<\/ol>\n<\/li>\n<li>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>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>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 style=\"list-style-type: lower-alpha\">\n<li>CCl<sub>2<\/sub>F<sub>2<\/sub>(<em>g<\/em>)<\/li>\n<li>CH<sub>3<\/sub>CH<sub>2<\/sub>F(<em>g<\/em>)<\/li>\n<\/ol>\n<\/li>\n<li>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>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>If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?<\/li>\n<li>If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?<\/li>\n<li id=\"ball-ch06_s04_qs01_qd01_qa11\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch06_s04_qs01_p19\" class=\"para\">A gas starts at the conditions 78.9 mL, 3.008 atm, and 56\u00b0C. Its conditions change to 35.6 mL and 2.55 atm. What is its final temperature?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch06_s04_qs01_qd01_qa12\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch06_s04_qs01_p21\" class=\"para\">The initial conditions of a sample of gas are 319 K, 3.087 L, and 591 torr. What is its final pressure if volume is changed to 2.222 L and temperature is changed to 299 K?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch06_s04_qs01_qd01_qa13\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch06_s04_qs01_p23\" class=\"para\">A gas starts with initial pressure of 7.11 atm, initial temperature of 66\u00b0C, and initial volume of 90.7 mL. If its conditions change to 33\u00b0C and 14.33 atm, what is its final volume?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q669834\">Selected Answers<\/span><\/p>\n<div id=\"q669834\" class=\"hidden-answer\" style=\"display: none\">\n<p>2. As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle\u2019s law.<\/p>\n<p>4. The answers are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha\">\n<li>The number of particles in the gas increases as the volume increases. This relationship may be written as <em>n<\/em> = constant \u00d7 <em>V<\/em>. It is a direct relationship.<\/li>\n<li>The temperature and pressure must be kept constant.<\/li>\n<\/ol>\n<p>6. The curve would be farther to the right and higher up, but the same basic shape.<\/p>\n<p>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.<\/p>\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 style=\"text-align: center\">[latex]\\large\\frac{{P}_{2}}{{T}_{2}}=\\frac{{P}_{1}}{{T1}_{}}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\large{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<p>12. Apply Charles\u2019s law to compute the volume of gas at the higher temperature:<\/p>\n<ul>\n<li><em>V<\/em><sub>1<\/sub> = 2.50 L<\/li>\n<li><em>T<\/em><sub>1<\/sub> = \u2013193 \u00b0C = 77.15 K<\/li>\n<li><em>V<\/em><sub>2<\/sub> = ?<\/li>\n<li><em>T<\/em><sub>2<\/sub> = 100 \u00b0C = 373.15 K<\/li>\n<\/ul>\n<p style=\"text-align: center\">[latex]\\large\\frac{{V}_{1}}{{T}_{1}}=\\frac{{V}_{2}}{{T}_{2}}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\large{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]<\/p>\n<p>14. <em>PV<\/em> = <em>nRT<\/em><\/p>\n<p style=\"text-align: center\">[latex]\\large V=\\frac{nRT}{P}=\\frac{8.80\\cancel{\\text{mol}}\\times 0.08206\\text{ L}\\cancel{\\text{atm}}{\\cancel{\\text{mol}}}^{{-1}}\\cancel{{\\text{K}}^{{-1}}}\\times 298.15\\cancel{\\text{K}}}{0.992\\cancel{\\text{atm}}}=217\\text{ L}[\/latex]<\/p>\n<p>16. [latex]\\large 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}{{-1}}^{}\\cancel{{\\text{K}}^{{-1}}}\\right)\\left(788.0\\cancel{\\text{K}}\\right)}=0.08190\\text{mol}=8.190\\times {10}^{{-2}}\\text{mol}[\/latex]<\/p>\n<p>[latex]\\large n\\times \\text{molar mass}=8.190\\times {10}^{{-2}}\\cancel{\\text{mol}}\\times 67.8052\\text{g}{\\cancel{\\text{mol}}}^{{-1}}=5.553\\text{g}[\/latex]<\/p>\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>m<\/em> = <em>n\u2133<\/em>, where \u2133 is the molar mass:<\/p>\n<p style=\"text-align: center\">[latex]\\large P,V,T\\,\\,\\,{\\xrightarrow{n=PV\\text{\/}RT}}\\,\\,\\,n,\\,\\,\\,{\\xrightarrow{m=n\\left(\\text{molar mass}\\right)}}\\,\\,\\,\\text{grams}[\/latex]<\/p>\n<p>or we can combine these equations to obtain:<\/p>\n<p style=\"text-align: center\">[latex]\\large\\text{mass}=m=\\frac{PV}{RT}\\times[\/latex]\u2133<\/p>\n<ol style=\"list-style-type: lower-alpha\">\n<li>[latex]\\large\\begin{array}{l}\\\\307\\cancel{\\text{torr}}\\times \\frac{1\\text{atm}}{760\\cancel{\\text{torr}}}=0.4039\\text{ atm }25^\\circ{\\text{ C}}=299.1 \\text{ 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}}^{{-1}}\\cancel{{\\text{K}}^{{-1}}}\\left(299.1\\cancel{\\text{K}}\\right)}\\times 44.01\\text{g}{\\text{mol}}^{{-1}}=7.24\\times {10}^{{-2}}\\text{g}\\end{array}[\/latex]<\/li>\n<li>[latex]\\large\\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}}^{{-1}}\\cancel{{\\text{K}}^{{-1}}}\\left(483\\cancel{\\text{K}}\\right)}\\times 28.05376\\text{ g}{\\text{mol}}^{{-1}}=23.1\\text{g}[\/latex]<\/li>\n<li>[latex]\\large\\begin{array}{l}\\\\ \\\\ 221\\cancel{\\text{mL}}\\times \\frac{1\\text{L}}{1000\\cancel{\\text{mL}}}=0.221\\text{L}-54^{\\circ}\\text{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}^{{-4}}\\text{atm}\\\\ \\text{Mass}=m=\\frac{3.03\\times {10}^{{-4}}\\cancel{\\text{atm}}\\left(0.221\\cancel{\\text{L}}\\right)}{0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{{-1}}\\cancel{{\\text{K}}^{{-1}}}\\left(219.15\\cancel{\\text{K}}\\right)}\\times 39.978\\text{ g}{\\text{mol}}^{{-1}}=1.5\\times {10}^{{-4}}\\text{g}\\end{array}[\/latex]<\/li>\n<\/ol>\n<p>20. [latex]\\large\\frac{{P}_{2}}{{T}_{2}}=\\frac{{P}_{1}}{{T}_{1}}[\/latex]<\/p>\n<p><em>T<\/em><sub>2<\/sub> = 49.5 + 273.15 = 322.65 K<\/p>\n<p>[latex]\\large{P}_{2}=\\frac{{P}_{1}{T}_{2}}{{T}_{1}}=149.6\\text{atm}\\times \\frac{322.65}{278.15}=173.5\\text{ atm}[\/latex]<\/p>\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]\\large 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}}^{{-1}}\\cancel{{\\text{K}}^{{-1}}}\\left(300.1\\cancel{\\text{K}}\\right)}=0.798\\text{mol}[\/latex] Mass of butane = 0.798 mol \u00d7 58.1234 g\/mol = 46.4 g<\/p>\n<p>24. For a gas exhibiting ideal behavior: <img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" 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\" width=\"518\" height=\"439\" alt=\"image\" \/><\/p>\n<p>26. The volume is as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha\">\n<li>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>g<\/em>) present. Use the ideal gas law <em>PV = nRT<\/em> to calculate the volume of CCl<sub>2<\/sub>F<sub>2<\/sub>(<em>g<\/em>):<br \/>\n[latex]\\large\\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]<br \/>\n<em>PV = nRT<\/em>, where <em>n<\/em> = # mol CCl<sub>2<\/sub>F<sub>2<br \/>\n<\/sub>[latex]\\large1\\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]<\/li>\n<li>[latex]\\large10.0\\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]<br \/>\n<em>PV = nRT<\/em>, with <em>n<\/em> = # mol CH<sub>3<\/sub>CH<sub>2<\/sub>F<br \/>\n1 atm \u00d7 <em>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<\/li>\n<\/ol>\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<p style=\"text-align: center\">[latex]\\large\\begin{array}{rcl}{}\\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<p>30. The pressure decreases by a factor of 3.<\/p>\n<p>31. 126 K, or \u2212147\u00b0C<\/p>\n<p>33. 40.6 mL<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>absolute zero: <\/strong>temperature at which the volume of a gas would be zero according to Charles\u2019s law.<\/p>\n<p><strong>Avogadro\u2019s law: <\/strong>volume of a gas at constant temperature and pressure is proportional to the number of gas molecules<\/p>\n<p><strong>Boyle\u2019s law: <\/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<p><strong>Charles\u2019s law: <\/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<p><strong>Gay-Lussac\u2019s law: <\/strong>(also, <strong>Amontons\u2019s law<\/strong>) pressure of a given number of moles of gas is directly proportional to its kelvin temperature when the volume is held constant<\/p>\n<p>&nbsp;<\/p>\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-1436\">\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>: OpenStaxCollege. <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><li>Introductory Chemistry- 1st Canadian Edition . <strong>Authored by<\/strong>: Jessie A. Key and David W. Ball. <strong>Provided by<\/strong>: BCCampus. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/opentextbc.ca\/introductorychemistry\/\">https:\/\/opentextbc.ca\/introductorychemistry\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Download this book for free at http:\/\/open.bccampus.ca<\/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":6181,"menu_order":2,"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\":\"OpenStaxCollege\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"},{\"type\":\"cc\",\"description\":\"Introductory Chemistry- 1st Canadian Edition \",\"author\":\"Jessie A. Key and David W. Ball\",\"organization\":\"BCCampus\",\"url\":\"https:\/\/opentextbc.ca\/introductorychemistry\/\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"Download this book for free at http:\/\/open.bccampus.ca\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1436","chapter","type-chapter","status-publish","hentry"],"part":196,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1436","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/users\/6181"}],"version-history":[{"count":11,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1436\/revisions"}],"predecessor-version":[{"id":1450,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1436\/revisions\/1450"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/parts\/196"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1436\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/media?parent=1436"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=1436"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/contributor?post=1436"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/license?post=1436"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}