{"id":2113,"date":"2015-04-22T20:52:05","date_gmt":"2015-04-22T20:52:05","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=2113"},"modified":"2015-08-31T21:56:34","modified_gmt":"2015-08-31T21:56:34","slug":"the-kinetic-molecular-theory-needs-formulas","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/chapter\/the-kinetic-molecular-theory-needs-formulas\/","title":{"raw":"The Kinetic-Molecular Theory","rendered":"The Kinetic-Molecular Theory"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>LEARNING OBJECTIVES<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n\t<li>State the postulates of the kinetic-molecular theory<\/li>\r\n\t<li>Use this theory\u2019s postulates to explain the gas laws<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idm127996416\">The gas laws that we have seen to this point, as well as the ideal gas equation, are empirical, that is, they have been derived from experimental observations. The mathematical forms of these laws closely describe the macroscopic behavior of most gases at pressures less than about 1 or 2 atm. Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships.<\/p>\r\n<p id=\"fs-idm165041184\">The <strong><span data-type=\"term\">kinetic molecular theory<\/span> <\/strong>(KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term \u201cmolecule\u201d will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.)<\/p>\r\n\r\n<ol id=\"fs-idm178411376\" data-number-style=\"arabic\">\r\n\t<li>Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.<\/li>\r\n\t<li>The molecules composing the gas are negligibly small compared to the distances between them.<\/li>\r\n\t<li>The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.<\/li>\r\n\t<li>Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are <em data-effect=\"italics\">elastic<\/em> (do not involve a loss of energy).<\/li>\r\n\t<li>The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.<\/li>\r\n<\/ol>\r\n<p id=\"fs-idm128213904\">The test of the KMT and its postulates is its ability to explain and describe the behavior of a gas. The various gas laws can be derived from the assumptions of the KMT, which have led chemists to believe that the assumptions of the theory accurately represent the properties of gas molecules. We will first look at the individual gas laws (Boyle\u2019s, Charles\u2019s, Amontons\u2019s, Avogadro\u2019s, and Dalton\u2019s laws) conceptually to see how the KMT explains them. Then, we will more carefully consider the relationships between molecular masses, speeds, and kinetic energies with temperature, and explain Graham\u2019s law.<\/p>\r\n\r\n<h2 data-type=\"title\">The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I<\/h2>\r\n<p id=\"fs-idm188928480\">Recalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows:<\/p>\r\n\r\n<ul id=\"fs-idm221875200\" data-bullet-style=\"bullet\">\r\n\t<li><em data-effect=\"italics\">Amontons\u2019s law.<\/em> If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure (Figure 1).<\/li>\r\n\t<li><em data-effect=\"italics\">Charles\u2019s law.<\/em> If the temperature of a gas is increased, a constant pressure may be maintained only if the volume occupied by the gas increases. This will result in greater average distances traveled by the molecules to reach the container walls, as well as increased wall surface area. These conditions will decrease the both the frequency of molecule-wall collisions and the number of collisions per unit area, the combined effects of which outweigh those of increased collision forces due to the greater kinetic energy at the higher temperature. The net result is a decrease in gas pressure.<\/li>\r\n\t<li><em data-effect=\"italics\">Boyle\u2019s law.<\/em> If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas (Figure 1).<\/li>\r\n\t<li><em data-effect=\"italics\">Avogadro\u2019s law.<\/em> At constant pressure and temperature, the frequency and force of molecule-wall collisions are constant. Under such conditions, increasing the number of gaseous molecules will require a proportional increase in the container volume in order to yield a decrease in the number of collisions per unit area to compensate for the increased frequency of collisions (Figure 1).<\/li>\r\n\t<li><em data-effect=\"italics\">Dalton\u2019s Law.<\/em> Because of the large distances between them, the molecules of one gas in a mixture bombard the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the (partial) pressures of the individual gases.<\/li>\r\n<\/ul>\r\n<figure id=\"CNX_Chem_09_04_KMT2\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"840\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212104\/CNX_Chem_09_04_KMT21.jpg\" alt=\"This figure shows 3 pairs of pistons and cylinders. In a, which is labeled, \u201cCharles\u2019s Law,\u201d the piston is positioned for the first cylinder so that just over half of the available volume contains 6 purple spheres with trails behind them. The trails indicate movement. Orange dashes extend from the interior surface of the cylinder where the spheres have collided. This cylinder is labeled, \u201cBaseline.\u201d In the second cylinder, the piston is in the same position, and the label, \u201cHeat\u201d is indicated in red capitalized text. Four red arrows with wavy stems are pointing upward to the base of the cylinder. The six purple spheres have longer trails behind them and the number of orange dashes indicating points of collision with the container walls has increased. A rectangle beneath the diagram states, \u201cTemperature increased, Volume constant equals Increased pressure.\u201d In b, which is labeled, \u201cBoyle\u2019s Law,\u201d the first, baseline cylinder shown is identical to the first cylinder in a. In the second cylinder, the piston has been moved, decreasing the volume available to the 6 purple spheres to half of the initial volume. The six purple spheres have longer trails behind them and the number of orange dashes indicating points of collision with the container walls has increased. This second cylinder is labeled, \u201cVolume decreased.\u201d A rectangle beneath the diagram states, \u201cVolume decreased, Wall area decreased equals Increased pressure.\u201d In c, which is labeled \u201cAvogadro\u2019s Law,\u201d the first, baseline cylinder shown is identical to the first cylinder in a. In the second cylinder, the number of purple spheres has changed from 6 to 12 and volume has doubled. This second cylinder is labeled \u201cIncreased gas.\u201d A rectangle beneath the diagram states, \u201cAt constant pressure, More gas molecules added equals Increased volume.\u201d\" width=\"840\" height=\"413\" data-media-type=\"image\/jpeg\" \/> Figure 1. (a) When gas temperature increases, gas pressure increases due to increased force and frequency of molecular collisions. (b) When volume decreases, gas pressure increases due to reduced frequency of molecular collisions. (c) When the amount of gas increases at a constant pressure, volume increases to yield a constant number of collisions per unit wall area.[\/caption]\r\n\r\n<\/figure>\r\n<h2 data-type=\"title\">Molecular Velocities and Kinetic Energy<\/h2>\r\n<p id=\"fs-idm178235328\">The previous discussion showed that the KMT qualitatively explains the behaviors described by the various gas laws. The postulates of this theory may be applied in a more quantitative fashion to derive these individual laws. To do this, we must first look at velocities and kinetic energies of gas molecules, and the temperature of a gas sample.<\/p>\r\n<p id=\"fs-idm49087808\">In a gas sample, individual molecules have widely varying speeds; however, because of the <em data-effect=\"italics\">vast<\/em> number of molecules and collisions involved, the molecular speed distribution and average speed are constant. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative numbers of molecules in a bulk sample of gas that possesses a given speed (Figure 2).<\/p>\r\n\r\n<figure id=\"CNX_Chem_09_05_MolSpeed1\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"650\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212106\/CNX_Chem_09_05_MolSpeed11.jpg\" alt=\"A graph is shown. The horizontal axis is labeled, \u201cVelocity v ( m divided by s ).\u201d This axis is marked by increments of 20 beginning at 0 and extending up to 120. The vertical axis is labeled, \u201cFraction of molecules.\u201d A positively or right-skewed curve is shown in red which begins at the origin and approaches the horizontal axis around 120 m per s. At the peak of the curve, a point is indicated with a black dot and is labeled, \u201cv subscript p.\u201d A vertical dashed line extends from this point to the horizontal axis at which point the intersection is labeled, \u201cv subscript p.\u201d Slightly to the right of the peak a second black dot is placed on the curve. This point is labeled, \u201cv subscript r m s.\u201d A vertical dashed line extends from this point to the horizontal axis at which point the intersection is labeled, \u201cv subscript r m s.\u201d The label, \u201cO subscript 2 at T equals 300 K\u201d appears in the open space to the right of the curve.\" width=\"650\" height=\"511\" data-media-type=\"image\/jpeg\" \/> Figure 2. The molecular speed distribution for oxygen gas at 300 K is shown here. Very few molecules move at either very low or very high speeds. The number of molecules with intermediate speeds increases rapidly up to a maximum, which is the most probable speed, then drops off rapidly. Note that the most probable speed, \u03bdp, is a little less than 400 m\/s, while the root mean square speed, urms, is closer to 500 m\/s.[\/caption]\r\n\r\n<\/figure>\r\n<p id=\"fs-idm188037232\">The kinetic energy (KE) of a particle of mass (<em data-effect=\"italics\">m<\/em>) and speed (<em data-effect=\"italics\">u<\/em>) is given by:<\/p>\r\n[latex]\\text{KE}=\\frac{1}{2}m{u}^{2}[\/latex]\r\n<p id=\"fs-idm178023104\">Expressing mass in kilograms and speed in meters per second will yield energy values in units of joules (J = kg m<sup>2<\/sup> s<sup>\u20132<\/sup>). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the <strong><span data-type=\"term\">root mean square velocity<\/span><\/strong> of a particle,<strong> <span data-type=\"term\"><em data-effect=\"italics\">u<\/em><sub>rms<\/sub><\/span><\/strong>, is defined as the square root of the average of the squares of the velocities with <em data-effect=\"italics\">n<\/em> = the number of particles:<\/p>\r\n[latex]{u}_{rms}=\\sqrt{\\overline{{u}^{2}}}=\\sqrt{\\frac{{u}_{1}^{2}+{u}_{2}^{2}+{u}_{3}^{2}+{u}_{4}^{2}+\\dots }{n}}[\/latex]\r\n<p id=\"fs-idm140862528\">The average kinetic energy, KE<sub>avg<\/sub>, is then equal to:<\/p>\r\n[latex]{\\text{KE}}_{\\text{avg}}=\\frac{1}{2}{mu}_{\\text{rms}}^{2}[\/latex]\r\n<p id=\"fs-idm52829008\">The KE<sub>avg<\/sub> of a collection of gas molecules is also directly proportional to the temperature of the gas and may be described by the equation:<\/p>\r\n[latex]{\\text{KE}}_{\\text{avg}}=\\frac{3}{2}RT[\/latex]\r\n<p id=\"fs-idm166238448\">where <em data-effect=\"italics\">R<\/em> is the gas constant and T is the kelvin temperature. When used in this equation, the appropriate form of the gas constant is 8.314 J\/K (8.314 kg m<sup>2<\/sup>s<sup>\u20132<\/sup>K<sup>\u20131<\/sup>). These two separate equations for KE<sub>avg<\/sub> may be combined and rearranged to yield a relation between molecular speed and temperature:<\/p>\r\n[latex]\\frac{1}{2}{mu}_{\\text{rms}}^{2}=\\frac{3}{2}RT[\/latex]\r\n\r\n[latex]{u}_{\\text{rms}}=\\sqrt{\\frac{3RT}{m}}[\/latex]\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1<\/h3>\r\n<h4 id=\"fs-idm217669472\"><strong><span data-type=\"title\">Calculation of <em data-effect=\"italics\">u<sub>rms<\/sub><\/em><\/span><\/strong><\/h4>\r\nCalculate the root-mean-square velocity for a nitrogen molecule at 30 \u00b0C.\r\n<h4 id=\"fs-idm124389728\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nConvert the temperature into Kelvin:\r\n\r\n[latex]30\\text{\\textdegree C}+273=\\text{303 K}[\/latex]\r\n<p id=\"fs-idm144782000\">Determine the mass of a nitrogen molecule in kilograms:<\/p>\r\n[latex]\\frac{28.0\\cancel{\\text{g}}}{\\text{1 mol}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{1 kg}}{1000\\cancel{\\text{g}}}=0.028\\text{kg\/mol}[\/latex]\r\n<p id=\"fs-idm160368176\">Replace the variables and constants in the root-mean-square velocity equation, replacing Joules with the equivalent kg m<sup>2<\/sup>s<sup>\u20132<\/sup>:<\/p>\r\n[latex]{u}_{\\text{rms}}=\\sqrt{\\frac{3RT}{m}}[\/latex]\r\n\r\n[latex]{u}_{rms}=\\sqrt{\\frac{3\\left(8.314\\text{J\/mol K}\\right)\\left(\\text{303 K}\\right)}{\\left(0.028\\text{kg\/mol}\\right)}}=\\sqrt{2.70\\times {10}^{5}{\\text{m}}^{2}{\\text{s}}^{-2}}=519\\text{m\/s}[\/latex]\r\n<h4 id=\"fs-idm87955168\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\r\nCalculate the root-mean-square velocity for an oxygen molecule at \u201323 \u00b0C.\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a0441 m\/s<\/div>\r\n<\/div>\r\n<p id=\"fs-idm298779008\">If the temperature of a gas increases, its KE<sub>avg<\/sub> increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall, that is, to the right. If temperature decreases, KE<sub>avg<\/sub> decreases, more molecules have lower speeds and fewer molecules have higher speeds, and the distribution shifts toward lower speeds overall, that is, to the left. This behavior is illustrated for nitrogen gas in Figure 3.<\/p>\r\n\r\n<figure id=\"CNX_Chem_09_05_MolSpeed2\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"650\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212107\/CNX_Chem_09_05_MolSpeed21.jpg\" alt=\"A graph with four positively or right-skewed curves of varying heights is shown. The horizontal axis is labeled, \u201cVelocity v ( m divided by s ).\u201d This axis is marked by increments of 500 beginning at 0 and extending up to 1500. The vertical axis is labeled, \u201cFraction of molecules.\u201d The label, \u201cN subscript 2,\u201d appears in the open space in the upper right area of the graph. The tallest and narrowest of these curves is labeled, \u201c100 K.\u201d Its right end appears to touch the horizontal axis around 700 m per s. It is followed by a slightly wider curve which is labeled, \u201c200 K,\u201d that is about three quarters of the height of the initial curve. Its right end appears to touch the horizontal axis around 850 m per s. The third curve is significantly wider and only about half the height of the initial curve. It is labeled, \u201c500 K.\u201d Its right end appears to touch the horizontal axis around 1450 m per s. The final curve is only about one third the height of the initial curve. It is much wider than the others, so much so that its right end has not yet reached the horizontal axis. This curve is labeled, \u201c1000 K.\u201d\" width=\"650\" height=\"511\" data-media-type=\"image\/jpeg\" \/> Figure 3. The molecular speed distribution for nitrogen gas (N<sub>2<\/sub>) shifts to the right and flattens as the temperature increases; it shifts to the left and heightens as the temperature decreases.[\/caption]\r\n\r\n<\/figure>\r\n<p id=\"fs-idm131874080\">At a given temperature, all gases have the same KE<sub>avg<\/sub> for their molecules. Gases composed of lighter molecules have more high-speed particles and a higher <em data-effect=\"italics\">u<sub>rms<\/sub><\/em>, with a speed distribution that peaks at relatively higher velocities. Gases consisting of heavier molecules have more low-speed particles, a lower <em data-effect=\"italics\">u<sub>rms<\/sub><\/em>, and a speed distribution that peaks at relatively lower velocities. This trend is demonstrated by the data for a series of noble gases shown in Figure 4.<\/p>\r\n\r\n<figure id=\"CNX_Chem_09_05_MolSpeed3\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"650\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212108\/CNX_Chem_09_05_MolSpeed31.jpg\" alt=\"A graph is shown with four positively or right-skewed curves of varying heights. The horizontal axis is labeled, \u201cVelocity v ( m divided by s ).\u201d This axis is marked by increments of 500 beginning at 0 and extending up to 3000. The vertical axis is labeled, \u201cFraction of molecules.\u201d The tallest and narrowest of these curves is labeled, \u201cX e.\u201d Its right end appears to touch the horizontal axis around 600 m per s. It is followed by a slightly wider curve which is labeled, \u201cA r,\u201d that is about half the height of the initial curve. Its right end appears to touch the horizontal axis around 900 m per s. The third curve is significantly wider and just over a third of the height of the initial curve. It is labeled, \u201cN e.\u201d Its right end appears to touch the horizontal axis around 1200 m per s. The final curve is only about one fourth the height of the initial curve. It is much wider than the others, so much so that its right reaches the horizontal axis around 2500 m per s. This curve is labeled, \u201cH e.\u201d\" width=\"650\" height=\"386\" data-media-type=\"image\/jpeg\" \/> Figure 4. Molecular velocity is directly related to molecular mass. At a given temperature, lighter molecules move faster on average than heavier molecules.[\/caption]\r\n\r\n<\/figure>\r\n<div class=\"textbox\">The<a href=\"http:\/\/phet.colorado.edu\/en\/simulation\/gas-properties\" target=\"_blank\"> PhET\u00a0gas simulator<\/a> may be used to examine the effect of temperature on molecular velocities. Examine the simulator\u2019s \u201cenergy histograms\u201d (molecular speed distributions) and \u201cspecies information\u201d (which gives average speed values) for molecules of different masses at various temperatures.<\/div>\r\n<h2 data-type=\"title\">The Kinetic-Molecular Theory Explains the Behavior of Gases, Part II<\/h2>\r\n<p id=\"fs-idp30560352\">According to Graham\u2019s law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates.<\/p>\r\n<p id=\"fs-idm65845776\">The rate of effusion of a gas depends directly on the (average) speed of its molecules:<\/p>\r\n[latex]\\text{effusion rate}\\propto {u}_{\\text{rms}}[\/latex]\r\n<p id=\"fs-idm219209680\">Using this relation, and the equation relating molecular speed to mass, Graham\u2019s law may be easily derived as shown here:<\/p>\r\n[latex]{u}_{\\text{rms}}=\\sqrt{\\frac{3RT}{m}}[\/latex]\r\n\r\n[latex]m=\\frac{3RT}{{u}_{rms}^{2}}=\\frac{3RT}{{\\overline{u}}^{2}}[\/latex]\r\n\r\n[latex]\\frac{\\text{effusion rate A}}{\\text{effusion rate B}}=\\frac{{u}_{rms\\text{A}}}{{u}_{rms\\text{B}}}=\\frac{\\sqrt{\\frac{3RT}{{m}_{\\text{A}}}}}{\\sqrt{\\frac{3RT}{{m}_{\\text{B}}}}}=\\sqrt{\\frac{{m}_{\\text{B}}}{{m}_{\\text{A}}}}[\/latex]\r\n<p id=\"fs-idm131494928\">The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham\u2019s law.<\/p>\r\n\r\n<div class=\"bcc-box bcc-success\">\r\n<h2>Key Concepts and Summary<\/h2>\r\n<p id=\"fs-idp236281408\">The kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average velocities determined by their absolute temperatures. The individual molecules of a gas exhibit a range of velocities, the distribution of these velocities being dependent on the temperature of the gas and the mass of its molecules.<\/p>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Key Equations<\/h3>\r\n<ul>\r\n\t<li>[latex]{u}_{rms}=\\sqrt{\\overline{{u}^{2}}}=\\sqrt{\\frac{{u}_{1}^{2}+{u}_{2}^{2}+{u}_{3}^{2}+{u}_{4}^{2}+\\dots }{n}}[\/latex]<\/li>\r\n\t<li>[latex]{\\text{KE}}_{\\text{avg}}=\\frac{3}{2}R\\text{T}[\/latex]<\/li>\r\n\t<li>[latex]{u}_{\\text{rms}}=\\sqrt{\\frac{3RT}{m}}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Chemistry End of Chapter Exercises<\/h3>\r\n<ol>\r\n\t<li id=\"fs-idm212848544\">Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape.<\/li>\r\n\t<li>Can the speed of a given molecule in a gas double at constant temperature? Explain your answer.<\/li>\r\n\t<li>Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:\r\n<ol>\r\n\t<li>The pressure of the gas is increased by reducing the volume at constant temperature.<\/li>\r\n\t<li>The pressure of the gas is increased by increasing the temperature at constant volume.<\/li>\r\n\t<li>The average velocity of the molecules is increased by a factor of 2.<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>The distribution of molecular velocities in a sample of helium is shown in Figure 9.34. If the sample is cooled, will the distribution of velocities look more like that of H<sub>2<\/sub> or of H<sub>2<\/sub>O? Explain your answer.<\/li>\r\n\t<li>What is the ratio of the average kinetic energy of a SO<sub>2<\/sub> molecule to that of an O<sub>2<\/sub> molecule in a mixture of two gases? What is the ratio of the root mean square speeds, <em data-effect=\"italics\">u<\/em><sub>rms<\/sub>, of the two gases?<\/li>\r\n\t<li>A 1-L sample of CO initially at STP is heated to 546 \u00b0C, and its volume is increased to 2 L.\r\n<ol>\r\n\t<li>What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?<\/li>\r\n\t<li>What is the effect on the average kinetic energy of the molecules?<\/li>\r\n\t<li>What is the effect on the root mean square speed of the molecules?<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>The root mean square speed of H<sub>2<\/sub> molecules at 25 \u00b0C is about 1.6 km\/s. What is the root mean square speed of a N<sub>2<\/sub> molecule at 25 \u00b0C?<\/li>\r\n\t<li>Answer the following questions:\r\n<ol>\r\n\t<li>Is the pressure of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?<\/li>\r\n\t<li>Is the density of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?<\/li>\r\n\t<li>At a pressure of 1 atm and a temperature of 20 \u00b0C, dry air has a density of 1.2256 g\/L. What is the (average) molar mass of dry air?<\/li>\r\n\t<li>The average temperature of the gas in a hot air balloon is 1.30 \u00d7 10<sup>2<\/sup> \u00b0F. Calculate its density, assuming the molar mass equals that of dry air.<\/li>\r\n\t<li>The lifting capacity of a hot air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?<\/li>\r\n\t<li>An average balloon has a diameter of 60 feet and a volume of 1.1 \u00d7 10<sup>5<\/sup> ft<sup>3<\/sup>. What is the lifting power of such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?<\/li>\r\n\t<li>A balloon carries 40.0 gallons of liquid propane (density 0.5005 g\/L). What volume of CO<sub>2<\/sub> and H<sub>2<\/sub>O gas is produced by the combustion of this propane?<\/li>\r\n\t<li>A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in kJ\/min) from the hot air in the bag during the flight?<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>Show that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas 2, [latex]\\frac{{R}_{1}}{{R}_{2}},[\/latex] is the same at 0 \u00b0C and 100 \u00b0C.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h4>Selected Answers<\/h4>\r\n2.\u00a0Yes. At any given instant, there are a range of values of molecular speeds in a sample of gas. Any single molecule can speed up or slow down as it collides with other molecules. The average velocity of all the molecules is constant at constant temperature.\r\n\r\n4.\u00a0H<sub>2<\/sub>O. Cooling slows the velocities of the He atoms, causing them to behave as though they were heavier.\r\n\r\n6.\u00a0Both the temperature and the volume are doubled for this gas (<em data-effect=\"italics\">n<\/em> constant), so <em data-effect=\"italics\">P<\/em> remains constant.\r\n\r\n(a) The number of collisions per unit area of the container wall is constant.\r\n\r\n(b) The average kinetic energy doubles; it is proportional to temperature.\r\n\r\n(c) The root mean square speed increases to [latex]\\sqrt{2}[\/latex] times its initial value; <em data-effect=\"italics\">u<\/em><sub>rms<\/sub> is proportional to [latex]\\sqrt{{\\text{KE}}_{\\text{avg}}}.[\/latex]\r\n\r\n8. (a) equal, because the balloon is free to expand until the pressures are equalized;\r\n\r\n(b) less than the density outside;\r\n\r\n(c) assume three-place accuracy throughout unless greater accuracy is stated:[latex]\\text{molar mass}=\\frac{DRT}{P}=1.2256\\text{g}\\cancel{{\\text{L}}^{\\text{-1}}}\\times \\frac{0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\times 293.15\\cancel{\\text{K}}}{1.00\\cancel{\\text{atm}}}=29.48{\\text{g mol}}^{\\text{-1}};[\/latex]\r\n\r\n(d) convert the temperature to \u00b0C; then use the ideal gas law:\r\n\r\n[latex]\\text{\\textdegree C}=\\frac{5}{9}\\left(\\text{F}-32\\right)=\\frac{5}{9}\\left(130-32\\right)=54.44\\text{\\textdegree C}=327.6\\text{K}[\/latex]\r\n\r\n[latex]D=\\frac{\\mathcal{M}P}{RT}=29.48\\text{g}\\cancel{{\\text{mol}}^{\\text{-1}}}\\times \\frac{1.00\\cancel{\\text{atm}}}{0.08206\\text{L}\\cancel{\\text{atm}}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\times 327.6\\cancel{\\text{K}}}=1.0966{\\text{g L}}^{\\text{-1}};[\/latex]\r\n\r\n(e) 1.2256 g\/L \u2013 1.09966 g\/L = 0.129 g\/L;\r\n\r\n(f) calculate the volume in liters, multiply the volume by the density difference to find the lifting capacity of the balloon, subtract the weight of the balloon after converting to pounds:\r\n\r\n[latex]1.1\\times 105{\\text{ft}}^{3}\\times {\\left(\\frac{\\text{12 in}}{\\text{91 ft}}\\right)}^{3}\\times {\\left(\\frac{\\text{2.54 cm}}{\\text{in}}\\right)}^{3}\\times \\frac{\\text{1 L}}{1000{\\text{cm}}^{3}}=3.11\\times {10}^{6}\\text{L}[\/latex]\r\n\r\n3.11 \u00d7 106 L \u00d7 0.129 g\/L = 4.01 \u00d7 10<sup>5<\/sup> g\r\n\r\n[latex]\\frac{4.01\\times {10}^{5}\\text{g}}{453.59{\\text{g lb}}^{\\text{-1}}}=884\\text{lb;}\\text{884 lb}-\\text{500 lb}=\\text{384 lb}[\/latex]\r\n\r\nnet lifting capacity = 384 lb;\r\n\r\n(g) First, find the mass of propane contained in 40.0 gal. Then calculate the moles of CO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) and H<sub>2<\/sub>O(<em data-effect=\"italics\">g<\/em>) produced from the balanced equation.\r\n\r\n[latex]40.0\\cancel{\\text{gal}}\\times \\frac{4\\left(0.9463\\text{L}\\right)}{1\\cancel{\\text{gal}}}=151.4\\text{L}[\/latex]\r\n\r\n[latex]151.4\\cancel{\\text{L}}\\times 0.5005\\text{g}{\\cancel{\\text{L}}}^{\\cancel{-1}}=75.8\\text{g}[\/latex]\r\n\r\nMolar mass of propane = 3(12.011) + 8(1.00794) = 36.033 + 8.064 = 44.097 g mol<sup>\u20131<\/sup>\r\n\r\n[latex]\\frac{75.8\\cancel{\\text{g}}}{44.097\\cancel{\\text{g}}{\\text{mol}}^{\\text{-1}}}=1.72\\text{mol}[\/latex]\r\n\r\nThe reaction is [latex]{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)+5{\\text{O}}_{2}\\left(g\\right)\\rightarrow 3{\\text{CO}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]\r\n\r\nFor each 1.72 mol propane, there are 3 \u00d7 1.72 mol = 5.15 mol of CO<sub>2<\/sub> and 4 \u00d7 1.72 mol = 6.88 mol H<sub>2<\/sub> O. The total volume at STP = 22.4 L \u00d7 12.04 = 270 L;\r\n\r\n(h) The total heat released is determined from the heat of combustion of the propane. Using the equation in part (g),[latex]\\begin{array}{ll}\\Delta {H}_{\\text{combustion}}^{ \\textdegree }\\hfill &amp; =3{\\Delta H}_{{\\text{CO}}_{2}\\left(g\\right)}^{\\textdegree }+4{\\Delta H}_{{\\text{H}}_{2}\\text{O}\\left(g\\right)}^{\\textdegree }-{\\Delta H}_{\\text{propane}}^{\\textdegree }\\hfill \\\\ \\hfill &amp; =3\\left(-393.51\\right)+4\\left(-241.82\\right)-\\left(-103.85\\right)\\hfill \\\\ \\hfill &amp; =-1180.52-967.28+103.85=-2043.96{\\text{kJ mol}}^{-1}\\hfill \\end{array}[\/latex]\r\n\r\nSince there is 1.72 mol propane, 1.72 \u00d7 2043.96 kJ mol<sup>-1<\/sup> = 3.52 \u00d7 10<sup>3<\/sup> kJ used for heating. This heat is used over 90 minutes, so [latex]\\frac{3.52\\times {10}^{3}\\text{kJ}}{\\text{90 min}}=39.1{\\text{kJ min}}^{\\text{-1}}[\/latex] is released.\r\n\r\n9. The rate at which a gas will diffuse, <em data-effect=\"italics\">R,<\/em> is proportional lo <em data-effect=\"italics\">u<\/em><sub>rms<\/sub>, the root mean square speed of its molecules. The square of this value, in turn, is proportional to the average kinetic energy. The average kinetic energy is:[latex]{\\overline{\\text{KE}}}_{\\text{avg}}=kT.[\/latex]For two different gases, 1 and 2, the constant of proportionality can be represented as <em data-effect=\"italics\">k<sub>1<\/sub><\/em> and <em data-effect=\"italics\">k<\/em><sub>2<\/sub>, respectively. Thus,[latex]\\frac{{R}_{1}}{{R}_{2}}=\\frac{{k}_{1}\\sqrt{T}}{{k}_{2}\\sqrt{T}}.[\/latex]As a result of this relationship, no matter at which temperature diffusion occurs, the temperature term will cancel out of the equation and the ratio of rates will be the same.\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Glossary<\/h3>\r\n<strong>kinetic molecular theory<\/strong>\r\ntheory based on simple principles and assumptions that effectively explains ideal gas behavior\r\n\r\n<strong>root mean square velocity (<em data-effect=\"italics\">u<\/em><sub>rms<\/sub>)<\/strong>\r\nmeasure of average velocity for a group of particles calculated as the square root of the average squared velocity\r\n\r\n<\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>LEARNING OBJECTIVES<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>State the postulates of the kinetic-molecular theory<\/li>\n<li>Use this theory\u2019s postulates to explain the gas laws<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idm127996416\">The gas laws that we have seen to this point, as well as the ideal gas equation, are empirical, that is, they have been derived from experimental observations. The mathematical forms of these laws closely describe the macroscopic behavior of most gases at pressures less than about 1 or 2 atm. Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships.<\/p>\n<p id=\"fs-idm165041184\">The <strong><span data-type=\"term\">kinetic molecular theory<\/span> <\/strong>(KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term \u201cmolecule\u201d will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.)<\/p>\n<ol id=\"fs-idm178411376\" data-number-style=\"arabic\">\n<li>Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.<\/li>\n<li>The molecules composing the gas are negligibly small compared to the distances between them.<\/li>\n<li>The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.<\/li>\n<li>Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are <em data-effect=\"italics\">elastic<\/em> (do not involve a loss of energy).<\/li>\n<li>The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.<\/li>\n<\/ol>\n<p id=\"fs-idm128213904\">The test of the KMT and its postulates is its ability to explain and describe the behavior of a gas. The various gas laws can be derived from the assumptions of the KMT, which have led chemists to believe that the assumptions of the theory accurately represent the properties of gas molecules. We will first look at the individual gas laws (Boyle\u2019s, Charles\u2019s, Amontons\u2019s, Avogadro\u2019s, and Dalton\u2019s laws) conceptually to see how the KMT explains them. Then, we will more carefully consider the relationships between molecular masses, speeds, and kinetic energies with temperature, and explain Graham\u2019s law.<\/p>\n<h2 data-type=\"title\">The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I<\/h2>\n<p id=\"fs-idm188928480\">Recalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows:<\/p>\n<ul id=\"fs-idm221875200\" data-bullet-style=\"bullet\">\n<li><em data-effect=\"italics\">Amontons\u2019s law.<\/em> If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure (Figure 1).<\/li>\n<li><em data-effect=\"italics\">Charles\u2019s law.<\/em> If the temperature of a gas is increased, a constant pressure may be maintained only if the volume occupied by the gas increases. This will result in greater average distances traveled by the molecules to reach the container walls, as well as increased wall surface area. These conditions will decrease the both the frequency of molecule-wall collisions and the number of collisions per unit area, the combined effects of which outweigh those of increased collision forces due to the greater kinetic energy at the higher temperature. The net result is a decrease in gas pressure.<\/li>\n<li><em data-effect=\"italics\">Boyle\u2019s law.<\/em> If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas (Figure 1).<\/li>\n<li><em data-effect=\"italics\">Avogadro\u2019s law.<\/em> At constant pressure and temperature, the frequency and force of molecule-wall collisions are constant. Under such conditions, increasing the number of gaseous molecules will require a proportional increase in the container volume in order to yield a decrease in the number of collisions per unit area to compensate for the increased frequency of collisions (Figure 1).<\/li>\n<li><em data-effect=\"italics\">Dalton\u2019s Law.<\/em> Because of the large distances between them, the molecules of one gas in a mixture bombard the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the (partial) pressures of the individual gases.<\/li>\n<\/ul>\n<figure id=\"CNX_Chem_09_04_KMT2\">\n<div style=\"width: 850px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212104\/CNX_Chem_09_04_KMT21.jpg\" alt=\"This figure shows 3 pairs of pistons and cylinders. In a, which is labeled, \u201cCharles\u2019s Law,\u201d the piston is positioned for the first cylinder so that just over half of the available volume contains 6 purple spheres with trails behind them. The trails indicate movement. Orange dashes extend from the interior surface of the cylinder where the spheres have collided. This cylinder is labeled, \u201cBaseline.\u201d In the second cylinder, the piston is in the same position, and the label, \u201cHeat\u201d is indicated in red capitalized text. Four red arrows with wavy stems are pointing upward to the base of the cylinder. The six purple spheres have longer trails behind them and the number of orange dashes indicating points of collision with the container walls has increased. A rectangle beneath the diagram states, \u201cTemperature increased, Volume constant equals Increased pressure.\u201d In b, which is labeled, \u201cBoyle\u2019s Law,\u201d the first, baseline cylinder shown is identical to the first cylinder in a. In the second cylinder, the piston has been moved, decreasing the volume available to the 6 purple spheres to half of the initial volume. The six purple spheres have longer trails behind them and the number of orange dashes indicating points of collision with the container walls has increased. This second cylinder is labeled, \u201cVolume decreased.\u201d A rectangle beneath the diagram states, \u201cVolume decreased, Wall area decreased equals Increased pressure.\u201d In c, which is labeled \u201cAvogadro\u2019s Law,\u201d the first, baseline cylinder shown is identical to the first cylinder in a. In the second cylinder, the number of purple spheres has changed from 6 to 12 and volume has doubled. This second cylinder is labeled \u201cIncreased gas.\u201d A rectangle beneath the diagram states, \u201cAt constant pressure, More gas molecules added equals Increased volume.\u201d\" width=\"840\" height=\"413\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. (a) When gas temperature increases, gas pressure increases due to increased force and frequency of molecular collisions. (b) When volume decreases, gas pressure increases due to reduced frequency of molecular collisions. (c) When the amount of gas increases at a constant pressure, volume increases to yield a constant number of collisions per unit wall area.<\/p>\n<\/div>\n<\/figure>\n<h2 data-type=\"title\">Molecular Velocities and Kinetic Energy<\/h2>\n<p id=\"fs-idm178235328\">The previous discussion showed that the KMT qualitatively explains the behaviors described by the various gas laws. The postulates of this theory may be applied in a more quantitative fashion to derive these individual laws. To do this, we must first look at velocities and kinetic energies of gas molecules, and the temperature of a gas sample.<\/p>\n<p id=\"fs-idm49087808\">In a gas sample, individual molecules have widely varying speeds; however, because of the <em data-effect=\"italics\">vast<\/em> number of molecules and collisions involved, the molecular speed distribution and average speed are constant. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative numbers of molecules in a bulk sample of gas that possesses a given speed (Figure 2).<\/p>\n<figure id=\"CNX_Chem_09_05_MolSpeed1\">\n<div style=\"width: 660px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212106\/CNX_Chem_09_05_MolSpeed11.jpg\" alt=\"A graph is shown. The horizontal axis is labeled, \u201cVelocity v ( m divided by s ).\u201d This axis is marked by increments of 20 beginning at 0 and extending up to 120. The vertical axis is labeled, \u201cFraction of molecules.\u201d A positively or right-skewed curve is shown in red which begins at the origin and approaches the horizontal axis around 120 m per s. At the peak of the curve, a point is indicated with a black dot and is labeled, \u201cv subscript p.\u201d A vertical dashed line extends from this point to the horizontal axis at which point the intersection is labeled, \u201cv subscript p.\u201d Slightly to the right of the peak a second black dot is placed on the curve. This point is labeled, \u201cv subscript r m s.\u201d A vertical dashed line extends from this point to the horizontal axis at which point the intersection is labeled, \u201cv subscript r m s.\u201d The label, \u201cO subscript 2 at T equals 300 K\u201d appears in the open space to the right of the curve.\" width=\"650\" height=\"511\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The molecular speed distribution for oxygen gas at 300 K is shown here. Very few molecules move at either very low or very high speeds. The number of molecules with intermediate speeds increases rapidly up to a maximum, which is the most probable speed, then drops off rapidly. Note that the most probable speed, \u03bdp, is a little less than 400 m\/s, while the root mean square speed, urms, is closer to 500 m\/s.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-idm188037232\">The kinetic energy (KE) of a particle of mass (<em data-effect=\"italics\">m<\/em>) and speed (<em data-effect=\"italics\">u<\/em>) is given by:<\/p>\n<p>[latex]\\text{KE}=\\frac{1}{2}m{u}^{2}[\/latex]<\/p>\n<p id=\"fs-idm178023104\">Expressing mass in kilograms and speed in meters per second will yield energy values in units of joules (J = kg m<sup>2<\/sup> s<sup>\u20132<\/sup>). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the <strong><span data-type=\"term\">root mean square velocity<\/span><\/strong> of a particle,<strong> <span data-type=\"term\"><em data-effect=\"italics\">u<\/em><sub>rms<\/sub><\/span><\/strong>, is defined as the square root of the average of the squares of the velocities with <em data-effect=\"italics\">n<\/em> = the number of particles:<\/p>\n<p>[latex]{u}_{rms}=\\sqrt{\\overline{{u}^{2}}}=\\sqrt{\\frac{{u}_{1}^{2}+{u}_{2}^{2}+{u}_{3}^{2}+{u}_{4}^{2}+\\dots }{n}}[\/latex]<\/p>\n<p id=\"fs-idm140862528\">The average kinetic energy, KE<sub>avg<\/sub>, is then equal to:<\/p>\n<p>[latex]{\\text{KE}}_{\\text{avg}}=\\frac{1}{2}{mu}_{\\text{rms}}^{2}[\/latex]<\/p>\n<p id=\"fs-idm52829008\">The KE<sub>avg<\/sub> of a collection of gas molecules is also directly proportional to the temperature of the gas and may be described by the equation:<\/p>\n<p>[latex]{\\text{KE}}_{\\text{avg}}=\\frac{3}{2}RT[\/latex]<\/p>\n<p id=\"fs-idm166238448\">where <em data-effect=\"italics\">R<\/em> is the gas constant and T is the kelvin temperature. When used in this equation, the appropriate form of the gas constant is 8.314 J\/K (8.314 kg m<sup>2<\/sup>s<sup>\u20132<\/sup>K<sup>\u20131<\/sup>). These two separate equations for KE<sub>avg<\/sub> may be combined and rearranged to yield a relation between molecular speed and temperature:<\/p>\n<p>[latex]\\frac{1}{2}{mu}_{\\text{rms}}^{2}=\\frac{3}{2}RT[\/latex]<\/p>\n<p>[latex]{u}_{\\text{rms}}=\\sqrt{\\frac{3RT}{m}}[\/latex]<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 1<\/h3>\n<h4 id=\"fs-idm217669472\"><strong><span data-type=\"title\">Calculation of <em data-effect=\"italics\">u<sub>rms<\/sub><\/em><\/span><\/strong><\/h4>\n<p>Calculate the root-mean-square velocity for a nitrogen molecule at 30 \u00b0C.<\/p>\n<h4 id=\"fs-idm124389728\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>Convert the temperature into Kelvin:<\/p>\n<p>[latex]30\\text{\\textdegree C}+273=\\text{303 K}[\/latex]<\/p>\n<p id=\"fs-idm144782000\">Determine the mass of a nitrogen molecule in kilograms:<\/p>\n<p>[latex]\\frac{28.0\\cancel{\\text{g}}}{\\text{1 mol}}\\phantom{\\rule{0.4em}{0ex}}\\times \\phantom{\\rule{0.4em}{0ex}}\\frac{\\text{1 kg}}{1000\\cancel{\\text{g}}}=0.028\\text{kg\/mol}[\/latex]<\/p>\n<p id=\"fs-idm160368176\">Replace the variables and constants in the root-mean-square velocity equation, replacing Joules with the equivalent kg m<sup>2<\/sup>s<sup>\u20132<\/sup>:<\/p>\n<p>[latex]{u}_{\\text{rms}}=\\sqrt{\\frac{3RT}{m}}[\/latex]<\/p>\n<p>[latex]{u}_{rms}=\\sqrt{\\frac{3\\left(8.314\\text{J\/mol K}\\right)\\left(\\text{303 K}\\right)}{\\left(0.028\\text{kg\/mol}\\right)}}=\\sqrt{2.70\\times {10}^{5}{\\text{m}}^{2}{\\text{s}}^{-2}}=519\\text{m\/s}[\/latex]<\/p>\n<h4 id=\"fs-idm87955168\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>Calculate the root-mean-square velocity for an oxygen molecule at \u201323 \u00b0C.<\/p>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a0441 m\/s<\/div>\n<\/div>\n<p id=\"fs-idm298779008\">If the temperature of a gas increases, its KE<sub>avg<\/sub> increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall, that is, to the right. If temperature decreases, KE<sub>avg<\/sub> decreases, more molecules have lower speeds and fewer molecules have higher speeds, and the distribution shifts toward lower speeds overall, that is, to the left. This behavior is illustrated for nitrogen gas in Figure 3.<\/p>\n<figure id=\"CNX_Chem_09_05_MolSpeed2\">\n<div style=\"width: 660px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212107\/CNX_Chem_09_05_MolSpeed21.jpg\" alt=\"A graph with four positively or right-skewed curves of varying heights is shown. The horizontal axis is labeled, \u201cVelocity v ( m divided by s ).\u201d This axis is marked by increments of 500 beginning at 0 and extending up to 1500. The vertical axis is labeled, \u201cFraction of molecules.\u201d The label, \u201cN subscript 2,\u201d appears in the open space in the upper right area of the graph. The tallest and narrowest of these curves is labeled, \u201c100 K.\u201d Its right end appears to touch the horizontal axis around 700 m per s. It is followed by a slightly wider curve which is labeled, \u201c200 K,\u201d that is about three quarters of the height of the initial curve. Its right end appears to touch the horizontal axis around 850 m per s. The third curve is significantly wider and only about half the height of the initial curve. It is labeled, \u201c500 K.\u201d Its right end appears to touch the horizontal axis around 1450 m per s. The final curve is only about one third the height of the initial curve. It is much wider than the others, so much so that its right end has not yet reached the horizontal axis. This curve is labeled, \u201c1000 K.\u201d\" width=\"650\" height=\"511\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. The molecular speed distribution for nitrogen gas (N<sub>2<\/sub>) shifts to the right and flattens as the temperature increases; it shifts to the left and heightens as the temperature decreases.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-idm131874080\">At a given temperature, all gases have the same KE<sub>avg<\/sub> for their molecules. Gases composed of lighter molecules have more high-speed particles and a higher <em data-effect=\"italics\">u<sub>rms<\/sub><\/em>, with a speed distribution that peaks at relatively higher velocities. Gases consisting of heavier molecules have more low-speed particles, a lower <em data-effect=\"italics\">u<sub>rms<\/sub><\/em>, and a speed distribution that peaks at relatively lower velocities. This trend is demonstrated by the data for a series of noble gases shown in Figure 4.<\/p>\n<figure id=\"CNX_Chem_09_05_MolSpeed3\">\n<div style=\"width: 660px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23212108\/CNX_Chem_09_05_MolSpeed31.jpg\" alt=\"A graph is shown with four positively or right-skewed curves of varying heights. The horizontal axis is labeled, \u201cVelocity v ( m divided by s ).\u201d This axis is marked by increments of 500 beginning at 0 and extending up to 3000. The vertical axis is labeled, \u201cFraction of molecules.\u201d The tallest and narrowest of these curves is labeled, \u201cX e.\u201d Its right end appears to touch the horizontal axis around 600 m per s. It is followed by a slightly wider curve which is labeled, \u201cA r,\u201d that is about half the height of the initial curve. Its right end appears to touch the horizontal axis around 900 m per s. The third curve is significantly wider and just over a third of the height of the initial curve. It is labeled, \u201cN e.\u201d Its right end appears to touch the horizontal axis around 1200 m per s. The final curve is only about one fourth the height of the initial curve. It is much wider than the others, so much so that its right reaches the horizontal axis around 2500 m per s. This curve is labeled, \u201cH e.\u201d\" width=\"650\" height=\"386\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. Molecular velocity is directly related to molecular mass. At a given temperature, lighter molecules move faster on average than heavier molecules.<\/p>\n<\/div>\n<\/figure>\n<div class=\"textbox\">The<a href=\"http:\/\/phet.colorado.edu\/en\/simulation\/gas-properties\" target=\"_blank\"> PhET\u00a0gas simulator<\/a> may be used to examine the effect of temperature on molecular velocities. Examine the simulator\u2019s \u201cenergy histograms\u201d (molecular speed distributions) and \u201cspecies information\u201d (which gives average speed values) for molecules of different masses at various temperatures.<\/div>\n<h2 data-type=\"title\">The Kinetic-Molecular Theory Explains the Behavior of Gases, Part II<\/h2>\n<p id=\"fs-idp30560352\">According to Graham\u2019s law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates.<\/p>\n<p id=\"fs-idm65845776\">The rate of effusion of a gas depends directly on the (average) speed of its molecules:<\/p>\n<p>[latex]\\text{effusion rate}\\propto {u}_{\\text{rms}}[\/latex]<\/p>\n<p id=\"fs-idm219209680\">Using this relation, and the equation relating molecular speed to mass, Graham\u2019s law may be easily derived as shown here:<\/p>\n<p>[latex]{u}_{\\text{rms}}=\\sqrt{\\frac{3RT}{m}}[\/latex]<\/p>\n<p>[latex]m=\\frac{3RT}{{u}_{rms}^{2}}=\\frac{3RT}{{\\overline{u}}^{2}}[\/latex]<\/p>\n<p>[latex]\\frac{\\text{effusion rate A}}{\\text{effusion rate B}}=\\frac{{u}_{rms\\text{A}}}{{u}_{rms\\text{B}}}=\\frac{\\sqrt{\\frac{3RT}{{m}_{\\text{A}}}}}{\\sqrt{\\frac{3RT}{{m}_{\\text{B}}}}}=\\sqrt{\\frac{{m}_{\\text{B}}}{{m}_{\\text{A}}}}[\/latex]<\/p>\n<p id=\"fs-idm131494928\">The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham\u2019s law.<\/p>\n<div class=\"bcc-box bcc-success\">\n<h2>Key Concepts and Summary<\/h2>\n<p id=\"fs-idp236281408\">The kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average velocities determined by their absolute temperatures. The individual molecules of a gas exhibit a range of velocities, the distribution of these velocities being dependent on the temperature of the gas and the mass of its molecules.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Key Equations<\/h3>\n<ul>\n<li>[latex]{u}_{rms}=\\sqrt{\\overline{{u}^{2}}}=\\sqrt{\\frac{{u}_{1}^{2}+{u}_{2}^{2}+{u}_{3}^{2}+{u}_{4}^{2}+\\dots }{n}}[\/latex]<\/li>\n<li>[latex]{\\text{KE}}_{\\text{avg}}=\\frac{3}{2}R\\text{T}[\/latex]<\/li>\n<li>[latex]{u}_{\\text{rms}}=\\sqrt{\\frac{3RT}{m}}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Chemistry End of Chapter Exercises<\/h3>\n<ol>\n<li id=\"fs-idm212848544\">Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape.<\/li>\n<li>Can the speed of a given molecule in a gas double at constant temperature? Explain your answer.<\/li>\n<li>Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:\n<ol>\n<li>The pressure of the gas is increased by reducing the volume at constant temperature.<\/li>\n<li>The pressure of the gas is increased by increasing the temperature at constant volume.<\/li>\n<li>The average velocity of the molecules is increased by a factor of 2.<\/li>\n<\/ol>\n<\/li>\n<li>The distribution of molecular velocities in a sample of helium is shown in Figure 9.34. If the sample is cooled, will the distribution of velocities look more like that of H<sub>2<\/sub> or of H<sub>2<\/sub>O? Explain your answer.<\/li>\n<li>What is the ratio of the average kinetic energy of a SO<sub>2<\/sub> molecule to that of an O<sub>2<\/sub> molecule in a mixture of two gases? What is the ratio of the root mean square speeds, <em data-effect=\"italics\">u<\/em><sub>rms<\/sub>, of the two gases?<\/li>\n<li>A 1-L sample of CO initially at STP is heated to 546 \u00b0C, and its volume is increased to 2 L.\n<ol>\n<li>What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?<\/li>\n<li>What is the effect on the average kinetic energy of the molecules?<\/li>\n<li>What is the effect on the root mean square speed of the molecules?<\/li>\n<\/ol>\n<\/li>\n<li>The root mean square speed of H<sub>2<\/sub> molecules at 25 \u00b0C is about 1.6 km\/s. What is the root mean square speed of a N<sub>2<\/sub> molecule at 25 \u00b0C?<\/li>\n<li>Answer the following questions:\n<ol>\n<li>Is the pressure of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?<\/li>\n<li>Is the density of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?<\/li>\n<li>At a pressure of 1 atm and a temperature of 20 \u00b0C, dry air has a density of 1.2256 g\/L. What is the (average) molar mass of dry air?<\/li>\n<li>The average temperature of the gas in a hot air balloon is 1.30 \u00d7 10<sup>2<\/sup> \u00b0F. Calculate its density, assuming the molar mass equals that of dry air.<\/li>\n<li>The lifting capacity of a hot air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?<\/li>\n<li>An average balloon has a diameter of 60 feet and a volume of 1.1 \u00d7 10<sup>5<\/sup> ft<sup>3<\/sup>. What is the lifting power of such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?<\/li>\n<li>A balloon carries 40.0 gallons of liquid propane (density 0.5005 g\/L). What volume of CO<sub>2<\/sub> and H<sub>2<\/sub>O gas is produced by the combustion of this propane?<\/li>\n<li>A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in kJ\/min) from the hot air in the bag during the flight?<\/li>\n<\/ol>\n<\/li>\n<li>Show that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas 2, [latex]\\frac{{R}_{1}}{{R}_{2}},[\/latex] is the same at 0 \u00b0C and 100 \u00b0C.<\/li>\n<\/ol>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h4>Selected Answers<\/h4>\n<p>2.\u00a0Yes. At any given instant, there are a range of values of molecular speeds in a sample of gas. Any single molecule can speed up or slow down as it collides with other molecules. The average velocity of all the molecules is constant at constant temperature.<\/p>\n<p>4.\u00a0H<sub>2<\/sub>O. Cooling slows the velocities of the He atoms, causing them to behave as though they were heavier.<\/p>\n<p>6.\u00a0Both the temperature and the volume are doubled for this gas (<em data-effect=\"italics\">n<\/em> constant), so <em data-effect=\"italics\">P<\/em> remains constant.<\/p>\n<p>(a) The number of collisions per unit area of the container wall is constant.<\/p>\n<p>(b) The average kinetic energy doubles; it is proportional to temperature.<\/p>\n<p>(c) The root mean square speed increases to [latex]\\sqrt{2}[\/latex] times its initial value; <em data-effect=\"italics\">u<\/em><sub>rms<\/sub> is proportional to [latex]\\sqrt{{\\text{KE}}_{\\text{avg}}}.[\/latex]<\/p>\n<p>8. (a) equal, because the balloon is free to expand until the pressures are equalized;<\/p>\n<p>(b) less than the density outside;<\/p>\n<p>(c) assume three-place accuracy throughout unless greater accuracy is stated:[latex]\\text{molar mass}=\\frac{DRT}{P}=1.2256\\text{g}\\cancel{{\\text{L}}^{\\text{-1}}}\\times \\frac{0.08206\\cancel{\\text{L}}\\cancel{\\text{atm}}{\\text{mol}}^{\\text{-1}}\\cancel{{\\text{K}}^{\\text{-1}}}\\times 293.15\\cancel{\\text{K}}}{1.00\\cancel{\\text{atm}}}=29.48{\\text{g mol}}^{\\text{-1}};[\/latex]<\/p>\n<p>(d) convert the temperature to \u00b0C; then use the ideal gas law:<\/p>\n<p>[latex]\\text{\\textdegree C}=\\frac{5}{9}\\left(\\text{F}-32\\right)=\\frac{5}{9}\\left(130-32\\right)=54.44\\text{\\textdegree C}=327.6\\text{K}[\/latex]<\/p>\n<p>[latex]D=\\frac{\\mathcal{M}P}{RT}=29.48\\text{g}\\cancel{{\\text{mol}}^{\\text{-1}}}\\times \\frac{1.00\\cancel{\\text{atm}}}{0.08206\\text{L}\\cancel{\\text{atm}}\\cancel{{\\text{mol}}^{\\text{-1}}}\\cancel{{\\text{K}}^{\\text{-1}}}\\times 327.6\\cancel{\\text{K}}}=1.0966{\\text{g L}}^{\\text{-1}};[\/latex]<\/p>\n<p>(e) 1.2256 g\/L \u2013 1.09966 g\/L = 0.129 g\/L;<\/p>\n<p>(f) calculate the volume in liters, multiply the volume by the density difference to find the lifting capacity of the balloon, subtract the weight of the balloon after converting to pounds:<\/p>\n<p>[latex]1.1\\times 105{\\text{ft}}^{3}\\times {\\left(\\frac{\\text{12 in}}{\\text{91 ft}}\\right)}^{3}\\times {\\left(\\frac{\\text{2.54 cm}}{\\text{in}}\\right)}^{3}\\times \\frac{\\text{1 L}}{1000{\\text{cm}}^{3}}=3.11\\times {10}^{6}\\text{L}[\/latex]<\/p>\n<p>3.11 \u00d7 106 L \u00d7 0.129 g\/L = 4.01 \u00d7 10<sup>5<\/sup> g<\/p>\n<p>[latex]\\frac{4.01\\times {10}^{5}\\text{g}}{453.59{\\text{g lb}}^{\\text{-1}}}=884\\text{lb;}\\text{884 lb}-\\text{500 lb}=\\text{384 lb}[\/latex]<\/p>\n<p>net lifting capacity = 384 lb;<\/p>\n<p>(g) First, find the mass of propane contained in 40.0 gal. Then calculate the moles of CO<sub>2<\/sub>(<em data-effect=\"italics\">g<\/em>) and H<sub>2<\/sub>O(<em data-effect=\"italics\">g<\/em>) produced from the balanced equation.<\/p>\n<p>[latex]40.0\\cancel{\\text{gal}}\\times \\frac{4\\left(0.9463\\text{L}\\right)}{1\\cancel{\\text{gal}}}=151.4\\text{L}[\/latex]<\/p>\n<p>[latex]151.4\\cancel{\\text{L}}\\times 0.5005\\text{g}{\\cancel{\\text{L}}}^{\\cancel{-1}}=75.8\\text{g}[\/latex]<\/p>\n<p>Molar mass of propane = 3(12.011) + 8(1.00794) = 36.033 + 8.064 = 44.097 g mol<sup>\u20131<\/sup><\/p>\n<p>[latex]\\frac{75.8\\cancel{\\text{g}}}{44.097\\cancel{\\text{g}}{\\text{mol}}^{\\text{-1}}}=1.72\\text{mol}[\/latex]<\/p>\n<p>The reaction is [latex]{\\text{C}}_{3}{\\text{H}}_{8}\\left(g\\right)+5{\\text{O}}_{2}\\left(g\\right)\\rightarrow 3{\\text{CO}}_{2}\\left(g\\right)+4{\\text{H}}_{2}\\text{O}\\left(g\\right)[\/latex]<\/p>\n<p>For each 1.72 mol propane, there are 3 \u00d7 1.72 mol = 5.15 mol of CO<sub>2<\/sub> and 4 \u00d7 1.72 mol = 6.88 mol H<sub>2<\/sub> O. The total volume at STP = 22.4 L \u00d7 12.04 = 270 L;<\/p>\n<p>(h) The total heat released is determined from the heat of combustion of the propane. Using the equation in part (g),[latex]\\begin{array}{ll}\\Delta {H}_{\\text{combustion}}^{ \\textdegree }\\hfill & =3{\\Delta H}_{{\\text{CO}}_{2}\\left(g\\right)}^{\\textdegree }+4{\\Delta H}_{{\\text{H}}_{2}\\text{O}\\left(g\\right)}^{\\textdegree }-{\\Delta H}_{\\text{propane}}^{\\textdegree }\\hfill \\\\ \\hfill & =3\\left(-393.51\\right)+4\\left(-241.82\\right)-\\left(-103.85\\right)\\hfill \\\\ \\hfill & =-1180.52-967.28+103.85=-2043.96{\\text{kJ mol}}^{-1}\\hfill \\end{array}[\/latex]<\/p>\n<p>Since there is 1.72 mol propane, 1.72 \u00d7 2043.96 kJ mol<sup>-1<\/sup> = 3.52 \u00d7 10<sup>3<\/sup> kJ used for heating. This heat is used over 90 minutes, so [latex]\\frac{3.52\\times {10}^{3}\\text{kJ}}{\\text{90 min}}=39.1{\\text{kJ min}}^{\\text{-1}}[\/latex] is released.<\/p>\n<p>9. The rate at which a gas will diffuse, <em data-effect=\"italics\">R,<\/em> is proportional lo <em data-effect=\"italics\">u<\/em><sub>rms<\/sub>, the root mean square speed of its molecules. The square of this value, in turn, is proportional to the average kinetic energy. The average kinetic energy is:[latex]{\\overline{\\text{KE}}}_{\\text{avg}}=kT.[\/latex]For two different gases, 1 and 2, the constant of proportionality can be represented as <em data-effect=\"italics\">k<sub>1<\/sub><\/em> and <em data-effect=\"italics\">k<\/em><sub>2<\/sub>, respectively. Thus,[latex]\\frac{{R}_{1}}{{R}_{2}}=\\frac{{k}_{1}\\sqrt{T}}{{k}_{2}\\sqrt{T}}.[\/latex]As a result of this relationship, no matter at which temperature diffusion occurs, the temperature term will cancel out of the equation and the ratio of rates will be the same.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Glossary<\/h3>\n<p><strong>kinetic molecular theory<\/strong><br \/>\ntheory based on simple principles and assumptions that effectively explains ideal gas behavior<\/p>\n<p><strong>root mean square velocity (<em data-effect=\"italics\">u<\/em><sub>rms<\/sub>)<\/strong><br \/>\nmeasure of average velocity for a group of particles calculated as the square root of the average squared velocity<\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2113\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":5,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2113","chapter","type-chapter","status-publish","hentry"],"part":3001,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/2113","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/users\/5"}],"version-history":[{"count":11,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/2113\/revisions"}],"predecessor-version":[{"id":5226,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/2113\/revisions\/5226"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/parts\/3001"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/2113\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/media?parent=2113"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapter-type?post=2113"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/contributor?post=2113"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/license?post=2113"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}