Chapter 17 Temperature and the Kinetic Theory of Gases

Chapter 7 emperature and the Kinetic heory of Gases Conceptual roblems rue or false: (a) he zeroth law of thermodynamics states that two objects in thermal equilibrium with each other must be in thermal equilibrium with a third object. (b) he ahrenheit and Celsius temperature scales differ only in the choice of the ice-point temperature. (c) he Celsius degree and the kelin are the same size. (a) alse. If two objects are in thermal equilibrium with a third, then they are in thermal equilibrium with each other. (b) alse. he ahrenheit and Celsius temperature scales also differ in the number of interals between the ice-point temperature and the steam-point temperature. (c) rue. How can you determine if two objects are in thermal equilibrium with each other when putting them into physical contact with each other would hae undesirable effects? (or example, if you put a piece of sodium in contact with water there would be a iolent chemical reaction.) Determine the Concept ut each in thermal equilibrium with a third body; that is, a thermometer. If each body is in thermal equilibrium with the third, then they are in thermal equilibrium with each other. [SS] Yesterday I woke up and it was 0 º in my bedroom, said ert to his old friend ort. hat s nothing, replied ort. y room was 5.0 ºC. Who had the colder room, ert or ort? icture the roblem We can decide which room was colder by conerting 0 to the equialent Celsius temperature. Using the ahrenheit-celsius conersion, conert 0 to the equialent Celsius temperature: 5 ( ) ( 0 ) 5 tc 9 t 9 6.7 C so ert' s room was colder. 7

74 Chapter 7 4 wo identical essels contain different ideal gases at the same pressure and temperature. It follows that (a) the number of gas molecules is the same in both essels, (b) the total mass of gas is the same in both essels, (c) the aerage speed of the gas molecules is the same in both essels, (d) None of the aboe. Determine the Concept Because the essels are identical (hae the same olume) and the two ideal gases are at the same pressure and temperature, the ideal-gas law ( V Nk ) tells us that the number of gas molecules must be the same in both is correct. essels. ( a) 5 [SS] igure 7-8 shows a plot of olume ersus absolute temperature for a process that takes a fixed amount of an ideal gas from point A to point B. What happens to the pressure of the gas during this process? Determine the Concept rom the ideal-gas law, we hae nr V. In the process depicted, both the temperature and the olume increase, but the temperature increases faster than does the olume. Hence, the pressure increases. 6 igure 7-9 shows a plot of pressure ersus absolute temperature for a process that takes a sample of an ideal gas from point A to point B. What happens to the olume of the gas during this process? Determine the Concept rom the ideal-gas law, we hae V nr. In the process depicted, both the temperature and the pressure increase, but the pressure increases faster than does the temperature. Hence, the olume decreases. 7 If a essel contains equal amounts, by mass, of helium and argon, which of the following are true? (a) (b) (c) (d) he partial pressure exerted by each of the two gases on the walls of the container is the same. he aerage speed of a helium atom is the same as that of an argon atom. he number of helium atoms and argon atoms in the essel are equal. None of the aboe. Determine the Concept (a) alse. he partial pressure exerted by each gas in the mixture is the pressure it would exert if it alone occupied the container. Because the densities of helium and argon are not the same, these gases occupy different olumes and, hence, their partial pressures are not the same. (b) alse. Assuming the gasses hae been in the essel for some time, their aerage kinetic energies would be the same. Because the densities of helium and

emperature and the Kinetic heory of Gases 75 argon are not the same, their aerage speeds must be different. (c) alse. We know that the olumes of the two gasses are equal (they both occupy the full olume of the container) and that their temperatures are equal. Because their pressures are different, the number of atoms of each gas must be different, too. (d) Because none of the aboe is true, (d) is true. 8 By what factor must the absolute temperature of a gas be increased to double the rms speed of its molecules? Determine the Concept We can use rms R a gas to the rms speed of its molecules. to relate the temperature of Express the dependence of the rms speed of the molecules of a gas on their absolute temperature: R rms where R is the gas constant, is the molar mass, and is the absolute temperature. Because rms, the temperature must be quadrupled in order to double the rms speed of the molecules. 9 wo different gases are at the same temperature. What can be said about the aerage translational kinetic energies of the molecules? What can be said about the rms speeds of the gas molecules? Determine the Concept he aerage kinetic energies are equal. he ratio of their rms speeds is equal to the square root of the reciprocal of the ratio of their molecular masses. 0 A essel holds a mixture of helium (He) and methane (CH 4 ). he ratio of the rms speed of the He atoms to that of the CH 4 molecules is (a), (b), (c) 4, (d) 6 icture the roblem We can express the rms speeds of the helium atoms and the methane molecules using R. rms Express the rms speed of the helium atoms: rms, He R He

76 Chapter 7 Express the rms speed of the methane molecules: rms.ch 4 R CH 4 Diide the first of these equations by the second to obtain: Use Appendix C to find the molar masses of helium and methane: rms, He CH 4 rms.ch 4 He rms, He rms.ch 4 6g/mol 4g/mol and (b) is correct. rue or false: If the pressure of a fixed amount of gas increases, the temperature of the gas must increase. alse. Whether the pressure changes also depends on whether and how the olume changes. In an isothermal process, the pressure can increase while the olume decreases and the temperature is constant. Why might the Celsius and ahrenheit scales be more conenient than the absolute scale for ordinary, nonscientific purposes? Determine the Concept or the Celsius scale, the ice point (0 C) and the boiling point of water at atm (00 C) are more conenient than 7 K and 7 K; temperatures in roughly this range are normally encountered. On the ahrenheit scale, the temperature of warm-blooded animals is roughly 00 ; this may be a more conenient reference than approximately 00 K. hroughout most of the world, the Celsius scale is the standard for nonscientific purposes. An astronomer claims that the temperature at the center of the Sun is about 0 7 degrees. Do you think that this temperature is in kelins, degrees Celsius, or doesn t it matter? Determine the Concept Because t C + 7.5 K and 0 7 >> 7, it doesn t matter. 4 Imagine that you hae a fixed amount of ideal gas in a container that expands to maintain constant pressure. If you double the absolute temperature of the gas, the aerage speed of the molecules (a) remains constant, (b) doubles, (c) quadruples, (d) increases by a factor of. Determine the Concept he aerage speed of the molecules in an ideal gas depends on the square root of the kelin temperature. Because, doubling a

emperature and the Kinetic heory of Gases 77 the temperature while maintaining constant pressure increases the aerage speed by a factor of. (d) is correct. 5 [SS] Suppose that you compress an ideal gas to half its original olume, while also haling its absolute temperature. During this process, the pressure of the gas (a) hales, (b) remains constant, (c) doubles, (d) quadruples. Determine the Concept rom the ideal-gas law, V nr, haling both the temperature and olume of the gas leaes the pressure unchanged. (b) is correct. 6 he aerage translational kinetic energy of the molecules of a gas depends on (a) the number of moles and the temperature, (b) the pressure and the temperature, (c) the pressure only, (d) the temperature only. Determine the Concept he aerage translational kinetic energy of the molecules of an ideal gas K a depends on its temperature according to K k. (d) is correct. 7 [SS] Which speed is greater, the speed of sound in a gas or the rms speed of the molecules of the gas? Justify your answer, using the appropriate formulas, and explain why your answer is intuitiely plausible. Determine the Concept he rms speed of molecules of an ideal gas is gien by R rms and the speed of sound in a gas is gien by γr sound. a he rms speed of the molecules of an ideal-gas is gien by: rms R he speed of sound in a gas is gien by: sound γr Diide the first of these equations by the second and simplify to obtain: rms sound R γr γ or a monatomic gas, γ.67 and: rms, monatomic sound.67.4

78 Chapter 7 or a diatomic gas, γ.40 and: rms, diatomic sound.40.46 In general, the rms speed is always somewhat greater than the speed of sound. Howeer, it is only the component of the molecular elocities in the direction of propagation that is releant to this issue. In addition, In a gas the mean free path is greater than the aerage intermolecular distance. 8 Imagine that you increase the temperature of a gas while holding its olume fixed. Explain in terms of molecular motion why the pressure of the gas on the walls of its container increases. Determine the Concept he pressure is a measure of the change in momentum per second of a gas molecule on collision with the wall of the container. When the gas is heated, the aerage elocity and the aerage momentum of the molecules increase and, as a consequence, the pressure exerted by the molecules increases. 9 Imagine that you compress a gas while holding it at fixed temperature (perhaps by immersing the container in cool water). Explain in terms of molecular motion why the pressure of the gas on the walls of its container increases. Determine the Concept If the olume decreases the pressure increases because more molecules hit a unit of area of the walls in a gien time. he fact that the temperature does not change tells us the molecular speed does not change with olume. 0 Oxygen has a molar mass of g/mol and nitrogen has a molar mass of 8 g/mol. he oxygen and nitrogen molecules in a room hae (a) (b) (c) (d) (e) (f) equal aerage translational kinetic energies, but the oxygen molecules hae a larger aerage speed than the nitrogen molecules hae. equal aerage translational kinetic energies, but the oxygen molecules hae a smaller aerage speed than the nitrogen molecules hae. equal aerage translational kinetic energies and equal aerage speeds. equal aerage speeds, but the oxygen molecules hae a larger aerage translational kinetic energy than the nitrogen molecules hae. equal aerage speeds, but the oxygen molecules hae a smaller aerage translational kinetic energy than the nitrogen molecules hae. None of the aboe. icture the roblem he aerage kinetic energies of the molecules are gien by K m k Assuming that the room s temperature distribution is a ( ). a

emperature and the Kinetic heory of Gases 79 uniform, we can conclude that the oxygen and nitrogen molecules hae equal aerage kinetic energies. Because the oxygen molecules are more massie, they must be moing slower than the nitrogen molecules. (b) is correct. [SS] Liquid nitrogen is relatiely cheap, while liquid helium is relatiely expensie. One reason for the difference in price is that while nitrogen is the most common constituent of the atmosphere, only small traces of helium can be found in the atmosphere. Use ideas from this chapter to explain why it is that only small traces of helium can be found in the atmosphere. Determine the Concept he aerage molecular speed of He gas at 00 K is about.4 km/s, so a significant fraction of He molecules hae speeds in excess of earth s escape elocity (. km/s). hus, they "leak" away into space. Oer time, the He content of the atmosphere decreases to almost nothing. Estimation and Approximation Estimate the total number of air molecules in your classroom. icture the roblem he number of air molecules in your classroom is gien by the ideal-gas law. rom the ideal-gas law, the number of molecules N in a gien olume V at pressure and temperature is gien by: Assuming that a typical classroom is a rectangular parallelepiped, its olume is gien by: V N k where k is Boltzmann s constant. V wh Substituting for V yields: N wh k Assume atmospheric pressure, room temperature and that the room is 5 m square and 5 m high to obtain: N 0.5 ka atm ( 5 m)( 5 m)( 5 m) atm (.8 0 J/K)( 9 K) 8 0

70 Chapter 7 Estimate the density of dry air at sea leel on a warm summer day. icture the roblem We can use the definition of mass density, the definition of the molar mass of a gas, and the ideal-gas law to estimate the density of air at sea leel on a warm day. he density of air at sea leel is gien by: he mass of the air molecules occupying a gien olume is the product of the number of moles n and the molar mass of the molecules. Substitute for m to obtain: rom the ideal-gas law, the number of moles in a gien olume depends on the pressure and temperature according to: Substitute in the expression for ρ and simplify to obtain: Assuming atmospheric pressure, a temperature of 7 C (9 ), and 9 g/mol for the molar mass of air, substitute numerical alues and ealuate ρ: m ρ V n ρ V n ρ ρ V R V V R atm R ( 9 g/mol) atm ( 8.4 J/mol K)( 00 K). kg/m 0.5 ka 4 A stoppered test tube that has a olume of 0.0 ml has.00 ml of water at its bottom. he water has a temperature of 00ºC and is initially at a pressure of.00 atm. he test tube is held oer a flame until the water has completely boiled away. Estimate the final pressure inside the test tube. icture the roblem Assuming the steam to be an ideal gas at a temperature of 7 K, we can use the ideal-gas law to estimate the pressure inside the test tube when the water is completely boiled away. Using the ideal-gas law, relate the pressure inside the test tube to its olume and the temperature: Nk V

emperature and the Kinetic heory of Gases 7 Relate the number of particles N to the mass of water, its molar mass, and Aogadro s number N A : Relate the mass of.00 ml of water to its density: m N N m N N A m ρv.00g A (.00 0 kg/m ) 6 (.00 0 m ) Substitute for m, N A, and (8 g/mol) and ealuate N: N 6.0 0 particles/mol (.00g) 8g/mol.46 0 particles Substitute numerical alues and ealuate : (.46 0 particles)(.8 0 J/K)( 7K).7 0 70 atm 7 6 0 m 0.0 ml ml atm N/m 0.5 0 N/m 5 [SS] In Chapter, we found that the escape speed at the surface of a planet of radius R is e gr, where g is the acceleration due to graity at the surface of the planet. If the rms speed of a gas is greater than about 5 to 0 percent of the escape speed of a planet, irtually all of the molecules of that gas will escape the atmosphere of the planet. (a) (b) (c) (d) At what temperature is rms for O equal to 5 percent of the escape speed for Earth? At what temperature is rms for H equal to 5 percent of the escape speed for Earth? emperatures in the upper atmosphere reach 000 K. How does this help account for the low abundance of hydrogen in Earth s atmosphere? Compute the temperatures for which the rms speeds of O and H are equal to 5 percent of the escape speed at the surface of the moon, where g is about one-sixth of its alue on Earth and R 78 km. How does this account for the absence of an atmosphere on the moon? icture the roblem We can find the escape temperatures for the earth and the moon by equating, in turn, 0.5 e and rms of O and H. We can compare these

7 Chapter 7 temperatures to explain the absence from Earth s upper atmosphere and from the surface of the moon. See Appendix C for the molar masses of O and H. (a) Express rms for O : R rms, O O where R is the gas constant, is the absolute temperature, and O is the molar mass of oxygen. Equate 0.5 e and rms, O : 0.5 gr earth R Sole for to obtain: 0.045gRearth () R Substitute numerical alues and ealuate for O : 6 ( )( 6.7 0 m)(.0 0 kg/mol) 0.045 9.8m/s.60 0 K 8.4 J/mol ( K) (b) Substitute numerical alues and ealuate for H : 6 ( )( 6.7 0 m)(.0 0 kg/mol) 0.045 9.8m/s ( K) 8.4 J/mol 0K (c) Because hydrogen is lighter than air it rises to the top of the atmosphere. Because the temperature is high there, a greater fraction of the molecules reach escape speed. (d) Express equation () at the surface of the moon: 0.045gmoonRmoon R 0.045( 6 gearth ) Rmoon R 0.005gearthRmoon R Substitute numerical alues and ealuate for O : 0.005 9.8m/s 6 ( )(.78 0 m)(.0 0 kg/mol) 60K 8.4 J/mol K

Substitute numerical alues and ealuate for H : emperature and the Kinetic heory of Gases 7 6 ( )(.78 0 m)(.0 0 kg/mol) 0K 0.005 9.8m/s 8.4 J/mol K Because g is less on the moon, the escape speed is lower. hus, a larger percentage of the molecules are moing at escape speed. 6 he escape speed for gas molecules in the atmosphere of ars is 5.0 km/s and the surface temperature of ars is typically 0ºC. Calculate the rms speeds for (a) H, (b) O, and (c) CO at this temperature. (d) Are H, O, and CO likely to be found in the atmosphere of ars? icture the roblem We can use rms R to calculate the rms speeds of H, O, and CO at 7 K and then compare these speeds to 0% of the escape elocity on ars to decide the likelihood of finding these gases in the atmosphere of ars. See Appendix C for molar masses. Express the rms speed of an atom as a function of the temperature: (a) Substitute numerical alues and ealuate rms for H : rms rms, H R ( K)( 7K) 8.4 J/mol.0 0 kg/mol.84 km/s (b) Ealuate rms for O : ( 8.4 J/mol K)( 7K) rms, O.0 0 46m/s kg/mol (c) Ealuate rms for CO : ( 8.4 J/mol K)( 7K) rms, CO 44.0 0 9m/s 5 esc 5 kg/mol (d) Calculate 0% of esc for ars: ( 5.0 km/s).0km/s Because is greater than rms CO and O but less than rms for H, O and CO, but not H should be present.

74 Chapter 7 7 [SS] he escape speed for gas molecules in the atmosphere of Jupiter is 60 km/s and the surface temperature of Jupiter is typically 50ºC. Calculate the rms speeds for (a) H, (b) O, and (c) CO at this temperature. (d) Are H, O, and CO likely to be found in the atmosphere of Jupiter? icture the roblem We can use rms R to calculate the rms speeds of H, O, and CO at K and then compare these speeds to 0% of the escape elocity on Jupiter to decide the likelihood of finding these gases in the atmosphere of Jupiter. See Appendix C for the molar masses of H, O, and CO. Express the rms speed of an atom as a function of the temperature: (a) Substitute numerical alues and ealuate rms for H : rms rms, H R ( K)( K) 8.4 J/mol.0 0 kg/mol.km/s (b) Ealuate rms for O : ( 8.4 J/mol K)( K) rms, O.0 0 0 m/s kg/mol (c) Ealuate rms for CO : ( 8.4 J/mol K)( K) rms, CO 44.0 0 64 m/s 5 esc 5 kg/mol (d) Calculate 0% of esc for Jupiter: ( 60 km/s) km/s Because e is greater than rms for O, CO and H, O, all three gasses should be found on Jupiter. 8 Estimate the aerage pressure on the front wall of a racquetball court, due to the collisions of the ball with the wall during a game. Use any reasonable numbers for the mass of the ball, its typical speed, and the dimensions of the court. Is the aerage pressure from the ball significant compared to that from the air? icture the roblem he aerage pressure exerted by the ball on the wall is the ratio of the aerage force it exerts on the wall to the area of the wall. he aerage force, in turn, is the rate at which the momentum of the ball changes during each collision with the wall. Assume that the mass of a racquetball is 00 g, that the court measures 5 m by 5 m, and that the speed of the racquetball is 0 m/s. We ll

emperature and the Kinetic heory of Gases 75 also assume that the interal Δt between collisions with the wall is s. he aerage pressure exerted on the wall by the racquetball is gien by: Assuming head-on collisions with the wall, the change in momentum of the ball during each collision is m and the aerage force exerted by the ball on the wall is: Substituting for a yields: Substitute numerical alues and ealuate a : Express the ratio of a and atm to obtain: a a a A wh where w and h are the width and height of the wall, respectiely. Δp m a Δt Δt where Δt is the elapsed time between collisions of the ball with the wall. a a or a atm m whδt (.kg)( 0 m/s) ( 5 m)( 5 m)( s) 0.a 0.a 6 0 5 0 a 6 a 0 atm 0.08 N/m and the aerage pressure from the ball is not significant compared to atmospheric pressure. 9 o a first approximation, the Sun consists of a gas of equal numbers of protons and electrons. (he masses of these particles can be found in Appendix B.) he temperature at the center of the Sun is about 0 7 K, and the density of the Sun is about 0 5 kg/m. Because the temperature is so high, the protons and electrons are separate particles (rather than being joined together to form hydrogen atoms). (a) Estimate the pressure at the center of the Sun. (b) Estimate the rms speeds of the protons and the electrons at the center of the Sun. icture the roblem We ll assume that we can model the plasma as an ideal gas, at least to a first approximation. hen we can apply the ideal gas law to an arbitrary olume of the material, say one cubic meter. (a) o the extent that the plasma acts like an ideal gas, the pressure at the center of the Sun is gien by: nr V

76 Chapter 7 he mass of our m olume of matter is 0 5 kg and this mass consists mostly of protons. One mole of protons has a mass of g, so in 0 5 kg, the number of moles of protons would be: Substitute numerical alues and ealuate : n m 5 0 kg 8 0 mol 0 kg protons protons protons or, counting electrons, 8 n n 0 mol protons 8 7 ( 0 mol)( 8.4 J/mol K)( 0 K) 0 6 0 m atm a 0.5 ka atm (b) o one significant figure, the aerage speed of the protons is the same as the rms speed: Substitute numerical alues and ealuate : rms, protons rms, protons rms, protons R protons 7 ( K)( 0 K) 8.4 J/mol 0.00 kg 5 5 0 m/s he rms speed of an electron at the center of the Sun is gien by: R rms, electrons electrons 000 R protons Substitute numerical alues and ealuate : rms, electrons rms, electrons 7 ( K)( 0 K) 8.4 J/mol 000 ( 0.00 kg) 0 7 m/s Remarks: he huge pressure we calculated in (a) is required to support the tremendous weight of the rest of the sun. he ery high speed we calculated for the plasma electrons is still an order-of-magnitude smaller than the speed of light. 0 You are designing a acuum chamber for fabricating reflectie coatings. Inside this chamber, a small sample of metal will be aporized so that its atoms trael in straight lines (the effects of graity are negligible during the brief time of flight) to a surface where they land to form a ery thin film. he sample of metal is 0 cm from the surface to which the metal atoms will adhere. How low must the pressure in the chamber be so that the metal atoms only rarely collide with air molecules before they land on the surface?

emperature and the Kinetic heory of Gases 77 icture the roblem We can use the expression for the mean free path of a molecule to eliminate the number of molecules per unit olume n V from the idealgas law. Assume that the air in the chamber is at room temperature (00 K) and that the aerage diameter of an air molecule is 4 0 0 m. Apply the ideal gas law to express the pressure in terms of the number of molecules per unit olume and the temperature: Nk V n V k he mean free path of an air molecule is gien by: λ n V π d n V λπ d Substitute for n V to obtain: Substitute numerical alues and ealuate : k λπ d (.8 0 J/K)( 00 K) 0 ( 0 cm) π ( 4 0 m) 0 ma [SS] In normal breathing conditions, approximately 5 percent of each exhaled breath is carbon dioxide. Gien this information and neglecting any difference in water-apor content, estimate the typical difference in mass between an inhaled breath and an exhaled breath. icture the roblem One breath (one s lung capacity) is about half a liter. he only thing that occurs in breathing is that oxygen is exchanged for carbon dioxide. Let s estimate that of the 0% of the air that is breathed in as oxygen, ¼ is exchanged for carbon dioxide. hen the mass difference between breaths will be 5% of a breath multiplied by the molar mass difference between oxygen and carbon dioxide and by the number of moles in a breath. Because this is an estimation problem, we ll use g/mol as an approximation for the molar mass of oxygen and 44 g/mol as an approximation for the molar mass of carbon dioxide. Express the difference in mass between an inhaled breath and an Δm m ( ) exhaled breath: CO O CO breath O f where fco mco n is the fraction of the air breathed in that is exchanged for carbon dioxide. he number of moles per breath is gien by: Vbreath n breath.4 L/mol

78 Chapter 7 Substituting for n breath yields: Δm f Vbreath ( ) CO O CO.4 L/mol Substitute numerical alues and ealuate Δm: Δm ( 0.05)( 44 g/mol g/mol) ( 0.5 L).4 L/mol 0 4 g emperature Scales A certain ski wax is rated for use between and 7.0 ºC. What is this temperature range on the ahrenheit scale? icture the roblem We can use the fact that 5 C 9 to set up a proportion that allows us to make easy interal conersions from either the Celsius or ahrenheit scale to the other. he proportion relating a temperature range on the ahrenheit scale of a temperature range on the Celsius scale is: Δt Δt C 9 Δt 5 C 9 Δt 5 C C Substitute numerical alues and ealuate Δt : Δt 9 5 C ( 7 C ( C) ) 9 Remarks: An equialent but slightly longer solution inoles conerting the two temperatures to their ahrenheit equialents and then subtracting these temperatures. [SS] he melting point of gold is 945.4 º. Express this temperature in degrees Celsius. icture the roblem We can use the ahrenheit-celsius conersion equation to find this temperature on the Celsius scale. Conert 945.4 to the equialent Celsius temperature: t C t 5 9 5 ( ) ( 945.4 ) 06 C 9 4 A weather report indicates that the temperature is expected to drop by 5.0 C oer the next four hours. By how many degrees on the ahrenheit scale will the temperature drop?

emperature and the Kinetic heory of Gases 79 icture the roblem We can use the fact that 5 C 9 to set up a proportion that allows us to make easy interal conersions from either the Celsius or ahrenheit scale to the other. he proportion relating a temperature range on the ahrenheit scale of a temperature range on the Celsius scale is: Δt Δt C 9 Δt 5 C 9 Δt 5 C C Substitute numerical alues and ealuate Δt : Δt 9 5 C ( 5.0 C ) 7.0 5 he length of the column of mercury in a thermometer is 4.00 cm when the thermometer is immersed in ice water at atm of pressure, and 4.0 cm when the thermometer is immersed in boiling water at atm of pressure. Assume that the length of the mercury column aries linearly with temperature. (a) Sketch a graph of the length of the mercury column ersus temperature (in degrees Celsius). (b) What is the length of the column at room temperature (.0ºC)? (c) If the mercury column is 5.4 cm long when the thermometer is immersed in a chemical solution, what is the temperature of the solution? icture the roblem We can use the equation of the graph plotted in (a) to (b) find the length of the mercury column at room temperature and (c) the temperature of the solution when the height of the mercury column is 5.4 cm. (a) A graph of the length of the mercury column ersus temperature (in degrees Celsius) is shown to the right. he equation of the line is: cm L 0.00 tc + 4.00 cm () C 4.0 4.00 L, cm 0 00 t C, C (b) Ealuate L (.0 C) cm L 0.00 C 8.40 cm (.0 C) + 4.00 cm (c) Sole equation () for t C to obtain: t C 4.00 cm L cm 0.00 C

70 Chapter 7 Substitute numerical alues and t 5.4 cm : ealuate ( ) C t C ( 5.4 cm) 5.4 cm 4.00 cm cm 0.00 C 07 C 6 he temperature of the interior of the Sun is about.0 0 7 K. What is this temperature in (a) Celsius degrees, (b) kelins, and (c) ahrenheit degrees? icture the roblem We can use the temperature conersion equations 9 t 5 tc + and tc 7.5K to conert 0 7 K to the ahrenheit and Celsius temperatures. Express the kelin temperature in terms of the Celsius temperature: (a) Sole for and ealuate t C : (b) Use the Celsius to ahrenheit conersion equation to ealuate t : t t C t C + 7.5K 7.5 K.0 0 9 5 7.0 0 K 7.5 K 7 C 7 (.0 0 C) 7.8 0 + 7 he boiling point of nitrogen, N, is 77.5 K. Express this temperature in degrees ahrenheit. icture the roblem While we could conert 77.5 K to a Celsius temperature and then conert the Celsius temperature to a ahrenheit temperature, an alternatie solution is to use the diagram to the right to set up a proportion for the direct conersion of the kelin temperature to its ahrenheit equialent. t K 7.5 7.5 77.5 Use the diagram to set up the proportion: t 7.5 K 77.5 K 7.5 K 7.5K or t 95.8 80 00

emperature and the Kinetic heory of Gases 7 Soling for t yields: 95.8 t 80 0 00 8 he pressure of a constant-olume gas thermometer is 0.400 atm at the ice point and 0.546 atm at the steam point. (a) Sketch a graph of pressure ersus Celsius temperature for this thermometer. (b) When the pressure is 0.00 atm, what is the temperature? (c) What is the pressure at 444.6 ºC (the boiling point of sulfur)? icture the roblem We can use the equation of the graph plotted in (a) to (b) find the temperature when the pressure is 0.00 atm and (c) the pressure when the temperature is 444.6ºC. (a) A graph of pressure (in atm) ersus temperature (in degrees Celsius) for this thermometer is shown to the right. he equation of this graph is: atm.46 0 tc C (b) Soling for t C yields: + 0.400 atm t 0.546 0.400 C, atm 0 00 0.400 atm atm.46 0 C t, C C Substitute numerical alues and t 0.00 atm : ealuate ( ) C t C ( 0.00 atm) 0.00 atm 0.400 atm atm.46 0 C 05 C Substitute numerical alues and ealuate ( 444.6 C) atm C : ( 444.6 C).46 0 ( 444.6 C) + 0.400 atm.05 atm 9 [SS] A constant-olume gas thermometer reads 50.0 torr at the triple point of water. (a) Sketch a graph of pressure s. absolute temperature for this thermometer. (b) What will be the pressure when the thermometer measures a temperature of 00 K? (c) What ideal-gas temperature corresponds to a pressure of 678 torr?

7 Chapter 7 icture the roblem We can use the equation of the graph plotted in (a) to (b) find the pressure when the temperature is 00 K and (c) the ideal-gas temperature when the pressure is 678 torr. (a) A graph of pressure (in torr) ersus temperature (in kelins) for this thermometer is shown to the right. he equation of this graph is:, torr 50.0 50.0 torr 7 K () 0 0 7, K (b) Ealuate when 00 K: 50.0 torr 7 K ( 00 K) ( 00 K) 54.9 torr (c) Sole equation () for to obtain: Ealuate ( 678 torr) : 7 K 50.0 torr 7 K 50.0 torr ( 678 torr) ( 678 torr).70 0 K 40 A constant-olume gas thermometer has a pressure of 0.0 torr when it reads a temperature of 7 K. (a) Sketch a graph of pressure s. absolute temperature for this thermometer. (b) What is its triple-point pressure? (c) What temperature corresponds to a pressure of 0.75 torr? icture the roblem We can use the equation of the graph plotted in (a) to (b) find the triple-point pressure and (c) the temperature when the pressure is 0.75 torr.

emperature and the Kinetic heory of Gases 7 (a) A graph of pressure ersus absolute temperature for this thermometer is shown to the right. he equation of this graph is: 0.0 torr 7 K (b) Sole for and ealuate the thermometer s triple-point (7.6 K) pressure: (), torr 0.0 0 0 0.0 torr 7 K.0 torr 7, K ( 7.6 K) ( 7.6 K) (c) Sole equation () for to obtain: Ealuate ( 0.75 torr) : 7 K 0.0 torr 7 K 0.0 torr ( 0.75 torr) ( 0.75 torr).8 K 4 At what temperature do the ahrenheit and Celsius temperature scales gie the same reading? icture the roblem We can find the temperature at which the ahrenheit and Celsius scales gie the same reading by setting t t C in the temperatureconersion equation. 5 5 Set t t C in t ( t ): t ( t ) C 9 9 Sole for and ealuate t : t C t 40.0 C 40.0 Remarks: If you e not already thought of doing so, you might use your graphing calculator to plot t C ersus t and t t C (a straight line at 45 ) on the same graph. heir intersection is at ( 40, 40). 4 Sodium melts at 7 K. What is the melting point of sodium on the Celsius and ahrenheit temperature scales?

74 Chapter 7 icture the roblem We can use the Celsius-to-absolute conersion equation to find 7 K on the Celsius scale and the Celsius-to-ahrenheit conersion equation to find the ahrenheit temperature corresponding to 7 K. Express the absolute temperature as a function of the Celsius temperature: Sole for and ealuate t C : t C t C + 7.5K 7.5 K 7K 7.5 K 98 C Use the Celsius-to-ahrenheit conersion equation to find t : t t 9 5 C 08 9 + 5 ( 97.9 ) + 4 he boiling point of oxygen at.00 atm is 90. K. What is the boiling point of oxygen at.00 atm on the Celsius and ahrenheit scales? icture the roblem We can use the Celsius-to-absolute conersion equation to find 90. K on the Celsius scale and the Celsius-to-ahrenheit conersion equation to find the ahrenheit temperature corresponding to 90. K. Express the absolute temperature as a function of the Celsius temperature: Sole for and ealuate t C : t C t C + 7.5K 7.5K 90.K 7.5K 8 C Use the Celsius-to-ahrenheit conersion equation to find t : t t 9 5 C 9 + 5 97 ( 8 ) + 44 On the Réaumur temperature scale, the melting point of ice is 0ºR and the boiling point of water is 80ºR. Derie expressions for conerting temperatures on the Réaumur scale to the Celsius and ahrenheit scales. icture the roblem We can use the following diagram to set up proportions that will allow us to conert temperatures on the Réaumur scale to Celsius and ahrenheit temperatures.

emperature and the Kinetic heory of Gases 75 00 C R 80 t C t t R 0 0 Referring to the diagram, set up a proportion to conert temperatures on the Réaumur scale to Celsius temperatures: Simplify to obtain: Referring to the diagram, set up a proportion to conert temperatures on the Réaumur scale to ahrenheit temperatures: tc 0 C tr 0 R 00 C 0 C 80 R 0 R t C t R 5 t C 4 tr 00 80 t tr 0 R 80 R 0 R Simplify to obtain: t t R 80 80 9 t t 4 R + 45 [SS] A thermistor is a solid-state deice widely used in a ariety of engineering applications. Its primary characteristic is that its electrical resistance aries greatly with temperature. Its temperature dependence is gien approximately by R R 0 e B/, where R is in ohms (Ω), is in kelins, and R 0 and B are constants that can be determined by measuring R at calibration points such as the ice point and the steam point. (a) If R 760 Ω at the ice point and 5 Ω at the steam point, find R 0 and B. (b) What is the resistance of the thermistor at t 98.6 º? (c) What is the rate of change of the resistance with temperature (dr/d) at the ice point and the steam point? (d) At which temperature is the thermistor most sensitie? icture the roblem We can use the temperature dependence of the resistance of the thermistor and the gien data to determine R 0 and B. Once we know these quantities, we can use the temperature-dependence equation to find the resistance at any temperature in the calibration range. Differentiation of R with respect to will allow us to express the rate of change of resistance with temperature at both the ice point and the steam point temperatures.

76 Chapter 7 (a) Express the resistance at the ice point as a function of temperature of the ice point: Express the resistance at the steam point as a function of temperature of the steam point: Diide equation () by equation () to obtain: Sole for B by taking the logarithm of both sides of the equation: 760 5 B 7 K Ω R0e () B 7K Ω R0e () Ω 48.0 e 5Ω 760 B 7 K B 7 K ln 48. B 7 and ln 48. B K 7 7.94 0 K 7 K.944 0 K Sole equation () for R 0 and substitute for B: R 0 760Ω B 7K e ( 760Ω).9 0 e.9 0 ( 760Ω).944 0 K 7K Ω Ω e B 7K (b) rom (a) we hae:.944 0 K ( ) R.9 0 Ω e Conert 98.6 to kelins to obtain: 0K Substitute for to obtain: R( 0 K) (.9 0 Ω).kΩ e.944 0 K 0 K (c) Differentiate R with respect to to obtain: dr d d d B ( R e ) 0 B R 0e B R e 0 B RB d d B

emperature and the Kinetic heory of Gases 77 Ealuate dr/d at the ice point: dr ( 760Ω)(.94 0 K) d ice point 89Ω / K ( 7.6 K) Ealuate dr/d at the steam point: dr ( 5Ω)(.94 0 K) d steam point ( 7.6 K) 4.Ω / K (d) he thermistor is more sensitie (has greater sensitiity) at lower temperatures. he Ideal-Gas Law 46 An ideal gas in a cylinder fitted with a piston (igure 7-0) is held at fixed pressure. If the temperature of the gas increases from 50 to 00 C, by what factor does the olume change? icture the roblem Let the subscript 50 refer to the gas at 50 C and the subscript 00 to the gas at 00 C. We can apply the ideal-gas law for a fixed amount of gas to find the ratio of the final and initial olumes. Apply the ideal-gas law for a fixed 00V00 50V50 amount of gas: 00 50 or, because 00 50, V 00 00 V 50 50 Substitute numerical alues and ealuate V 00 /V 50 : V V 50 ( 7.5+ 00) ( 7.5 + 50) K K 00 or a 5% increase in olume..5 47 [SS] A 0.0-L essel contains gas at a temperature of 0.00ºC and a pressure of 4.00 atm. How many moles of gas are in the essel? How many molecules? icture the roblem We can use the ideal-gas law to find the number of moles of gas in the essel and the definition of Aogadro s number to find the number of molecules.

78 Chapter 7 Apply the ideal-gas law to the gas: Substitute numerical alues and ealuate n: V V nr n R ( 4.00 atm)( 0.0L) ( 8.06 0 L atm/mol K)( 7K) n.786 mol.79 mol Relate the number of molecules N in the gas in terms of the number of moles n: N nn A Substitute numerical alues and ealuate N: N 4 (.786 mol)( 6.0 0 molecules/mol).08 0 molecules 48 A pressure as low as.00 0 8 torr can be achieed using an oil diffusion pump. How many molecules are there in.00 cm of a gas at this pressure if its temperature is 00 K? icture the roblem We can use the ideal-gas law to relate the number of molecules in the gas to its pressure, olume, and temperature. Sole the ideal-gas law for the number of molecules in a gas as a function of its pressure, olume, and temperature: V N k Substitute numerical alues and ealuate N: N 8 6 (.00 0 torr)(. a/torr)(.00 0 m ) (.8 0 J/K)( 00K). 0 8 49 You copy the following paragraph from a artian physics textbook: snorf of an ideal gas occupies a olume of.5 zaks. At a temperature of glips, the gas has a pressure of.5 klads. At a temperature of 0 glips, the same gas now has a pressure of 8.7 klads. Determine the temperature of absolute zero in glips. icture the roblem Because the gas is ideal, its pressure is directly proportional to its temperature. Hence, a graph of ersus will be linear and the linear equation relating and can be soled for the temperature corresponding to zero pressure. We ll assume that the data was taken at constant olume.

emperature and the Kinetic heory of Gases 79 A graph of, in klads, as function of, in glips, is shown to the right. he equation of this graph is:, klads.5.8 klads + 9. 9 klads glips 8.7 0 0 0, glips When 0: Sole for 0 to obtain:.8 klads 0 0 9.9 klads glips + 0 8 glips 50 A motorist inflates the tires of her car to a gauge pressure of 80 ka on a day when the temperature is 8.0ºC. When she arries at her destination, the tire pressure has increased to 45 ka. What is the temperature of the tires if we assume that (a) the tires do not expand or (b) that the tires expand so the olume of the enclosed air increases by 7 percent? icture the roblem Let the subscript refer to the tires when their gauge pressure is 80 ka and the subscript to conditions when their gauge pressure is 45 ka. Assume that the air in the tires behaes as an ideal gas. hen, we can apply the ideal-gas law for a fixed amount of gas to relate the temperatures to the pressures and olumes of the tires. (a) Apply the ideal-gas law for a fixed amount of gas to the air in the tires: Sole for : V V () where the temperatures and pressures are absolute. because V V. Substitute numerical alues to obtain: ( 65K) 45ka + 0 ka 80 ka + 0 ka 6.K 5 C

740 Chapter 7 (b) Use equation () with V.07 V. Sole for : V 07 V (. ) Substitute numerical alues and ealuate : (.07)( 6.K) 76 C 49.K 5 A room is 6.0 m by 5.0 m by.0 m. (a) If the air pressure in the room is.0 atm and the temperature is 00 K, find the number of moles of air in the room. (b) If the temperature increases by 5.0 K and the pressure remains constant, how many moles of air leae the room? icture the roblem We can apply the ideal-gas law to find the number of moles of air in the room as a function of the temperature. (a) Use the ideal-gas law to relate the number of moles of air in the room to the pressure, olume, and temperature of the air: Substitute numerical alues and ealuate n: V n () R n ( 0.5ka)( 6.0m)( 5.0m)(.0m) ( 8.4 J/mol K)( 00K).66 0 mol.7 0 mol (b) Letting n represent the number of moles in the room when the temperature rises by 5 K, express the number of moles of air that leae the room: Apply the ideal-gas law to obtain: Diide equation () by equation () to obtain: Substitute for n to obtain: Δ n n n' V n' () R' n' and n' n n ' ' Δ n n n n ' '

Substitute numerical alues and emperature and the Kinetic heory of Gases 74 Δ n (.66 0 mol) ealuate Δn: 60mol 00K 05K 5 Image that 0.0 g of liquid helium, initially at 4.0 K, eaporate into an empty balloon that is kept at.00 atm pressure. What is the olume of the balloon at (a) 5.0 K and (b) 9 K? icture the roblem Let the subscript refer to helium gas at 4. K and the subscript to the gas at 9 K. We can apply the ideal-gas law to find the olume of the gas at 4. K and a fixed amount of gas to find its olume at 9 K. See Appendix C for the molar mass of helium. (a) Apply the ideal-gas law to the helium gas to express its olume: nr m R V mr Substitute numerical alues and ealuate V : ( 0.0 g)( 0.0806 L atm/mol K)( 5.0K) ( 4.00 g/mol)(.00atm) V 5.5 L 5. L (b) Apply the ideal-gas law for a fixed amount of gas and sole for the olume of the helium gas at 9 K: Substitute numerical alues and ealuate V : V V and, because, V V V 9K 5.0K ( 5.5L) 60.L 5 A closed container with a olume of 6.00 L holds 0.0 g of liquid helium at 5.0 K and enough air to fill the rest of its olume at a pressure of.00 atm. he helium then eaporates and the container warms to room temperature (9 K). What is the final pressure inside the container? icture the roblem We can apply the law of partial pressures to find the final pressure inside the container. See Appendix C for the molar mass of helium and of air.

74 Chapter 7 he final pressure inside the container is the sum of the partial pressures of helium gas and air: + () final He gas air he pressure exerted by the air molecules at room temperature is gien by the ideal-gas law: air nairr V ρairr air mairr V air ρairvr V air he pressure exerted by the helium molecules at room temperature is also gien by the ideal-gas law: nher V He gas mher V He Substituting in equation () yields: final mher ρairr + V He m He ρair + V He air air R Substitute numerical alues and ealuate : final 0.0 g g 0 m 4.00 6.00 L mol L.4 0 6 a atm 0.5 ka kg.9 m J + 8.4 g 8.8 mol K mol.atm ( 9 K) 54 An automobile tire is filled to a gauge pressure of 00 ka when its temperature is 0ºC. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.) After the car has been drien at high speeds, the tire temperature increases to 50ºC. (a) Assuming that the olume of the tire does not change and that air behaes as an ideal gas, find the gauge pressure of the air in the tire. (b) Calculate the gauge pressure if the tire expands so the olume of the enclosed air increases by 0 percent. icture the roblem Let the subscript refer to the tire when its temperature is 0 C and the subscript to conditions when its temperature is 50 C. We can apply the ideal-gas law for a fixed amount of gas to relate the temperatures to the pressures of the air in the tire.

emperature and the Kinetic heory of Gases 74 (a) Apply the ideal-gas law for a fixed amount of gas and sole for pressure at the higher temperature: V () and V because V V Substitute numerical alues to obtain: (b) Sole equation () for : Because V.0 V : K 9K ka and ( 00 ka + 0ka) ka 0ka, gauge V V V.0V.0 ka Substitute numerical alues and K.0 9K ealuate and,gauge : ( )( ) ( 00 ka 0ka ) + 0 ka and 0 ka 0ka, gauge 0ka 55 [SS] After nitrogen (N ) and oxygen (O ), the most abundant molecule in Earth's atmosphere is water, H O. Howeer, the fraction of H O molecules in a gien olume of air aries dramatically, from practically zero percent under the driest conditions to as high as 4 percent where it is ery humid. (a) At a gien temperature and pressure, would air be denser when its water apor content is large or small? (b) What is the difference in mass, at room temperature and atmospheric pressure, between a cubic meter of air with no water apor molecules, and a cubic meter of air in which 4 percent of the molecules are water apor molecules? icture the roblem (a) At a gien temperature and pressure, the ideal-gas law tells us that the total number of molecules per unit olume, N/V, is constant. he denser gas will, therefore, be the one in which the aerage mass per molecule is greater. (b) We can apply the ideal-gas law and use the relationship between the masses of the dry and humid air and their molar masses to find the difference in

744 Chapter 7 mass, at room temperature and atmospheric pressure, between a cubic meter of air containing no water apor, and a cubic meter of air containing 4% water apor. See Appendix C for the molar masses of N, O, and H O. (a) he molecular mass of N is 8.04 amu (see Appendix C), that of O is.999 amu, and that of H O is 8.05 amu. Because H O molecules are lighter than the predominant molecules in air, a gien olume of air will be less when its water apor content is lower. (b) Express the difference in mass between a cubic meter of air containing no water apor, and a cubic meter of air containing 4% water apor: he masses of the dry air and humid air are related to their molar masses according to: Δm m dry dry. 04 m m 0 n humid dry and m 0. n humid 04 humid Substitute for m dry and m humid and Δm 0.04n simplify to obtain: 0.04n( ) dry dry 0.04n humid humid rom the ideal-gas law we hae: atmv n R and so Δm can be written as 0.04atmV Δm dry R ( ) humid Substitute numerical alues (8.8 g/mol is an 80% nitrogen and 0% oxygen weighted aerage) and ealuate Δm: Δm ( 0.04)( 0.5 ka)(.00 m ) ( 8.4 J/mol K)( 00 K) ( 8.8g/mol 8.05 g/mol) 8 g 56 A scuba dier is 40 m below the surface of a lake, where the temperature is 5.0ºC. He releases an air bubble that has a olume of 5 cm. he bubble rises to the surface, where the temperature is 5ºC. Assume that the air in the bubble is always in thermal equilibrium with the surrounding water, and assume that there is no exchange of molecules between the bubble and the surrounding water. What is the olume of the bubble right before it breaks the surface? Hint: Remember that the pressure also changes.

emperature and the Kinetic heory of Gases 745 icture the roblem Let the subscript refer to the conditions at the bottom of the lake and the subscript to the surface of the lake and apply the ideal-gas law for a fixed amount of gas. Apply the ideal-gas law for a fixed V V V V amount of gas: he pressure at the bottom of the lake is the sum of the pressure at its surface (atmospheric) and the pressure due to the depth of the lake: atm + ρgh Substituting for yields: V ( + ρgh) Substitute numerical alues and ealuate V : V V atm ( 5 cm )( 98 K) [ 0.5 ka + (.00 0 kg/m )( 9.8 m/s )( 40 m) ] 78 cm ( 78 K)( 0.5 ka) 57 [SS] A hot-air balloon is open at the bottom. he balloon has a olume of 446 m is filled with air with an aerage temperature of 00 C. he air outside the balloon has a temperature of 0.0 C and a pressure of.00 atm. How large a payload (including the enelope of the balloon itself) can the balloon lift? Use 9.0 g/mol for the molar mass of air. (Neglect the olume of both the payload and the enelope of the balloon.) icture the roblem Assume that the olume of the balloon is not changing. hen the air inside and outside the balloon must be at the same pressure of about.00 atm. he contents of the balloon are the air molecules inside it. We can use Archimedes principle to express the buoyant force on the balloon and we can find the weight of the air molecules inside the balloon. You ll need to determine the molar mass of air. See Appendix C for the molar masses of oxygen and nitrogen. Express the net force on the balloon and its contents: Using Archimedes principle, express the buoyant force on the balloon: B () net w air inside the balloon B wdisplaced fluid mdisplaced fluidg or B ρ o Vballoong where ρ o is the density of the air outside the balloon.

746 Chapter 7 Express the weight of the air inside the balloon: wair inside the balloon ρivballoong where ρ i is the density of the air inside the balloon. Substitute in equation () for B and ρ V g ρ V w air inside the balloon to obtain: ( ) V g net o ρ o balloon ρ i i balloon balloon g () Express the densities of the air molecules in terms of their number densities, molecular mass, and Aogadro s number: Using the ideal-gas law, relate the number density of air N/V to its temperature and pressure: Substitute for V N to obtain: ρ NA N V V Nk and ρ NA k N V k Substitute in equation () and simplify to obtain: net N Ak o N Ak o V i balloon g ( π d )g 6 i Assuming that the aerage molar mass of air is 8.8 g/mol, substitute numerical alues and ealuate net : net [ ]( 9.8m/s ) 97 K 48K 6 ( 6.0 0 particles/mol)(.8 0 J/K) ( 8.8g/mol)( 0.5 ka) π ( 5.0m).0 kn 58 A helium balloon is used to lift a load of 0 N. he weight of the enelope of the balloon is 50.0 N and the olume of the helium when the balloon is fully inflated is.0 m. he temperature of the air is 0ºC and the atmospheric pressure is.00 atm. he balloon is inflated with a sufficient amount of helium gas that the net upward force on the balloon and its load is 0.0 N. Neglect any effects due to the changes of temperature as the altitude changes. (a) How many moles of helium gas are contained in the balloon? (b) At what altitude will the balloon be fully inflated? (c) Does the balloon eer reach the altitude at which it is fully inflated? (d) If the answer to (c) is Yes, what is the maximum altitude attained by the balloon?

emperature and the Kinetic heory of Gases 747 icture the roblem (a) We can find the number of moles of helium gas in the balloon by applying the ideal-gas law to relate n to the pressure, olume, and temperature of the helium and Archimedes principle to find the olume of the helium. In art (b), we can apply the result of roblem -9 to relate atmospheric pressure to altitude and use the ideal-gas law to determine the pressure of the gas when the balloon is fully inflated. In art (c), we ll find the net force acting on the balloon at the altitude at which it is fully inflated in order to decide whether it can rise to that altitude. (a) Apply the ideal-gas law to the helium in the balloon and sole for n: Relate the net force on the balloon to its weight: Use Archimedes principle to express the buoyant force on the balloon in terms of the olume of the balloon: Substitute to obtain: V n () R ρ B wskin wload whe B w ρ Vg displacedair air ρ 0 N air Vg wskin wload HeVg 0.0 N Sole for the olume of the helium: V 0.0 N + wskin + w ( ρ ρ )g air He load Substituting for V in equation () yields: n ( 0.0 N + w + w ) R ( ρ ρ )g air skin He load Substitute numerical alues and ealuate n: n (.00atm)( 0.0 N + 50.0 N + 0 N) ( 8.06 0 L atm/mol K)( 7K)(.9kg/m 0.79 kg/m )( 9.8m/s ) 776 mol L 0 m (b) Using the result of roblem -9, express the ariation in atmospheric pressure with altitude: Ch ( h) e 0 where C 0. km.

748 Chapter 7 Soling for h yields: Using the ideal-gas law, express : Substitute for in equation () and simplify to obtain: 0 h ln C nr V ( h) 0 0 V h ln ln C nr C nr V () Substitute numerical alues and ealuate h: h 0. km ln 4.84 km 4.8 km (.00 atm).0m L 0 m ( 776 mol)( 8.06 0 L atm/mol K)( 7K) (c) Express the condition that must be satisfied if the balloon is to reach its fully inflated altitude: he total weight is the sum of the weights of the load, balloon skin, and helium: Express the buoyant force on the balloon at h 4.84 km: Express the dependence of the density of the air on atmospheric pressure: Substitute for ρ air,h in equation (4) to obtain: w 0 () net B tot w tot wload + wskin + whe or, because whe ρhevg, w w + w + ρ Vg tot load skin He B ρ air, hvg (4) ρ ρ air, h ρ (5) air air, h 0 ρ air B ρairvg 0 0

emperature and the Kinetic heory of Gases 749 Substitute for w tot and B in equation () and simplify to obtain: net Vg ρ 0 ( ρ air He ) wload wskin Ch or, because ( ) net h 0 e, ( ρ air He ) wload wskin Ch e Vg ρ Substitute numerical alues and ealuate net : net e ( 0. km )( 4.84 km ) (.0 m )( 9.8 m/s )(.9 kg/m 0.79 kg/m ) 0 N 50 N 0 N Because net 0 N > 0, the balloon will rise higher than the altitude at which it is fully inflated. (d) he balloon will rise until the net force acting on it is zero. Because the buoyant force depends on the density of the air, the balloon will rise until the density of the air has decreased sufficiently for the buoyant force to just equal the total weight of the balloon. Substitute equation (5) in equation ρair h ln () to obtain: C ρair, h Using equation (4), express the density of the air in terms of B : Substituting for ρair, h and simplifying yields: ρ air, h B Vg ρ air Vgρ h ln ln C B C B Vg air Substitute numerical alues and ealuate h: h (.0 m )( 9.8 m/s )(.9 kg/m ) ln 0. km 90.5 N 6.0km Kinetic heory of Gases 59 [SS] (a) One mole of argon gas is confined to a.0-liter container at a pressure of 0 atm. What is the rms speed of the argon atoms? (b) Compare your answer to the rms speed for helium atoms under the same conditions.

750 Chapter 7 icture the roblem We can express the rms speeds of argon and helium atoms by combining V nr and R to obtain an expression for rms in rms terms of, V, and. See Appendix C for the molar masses of argon and helium. Express the rms speed of an atom as a function of the temperature and its molar mass: rms R rom the ideal-gas law we hae: V R n Substitute for R to obtain: rms V n (a) Substitute numerical alues and ealuate rms for argon atoms: ( )( 0.5ka/atm)(.0 0 m ) ( mol)( 9.948 0 kg/mol) 0atm rms, Ar 0.8km/s (b) Substitute numerical alues and ealuate rms for helium atoms: ( )( 0.5ka/atm)(.0 0 m ) ( mol)( 4.00 0 kg/mol) 0atm rms, He 0.87 km/s he rms speed of argon atoms is slightly less than one third the rms speed of helium atoms. 60 ind the total translational kinetic energy of the molecules of.0 L of oxygen gas at a temperature of 0.0ºC and a pressure of.0 atm. icture the roblem We can express the total translational kinetic energy of the oxygen gas by combining K nr and the ideal-gas law to obtain an expression for K in terms of the pressure and olume of the gas. Relate the total translational kinetic energy of translation to the temperature of the gas: Using the ideal-gas law, substitute for nr to obtain: K K nr V

emperature and the Kinetic heory of Gases 75 Substitute numerical alues and ealuate K: K ( 0.5 ka) 0 m.0 L L 0.5 kj 6 Estimate the rms speed and the aerage kinetic energy of a hydrogen atom in a gas at a temperature of.0 0 7 K. (At this temperature, which is approximately the temperature in the interior of a star, hydrogen atoms are ionized and become protons.) icture the roblem Because we re gien the temperature of the hydrogen atom and know its molar mass, we can find its rms speed using rms R and its aerage kinetic energy from K a k. See Appendix C for the molar mass of hydrogen. Relate the rms speed of a hydrogen atom to its temperature and molar mass: rms R H Substitute numerical alues and ealuate rms : rms 7 ( K)(.0 0 K) 8.4 J/mol.0079 0 5.0 0 5 m/s kg/mol Express the aerage kinetic energy of the hydrogen atom as a function of its temperature: Substitute numerical alues and ealuate K a : Ka K a k 7 (.8 0 J/K)(.0 0 K). 0 6 J 6 Liquid helium has a temperature of only 4.0 K and is in equilibrium with its apor at atmospheric pressure. Calculate the rms speed of a helium atom in the apor at this temperature, and comment on the result. icture the roblem he rms speed of helium atoms is gien by See Appendix C for the molar mass of helium. R rms. He

75 Chapter 7 he rms speed of helium atoms is gien by: Substitute numerical alues and ealuate rms : rms rms R He ( K)( 4.0 K) 8.4 J/mol 4.00 g/mol 6 m/s hermal speeds, een at temperatures as low as 4.0 K, are ery large compared to most of the speeds we experience directly. 6 Show that the mean free path for a molecule in an ideal gas at temperature and pressure is gien by λ k π d. icture the roblem We can combine λ and V nr to express n π d the mean free path for a molecule in an ideal gas in terms of the pressure and temperature. Express the mean free path of a λ molecule in an ideal gas: n π d () where N nn A n V V Sole the ideal-gas law for the olume of the gas: Substitute for V in the expression for n to obtain: nr V nna n nr k Substitute for n in equation () to k λ obtain: π d 64 State-of-the-art acuum equipment can attain pressures as low as 7.0 0 a. Suppose that a chamber contains helium at this pressure and at room temperature (00 K). Estimate the mean free path and the collision time for helium in the chamber. Assume the diameter of a helium atom is.0 0 0 m. icture the roblem We can find the collision time from the mean free path and the aerage (rms) speed of the helium molecules. We can use the result of roblem

emperature and the Kinetic heory of Gases 75 6 to find the mean free path of the molecules and rms R to find the aerage speed of the molecules. See Appendix C for the molar mass of helium. Express the collision time in terms of the mean free path for and the aerage speed of a helium molecule: λ λ τ () a rms rom roblem 6, the mean free k λ path of the gas is gien by: π d Substitute numerical alues and ealuate the mean free path λ: λ (.8 0 J/K)( 00 K) 0 ( 7.0 0 a) π (.0 0 m). 0 9 m. 0 9 m Express the rms speed of the helium molecules: rms R He Substitute for rms in equation () and simplify to obtain: λ He τ λ R R He Substitute numerical alues and ealuate τ : τ 9 (. 0 m) 4.007 g/mol 8.4 J/mol ( K)( 00 K) 9.7 0 5 s 65 [SS] Oxygen (O ) is confined to a cube-shaped container 5 cm on an edge at a temperature of 00 K. Compare the aerage kinetic energy of a molecule of the gas to the change in its graitational potential energy if it falls 5 cm (the height of the container). icture the roblem We can use K k and ΔU mgh gh NA to express the ratio of the aerage kinetic energy of a molecule of the gas to the change in its graitational potential energy if it falls from the top of the container to the bottom. See Appendix C for the molar mass of oxygen. Express the aerage kinetic energy of a molecule of the gas as a function of its temperature: K a k

754 Chapter 7 Letting h represent the height of the container, express the change in the potential energy of a molecule as it falls from the top of the container to the bottom: ΔU mgh O N A gh Express the ratio of K a to ΔU and simplify to obtain: K a ΔU k O gh N N A O A k gh Substitute numerical alues and ealuate K a /ΔU: K a ΔU ( 0 particles/mol)(.8 0 J/K)( 00K) (.0 0 kg/mol)( 9.8m/s )( 0.5m) 6.0 7.9 0 4 *he Distribution of olecular Speeds 66 Use calculus to show that f(), gien by Equation 7-6, has its maximum alue at a speed k / m. icture the roblem Equation 7-6 gies the axwell-boltzmann speed distribution. Setting its deriatie with respect to equal to zero will tell us where the function s extreme alues lie. Differentiate Equation 7-6with respect to : df d d d 4 π 4 π m k m k e m k m e k m k Set df/d 0 for extrema and sole m for : 0 k k m Examination of the graph of f() makes it clear that this extreme alue is, in fact, a maximum. See igure 7-7 and note that it is concae downward at k m. Remarks: An alternatie to the examination of f() in order to conclude that k m maximizes the axwell-boltzmann speed distribution function is to show that d f/d < 0 at k m.

emperature and the Kinetic heory of Gases 755 67 [SS] he fractional distribution function f() is defined in Equation 7-6. Because f() d gies the fraction of molecules that hae speeds in the range between and + d, the integral of f() d oer all the possible ranges of speeds must equal. Gien that the integral e a d π 4 a /, show that () f d, where f() is gien by Equation 7-6. 0 icture the roblem We can show that f() is normalized by using the gien integral to integrate it oer all possible speeds. Express the integral of Equation 7-6: f () d 0 0 4 m π k 0 m k e d Let a m k to obtain: 0 f 4 () d a π 0 a e d Use the gien integral to obtain: π f 4 0 π hat is, f() is normalized. 4 ( ) d a a 68 Gien that the integral e a d 0, calculate the aerage a speed a of molecules in a gas using the axwell Boltzmann distribution function. icture the roblem In roblem 67 we showed that f() is normalized. Hence we can ealuate a using f () d. 0 he aerage speed of the molecules in the gas is gien by: a 0 f () 4 π d m k 0 m k e d Substitute a m k : a 4 a π 0 a e d

756 Chapter 7 Use the gien integral to obtain: a 4 a π a π a π k m 69 he translational kinetic energies of the molecules of a gas are distributed according to the axwell-boltzmann energy distribution, Equation 7-8. (a) Determine the most probable alue of the translational kinetic energy (in terms of the temperature ) and compare this alue to the aerage alue. (b) Sketch a graph of the translational kinetic energy distribution [f(e) ersus E] and label the most probable energy and the aerage energy. (Do not worry about calibrating the ertical scale of the graph.) (c) Your teacher says, Just looking at the graph f(e) ersus E of allows you to see that the aerage translational kinetic energy is considerably greater than the most probable translational kinetic energy. What feature(s) of the graph support her claim? icture the roblem We can set the deriatie of f(e) with respect to E equal to zero in order to determine the most probable alue of the kinetic energy of the gas molecules. (a) he axwell-boltzmann energy π k E k distribution function is: ( ) f E Differentiate this expression with respect to E to obtain: d f de π k k Ee E k E kt ( E) E e + E e Set this deriatie equal to zero for extrema alues and simplify to obtain: Soling for E peak yields: he aerage alue of the energy of the gas molecules is gien by: E peak k Epeak kt E peake + Epeak e E peak Ea k k k 0 Express the ratio of E peak to E a : E E peak a k k E peak E a (b) he following graph of the energy distribution was plotted using a spreadsheet program. Note that k was set equal to and that, as predicted by our result in (a), the peak alue occurs for E 0.5.

emperature and the Kinetic heory of Gases 757 f (E ) 0.50 0.45 0.40 0.5 0.0 0.5 0.0 0.5 0.0 0.05 0.00 0.0 0.5.0.5.0.5.0.5 4.0 4.5 5.0 E (c) he graph rises from zero to the peak much more rapidly than it falls off to the right of the peak. Because the distribution is so strongly skewed to the right of the peak, the outlying molecules with relatiely high energies pull the aerage (k/) far to the right of the most probable alue (k/). General roblems 70 ind the temperature at which the rms speed of a molecule of hydrogen gas equals 4 m/s. icture the roblem We can use rms R to relate the temperature of the H molecules to their rms speed. See Appendix C for the molar mass of hydrogen. Relate the rms speed of the hydrogen molecules to its temperature: R rms H H R rms Substitute numerical alues and ealuate : (.06 0 kg/mol)( 4m/s) ( 8.4 J/mol K) 9.5K 7 (a) If.0 mol of a gas in a cylindrical container occupies a olume of 0 L at a pressure of.0 atm, what is the temperature of the gas in kelins? (b) he cylinder is fitted with a piston so that the olume of the gas (igure 7-0) can ary. When the gas is heated at constant pressure, it expands to a olume of 0 L. What is the temperature of the gas in kelins? (c) Next, the olume is fixed at 0 L, and the gas s temperature is increased to 50 K. What is the pressure of the gas now?

758 Chapter 7 icture the roblem We can use the ideal-gas law to find the initial temperature of the gas and the ideal-gas law for a fixed amount of gas to relate the olumes, pressures, and temperatures resulting from the gien processes. (a) Apply the ideal-gas law to express the temperature of the gas: Substitute numerical alues and ealuate : V nr ( 0.5 ka)( 0 0 m ) (.0 mol)( 8.4 J/mol K).9 K. 0 K (b) Use the ideal-gas law for a fixed amount of gas to relate the temperatures and olumes: V V or, because, V V Sole for and ealuate : (.9K) V V.4 0 K 4.7 K (c) Use the ideal-gas law for a fixed amount of gas to relate the temperatures and pressures: Sole for : Substitute numerical alues and ealuate : V V or, because V V, 50K 4.7 K (.0atm).4atm 7 (a) he olume per molecule of a gas is the reciprocal of the number density (the number of molecules per unit olume). (a) ind the aerage olume per molecule for dry air at room temperature and atmospheric pressure. (b) ake the cube root of your answer to art (a) to obtain a rough estimate of the aerage distance d between air molecules. (c) ind or estimate the aerage diameter D of an air molecule, and compare it to your answer to art (b). (d) Sketch the molecules in a cube-shaped olume of air, with the edge length of the cube equal

emperature and the Kinetic heory of Gases 759 to d. ake your figure to scale and place the molecules in what you think is a typical configuration. (e) Use your picture to explain why the mean free path of an air molecule is much greater than the aerage distance between molecules. icture the roblem (a) We can use the ideal-gas law to find the aerage olume per molecule. (b) If one were to diide a container of air into little cubes, with one cube per molecule, then this distance would be the width of each cube or, equialently, the distance from the center of one cube to the centers of the neighboring cubes. On aerage, we can imagine that the molecules are at the centers of their respectie cubes, so this distance is also the aerage distance between neighboring molecules. (a) he ideal-gas law relates the number of molecules N in a gas to the olume V they occupy: V Nk V N k Substitute numerical alues and ealuate V/N: V N (.8 0 J/K)( 9 K).99 0 0.5 ka 6 m (b) aking the cube root of V/N yields: d.99 0 m 6.4 nm (c) Example 7-0 gies the aerage diameter of an air molecule as: D.75 0 0 m 0.75 nm or about /0 the aerage distance between molecules. (d) A sketch of the molecules in a cube-shaped olume of air, with the edge length of the cube equal to d, follows. he random distribution of the molecules is a typical configuration. d d d

760 Chapter 7 (e) If a particular molecule in the diagram is moing in a random direction, its chance of colliding with a neighbor is ery small because it can miss in either of the two directions perpendicular to its motion. So the mean free path, or aerage distance between collisions, should be many times larger than the distance to the nearest neighbor. 7 [SS] he axwell-boltzmann distribution applies not just to gases, but also to the molecular motions within liquids. he fact that not all molecules hae the same speed helps us understand the process of eaporation. (a) Explain in terms of molecular motion why a drop of water becomes cooler as molecules eaporate from the drop s surface. (Eaporatie cooling is an important mechanism for regulating our body temperatures, and is also used to cool buildings in hot, dry locations.) (b) Use the axwell-boltzmann distribution to explain why een a slight increase in temperature can greatly increase the rate at which a drop of water eaporates. Determine the Concept (a) o escape from the surface of a droplet of water, molecules must hae enough translational kinetic energy to oercome the attractie forces from their neighbors. herefore the molecules that escape will be those that are moing faster, leaing the slower molecules behind. he slower molecules hae less kinetic energy, so the temperature of the droplet, which is proportional to the aerage translational kinetic energy per molecule, decreases. (b) As long as the temperature isn t too high, the molecules that eaporate from a surface will be only those with the most extreme speeds, at the high-energy tail of the axwell-boltzmann distribution. Within this part of the distribution, increasing the temperature only slightly can greatly increase the percentage of molecules with speeds aboe a certain threshold. or example, suppose that we set an initial threshold at E 5k, then imagine increasing the temperature by 0% so.. At the threshold, the ratio of the new energy distribution to the old one is ( ) ( ) an increase of almost 7%. E + E 5 k k e e (.). 5 e e. 66 74 A cubic metal box that has 0-cm long edges contains air at a pressure of.0 atm and a temperature of 00 K. he box is sealed so that the enclosed olume remains constant, and it is heated to a temperature of 400 K. ind the force due to the internal air pressure on each wall of the box.

emperature and the Kinetic heory of Gases 76 icture the roblem We can use the definition of pressure to express the net force on each wall of the box in terms of its area and the pressure differential between the inside and the outside of the box. We can apply the ideal-gas law for a fixed amount of gas to find the pressure inside the box after it has been heated. Using the definition of pressure, express the net force on each wall of the box: AΔ A ( ) inside atm () Use the ideal-gas law for a fixed amount of gas to relate the initial and final pressures of the gas: atmv initial initial insidev final final or, because V initial V final, atm inside initial final Sole for inside to obtain: inside final initial atm Substituting in equation () and simplifying yields: A A final initial final initial atm atm atm Substitute numerical alues and ealuate : ( 0.0 m) ( 0.5 ka).4 kn 400 K 00 K 75 [SS] In attempting to create liquid hydrogen for fuel, one of the proposals is to conert plain old water (H O) into H and O gases by electrolysis. How many moles of each of these gases result from the electrolysis of.0 L of water? icture the roblem We can use the molar mass of water to find the number of moles in.0 L of water. Because there are two hydrogen atoms in each molecule of water, there must be as many hydrogen molecules in the gas formed by electrolysis as there were molecules of water and, because there is one oxygen atom in each molecule of water, there must be half as many oxygen molecules in the gas formed by electrolysis as there were molecules of water. See Appendix C for the molar masses of hydrogen and oxygen.

76 Chapter 7 Express the electrolysis of water into H and O : n n + n H O H O Express the number of moles in g.0 L 000.0 L of water: n L H O 0mol 8.0g/mol Because there are two hydrogen atoms for each water molecule: n H 0 mol Because there is one oxygen atom for each water molecule: n ( 0 mol) 55mol n H O O 76 A hollow 40-cm-long cylinder of negligible mass rests on its side on a horizontal frictionless table. he cylinder is diided into two equal sections by a ertical non-porous membrane. One section contains nitrogen and the other contains oxygen. he pressure of the nitrogen is twice that of the oxygen. How far will the cylinder moe if the membrane breaks? icture the roblem he diagram shows the cylinder before remoal of the membrane. We ll assume that the gases are at the same temperature. he approximate location of the center of mass (C) is indicated. We can find the distance the cylinder moes by finding the location of the C after the membrane is remoed. See Appendix C for the molar masses of oxygen and nitrogen. N O 0 C 40 x, cm Express the distance the cylinder will moe in terms of the moement of the center of mass when the membrane is remoed: Apply the ideal-gas law to both collections of molecules to obtain: Δ x x cm, x after N V n k N N and V n k O O O cm,before

emperature and the Kinetic heory of Gases 76 Diide the first of these equations by the second to obtain: Express the mass of O in terms of its molar mass and the number of moles of oxygen: Express the mass of N in terms of its molar mass and the number of moles of nitrogen: N O n n N O or, because N O, O n N n N n O n O O m n N O O O m n O N Using the definition of center of mass, express the center of mass before the membrane is remoed: x cm,before i i n x m i m i i n ( N ) ( N ) xcm,n + n( O ) ( O ) n( N ) ( N ) + n( O ) ( O ) ( O ) ( N ) xcm,n + n( O ) ( O ) n( O ) ( N ) + n( O ) ( O ) ( N ) xcm,n + ( O ) x cm,o ( N ) + ( O ) x cm,o x cm,o Substitute numerical alues and ealuate x cm,before : x ( )( 8.0g) + ( 0cm)(.00g) ( 8.0g) +.00g 0cm cm, before Locate the center of mass after the membrane is remoed: x ( 0cm)( 8.0g) + ( 0cm)(.00g) ( 8.0g) +.00g cm, after 7.7 cm 0.00cm Substitute to obtain: Δ x 0.00 cm 7.7 cm.7cm

764 Chapter 7 Because momentum must be consered during this process and the center of mass moed to the right, the cylinder moed.7 cm to the left. 77 A cylinder of fixed olume contains a mixture of helium gas (He) and hydrogen gas (H ) at a temperature and pressure. If the temperature is doubled to, the pressure would also double, except for the fact that at temperature the H is essentially 00 percent dissociated into H. In reality, at pressure the temperature is. If the mass of the hydrogen in the cylinder is m, what is the mass of the nitrogen in the cylinder? icture the roblem We can apply the ideal-gas law to the two processes to find the number of moles of hydrogen in terms of the number of moles of nitrogen in the gas. Using the definition of molar mass, we can relate the mass of each gas to the number of moles of each gas and their molar masses. See Appendix C for the molar masses of nitrogen gas and hydrogen gas. Apply the ideal-gas law to the first case: Apply the ideal-gas law to the second case: Diide the second of these equations by the first and simplify to express nh in terms of n N : Relate the m N to n N : Relate the m to n H : [ nn nh ] R V + [ nn nh ] + H V N R n n () m N n n N N N ( 8.0g/mol) and mn n N 8.0g/mol m n n H H H (.06 g/mol) and mh n H.06 g/mol

emperature and the Kinetic heory of Gases 765 Substitute in equation () and sole for m N : m.06g/mol and m N 7m m N 8.0g/mol 78 he mean free path for O molecules at a temperature of 00 K and at.00 atm pressure is 7.0 0 8 m. Use this data to estimate the size of an O molecule. icture the roblem Because the O molecule resembles spheres stuck together, which in cross section look something like two circles, we can estimate the radius of the molecule from the formula for the area of a circle. We can express the area, and hence the radius, of the circle in terms of the mean free path and the number density of the molecules and use the ideal-gas law to express the number density. Express the area of two circles of diameter d that touch each other: πd πd A 4 d A () π Relate the mean free path of the molecules to their number density and cross-sectional area: λ A n A n λ Substitute for A in equation () to obtain: d πn λ Use the ideal-gas law to relate the number density of the O molecules to their temperature and pressure: V Nk or N n V k Substituting for n yields: d k πλ Substitute numerical alues and ealuate d: d ( 0 J/K)( 00K).8 π 8 ( 0.5 ka)( 7.0 0 m) 0.605nm 79 [SS] Current experiments in atomic trapping and cooling can create low-density gases of rubidium and other atoms with temperatures in the nanokelin (0 9 K) range. hese atoms are trapped and cooled using magnetic

766 Chapter 7 fields and lasers in ultrahigh acuum chambers. One method that is used to measure the temperature of a trapped gas is to turn the trap off and measure the time it takes for molecules of the gas to fall a gien distance. Consider a gas of rubidium atoms at a temperature of 0 nk. Calculate how long it would take an atom traeling at the rms speed of the gas to fall a distance of 0.0 cm if (a) it were initially moing directly downward and (b) if it were initially moing directly upward. Assume that the atom doesn t collide with any others along its trajectory. icture the roblem Choose a coordinate system in which downward is the positie direction. We can use a constant-acceleration equation to relate the fall distance to the initial elocity of the molecule, the acceleration due to graity, the fall time, and rms k m to find the initial elocity of the rubidium molecules. (a) Using a constant-acceleration equation, relate the fall distance to the initial elocity of a molecule, the acceleration due to graity, and the fall time: y + 0t gt () Relate the rms speed of rubidium atoms to their temperature and mass: rms k m Rb Substitute numerical alues and ealuate rms : rms ( 0 J/K)( 0nK).8 7 ( 85.47 u)(.6606 0 kg/u) 5.98 0 m/s Letting rms 0, substitute in equation () to obtain: ( 5.98 0 m/s) t + ( 9.8m/s ) 0.00m t Use the quadratic formula or your graphing calculator to sole this equation for its positie root: t 0.48s 4 ms (b) If the atom is initially moing upward: rms 0 5.98 0 m/s

emperature and the Kinetic heory of Gases 767 Substitute in equation () to obtain: ( 5.98 0 m/s) t + ( 9.8m/s ) 0.00m t Use the quadratic formula or your graphing calculator to sole this equation for its positie root: t 0.464 s 46 ms 80 A cylinder is filled with 0.0 mol of an ideal gas at standard temperature and pressure, and a.4-kg piston seals the gas in the cylinder (igure 7-) with a frictionless seal. he trapped column of gas is.4-m high. he piston and cylinder are surrounded by air, also at standard temperature and pressure. he piston is released from rest and starts to fall. he motion of the piston ceases after the oscillations stop with the piston and the trapped air in thermal equilibrium with the surrounding air. (a) ind the height of the gas column. (b) Suppose that the piston is pushed down below its equilibrium position by a small amount and then released. Assuming that the temperature of the gas remains constant, find the frequency of ibration of the piston. icture the roblem (a) Let A be the cross-sectional area of the cylinder. We can use the ideal-gas law to find the height of the piston under equilibrium conditions. In (b), we can apply Newton s nd law and the ideal-gas law for a fixed amount of gas to the show that, for small displacements from its equilibrium position, the piston executes simple harmonic motion. (a) Express the pressure inside the cylinder: Apply the ideal-gas law to obtain a second expression for the pressure of the gas in the cylinder: Equating these two expressions yields: + in atm mg A nr nr in () V ha mg atm + A nr ha Sole for h to obtain: nr (.4m) h A atm.4m mg + A + mg atm A atm A + mg atm

768 Chapter 7 At S, 0.0 mol of gas occupies.4 L. herefore: (.4 m) A.4 0 m and A 9. 0 4 m Substitute numerical alues and h +.4m ealuate h: (.4kg)( 9.8m/s ) 4 ( 9. 0 m )( 0.5 ka).096 m.m (b) Relate the frequency of ibration of the piston to its mass and a stiffness constant: Letting y be the displacement from equilibrium, apply y may to the piston in its equilibrium position: or a small displacement y aboe equilibrium: k f () π m where m is the mass of the piston and k is a constant of proportionality. in A mg atma ' in A mg atm A ma y or ' in A in A ma y () 0 Using the ideal-gas law for a fixed amount of gas and constant temperature, relate ' to : in in ' in V' or ' V in ( V Ay) V + ' in in in in V V + Ay Substituting for V and simplifying yields: Substitute in equation () to obtain: Simplify equation (4) to obtain: ' in Ah A in A in A Ah + Ay y + h y in A + in A ma y h or, for y << h, y in A in A ma y (4) h in A y h ma y

emperature and the Kinetic heory of Gases 769 Substitute in equation () to obtain: Soling for a y yields: nr y nr A ma y y ma y Ah h h nr mh a y y or k a y y, the condition for SH m k nr where m mh Substitute in equation () to obtain: f π nr mh Substitute numerical alues and ealuate f: f π ( 0.0 mol)( 8.4 J/mol K)( 00K) (.4kg)(.096 m).0hz 8 During this problem, you will use a spreadsheet to study the distribution of molecular speeds in a gas. igure 7- should help you get started. (a) Enter the alues for constants R,, and as shown. hen in column A, enter alues of speed ranging from 0 to 00 m/s, in increments of m/s. (his spreadsheet will be long.) In cell B7, enter the formula for the axwell- Boltzmann fraction fractional speed distribution. his formula contains parameters, R, and. Substitute A7 for, B$ or R, B$ for and B$ for. hen use the ILL DOWN command to enter the formula in the cells below B7. Create a graph of f() ersus using the data in columns A and B. (b) Explore how the graph changes as you increase and decrease the temperature, and describe the results. (c) Add a third column in which each cell contains the cumulatie sum of all f() alues, multiplied by the interal size d (which equals ), in the rows aboe and including the row in question. What is the physical interpretation of the numbers in this column? (d) or nitrogen gas at 00 K, what percentage of the molecules has speeds less than 00 m/s? (e) or nitrogen gas at 00 K, what percentage of the molecules has speeds greater than 700 m/s? (a) he first few rows of the spreadsheet are shown below. Note that the column for art (c) is included. A B C R 8. J/mol-K 0.08 kg/mol 00 K 4 5 f() sum f()d

770 Chapter 7 6 (m/s) (s/m) (unitless) 7 0 0.00E+00 0.00E+00 8.00E 08.00E 08 9.0E 07.50E 07 0.70E 07 4.0E 07 4 4.80E 07 9.0E 07 5 7.5E 07.65E 06 A graph of f() for nitrogen at 00 K follows: 0.005 0.000 f ( ), s/m 0.005 0.000 0.0005 0.0000 0 00 400 600 800 000 00, m/s (b) As the temperature is increased the horizontal position of the peak moes to the right in proportion to the square root of the temperature, while the height of the peak drops by the same factor, presering the total area under the graph (which must be.0). (c) Each number in column C of the spreadsheet (shown in (a)) is approximately equal to the integral of f() from zero up to the corresponding alue. his integral represents the probability of a molecule haing a speed less than or equal to this alue of.

emperature and the Kinetic heory of Gases 77..0 0.8 f ( ) d 0.6 0.4 0. 0.0 0 00 400 600 800 000 00, m/s (d) Looking in cell C07, we find that the probability (at 00 K) of a nitrogen molecule haing a speed less than 00 m/s is 0. 0706, or about 7%. Note that this alue is consistent with the graph of f ( )d shown immediately aboe. (e) Looking in cell C707, we find that the probability of a nitrogen molecule haing a speed less than 700 m/s is approximately 0.86, so the probability of it haing a speed greater than this would be 0.86 or 0. 8, or a little under 4%.

77 Chapter 7