chapter17 Physics

Chapter 17 Temperature and the Kinetic Theory of GasesConceptual Problems 1 (a) (b) (c) • True or false: The zeroth law of thermodynamics states that two objects in thermal equilibrium with each other must be in thermal equilibrium with a third object. The Fahrenheit and Celsius temperature scales differ only in the choice of the ice-point temperature. The Celsius degree and the kelvin are the same size. (a) False. If two objects are in thermal equilibrium with a third, then they are in thermal equilibrium with each other. (b) False. The Fahrenheit and Celsius temperature scales also differ in the number of intervals between the ice-point temperature and the steam-point temperature. (c) True. 2 • How can you determine if two objects are in thermal equilibrium with each other when putting them into physical contact with each other would have undesirable effects? (For example, if you put a piece of sodium in contact with water there would be a violent chemical reaction.) Determine the Concept Put 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. 3 • [SSM] ″Yesterday I woke up and it was 20 ºF in my bedroom,″ said Mert to his old friend Mort. ″That’s nothing,″ replied Mort. ″My room was –5.0 ºC.″ Who had the colder room, Mert or Mort? Picture the Problem We can decide which room was colder by converting 20°F to the equivalent Celsius temperature. Using the Fahrenheit-Celsius conversion, convert 20°F to the equivalent Celsius temperature: tC = 5 9 5 (t F − 32°) = 9 (20° − 32°) = −6.7°C so Mert' s room was colder. 1713 1714 Chapter 17 4 • Two identical vessels contain different ideal gases at the same pressure and temperature. It follows that (a) the number of gas molecules is the same in both vessels, (b) the total mass of gas is the same in both vessels, (c) the average speed of the gas molecules is the same in both vessels, (d) None of the above. Determine the Concept Because the vessels are identical (have the same volume) and the two ideal gases are at the same pressure and temperature, the ideal-gas law ( PV = NkT ) tells us that the number of gas molecules must be the same in both vessels. (a ) is correct. 5 • [SSM] Figure 17-18 shows a plot of volume versus 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 From the ideal-gas law, we have P = nR T V . In the process depicted, both the temperature and the volume increase, but the temperature increases faster than does the volume. Hence, the pressure increases. 6 • Figure 17-19 shows a plot of pressure versus absolute temperature for a process that takes a sample of an ideal gas from point A to point B. What happens to the volume of the gas during this process? Determine the Concept From the ideal-gas law, we have V = nR T P. In the process depicted, both the temperature and the pressure increase, but the pressure increases faster than does the temperature. Hence, the volume decreases. 7 • If a vessel contains equal amounts, by mass, of helium and argon, which of the following are true? (a) (b) (c) (d) The partial pressure exerted by each of the two gases on the walls of the container is the same. The average speed of a helium atom is the same as that of an argon atom. The number of helium atoms and argon atoms in the vessel are equal. None of the above. Determine the Concept (a) False. The 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 volumes and, hence, their partial pressures are not the same. (b) False. Assuming the gasses have been in the vessel for some time, their average kinetic energies would be the same. Because the densities of helium and Temperature and the Kinetic Theory of Gases 1715 argon are not the same, their average speeds must be different. (c) False. We know that the volumes of the two gasses are equal (they both occupy the full volume 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 above 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 vrms = 3RT M to relate the temperature of a gas to the rms speed of its molecules. Express the dependence of the rms speed of the molecules of a gas on their absolute temperature: 3RT M where R is the gas constant, M is the molar mass, and T is the absolute temperature. vrms = Because v rms ∝ T , the temperature must be quadrupled in order to double the rms speed of the molecules. 9 • Two different gases are at the same temperature. What can be said about the average translational kinetic energies of the molecules? What can be said about the rms speeds of the gas molecules? Determine the Concept The average kinetic energies are equal. The ratio of their rms speeds is equal to the square root of the reciprocal of the ratio of their molecular masses. 10 • A vessel holds a mixture of helium (He) and methane (CH4). The ratio of the rms speed of the He atoms to that of the CH4 molecules is (a) 1, (b) 2, (c) 4, (d) 16 Picture the Problem We can express the rms speeds of the helium atoms and the methane molecules using vrms = 3RT M . Express the rms speed of the helium atoms: vrms, He = 3RT M He 15 K and 107 >> 273. the temperature of warm-blooded animals is roughly 100°F. the pressure can increase while the volume decreases and the temperature is constant. 14 • Imagine that you have a fixed amount of ideal gas in a container that expands to maintain constant pressure. Throughout most of the world. CH 4 = 16 g/mol =2 4 g/mol and (b) is correct. Whether the pressure changes also depends on whether and how the volume changes. or doesn’t it matter? Determine the Concept Because T = tC + 273. degrees Celsius. Do you think that this temperature is in kelvins. it doesn’t matter. the ice point (0°C) and the boiling point of water at 1 atm (100°C) are more convenient than 273 K and 373 K. this may be a more convenient reference than approximately 300 K. temperatures in roughly this range are normally encountered. CH 4 = vrms. 11 • True or false: If the pressure of a fixed amount of gas increases. Determine the Concept The average speed of the molecules in an ideal gas depends on the square root of the kelvin temperature. (d) increases by a factor of 2 . 13 • An astronomer claims that the temperature at the center of the Sun is 7 about 10 degrees. (b) doubles. nonscientific purposes? Determine the Concept For the Celsius scale. the average speed of the molecules (a) remains constant. doubling . (c) quadruples. On the Fahrenheit scale. He vrms. If you double the absolute temperature of the gas. False. the temperature of the gas must increase. Because vav ∝ T . the Celsius scale is the standard for nonscientific purposes. 12 • Why might the Celsius and Fahrenheit scales be more convenient than the absolute scale for ordinary. He vrms. CH 4 = 3RT M CH 4 M CH 4 M He Divide the first of these equations by the second to obtain: Use Appendix C to find the molar masses of helium and methane: vrms. In an isothermal process.1716 Chapter 17 Express the rms speed of the methane molecules: vrms. (d) quadruples. (b) remains constant. Determine the Concept The average translational kinetic energy of the molecules of an ideal gas Kav depends on its temperature T according to K av = 3 kT . the pressure of the gas (a) halves.Temperature and the Kinetic Theory of Gases 1717 the temperature while maintaining constant pressure increases the average speed by a factor of 2 . while also halving its absolute temperature. 15 • [SSM] Suppose that you compress an ideal gas to half its original volume.67 and: vrms. (d) the temperature only. 16 • The average translational kinetic energy of the molecules of a gas depends on (a) the number of moles and the temperature. (c) the pressure only. (b) the pressure and the temperature.34 1. and explain why your answer is intuitively plausible. monatomic vsound = 3 = 1. Determine the Concept From the ideal-gas law. PV = nRT . the speed of sound in a gas or the rms speed of the molecules of the gas? Justify your answer. (d ) is correct. (b) is correct. (c) doubles. 17 •• [SSM] Which speed is greater. Determine the Concept The rms speed of molecules of an ideal gas is given by 3RT γRT and the speed of sound in a gas is given by vsound = . using the appropriate formulas. vrms = M M The rms speed of the molecules of an ideal-gas is given by: The speed of sound in a gas is given by: Divide the first of these equations by the second and simplify to obtain: vrms = 3RT M vsound = γRT M vrms = vsound 3RT M = 3 γ γRT M For a monatomic gas. γ = 1.67 . halving both the temperature and volume of the gas leaves the pressure unchanged. (d ) is 2 correct. During this process. When the gas is heated. the pressure exerted by the molecules increases. 19 •• Imagine that you compress a gas while holding it at fixed temperature (perhaps by immersing the container in cool water). 20 •• Oxygen has a molar mass of 32 g/mol and nitrogen has a molar mass of 28 g/mol. equal average speeds. the average velocity and the average momentum of the molecules increase and. but the oxygen molecules have a smaller average translational kinetic energy than the nitrogen molecules have. Determine the Concept The pressure is a measure of the change in momentum per second of a gas molecule on collision with the wall of the container. equal average translational kinetic energies and equal average speeds. None of the above. Explain in terms of molecular motion why the pressure of the gas on the walls of its container increases. γ = 1.40 For a diatomic gas. In addition. The fact that the temperature does not change tells us the molecular speed does not change with volume. Picture the Problem The average kinetic energies of the molecules are given by K av = (1 mv 2 )av = 3 kT . but the oxygen molecules have a larger average translational kinetic energy than the nitrogen molecules have. the rms speed is always somewhat greater than the speed of sound. Determine the Concept If the volume decreases the pressure increases because more molecules hit a unit of area of the walls in a given time. equal average translational kinetic energies. as a consequence. it is only the component of the molecular velocities in the direction of propagation that is relevant to this issue. Assuming that the room’s temperature distribution is 2 2 . equal average speeds.46 1. but the oxygen molecules have a smaller average speed than the nitrogen molecules have. However. 18 •• Imagine that you increase the temperature of a gas while holding its volume fixed. Explain in terms of molecular motion why the pressure of the gas on the walls of its container increases. diatomic vsound = 3 = 1. In a gas the mean free path is greater than the average intermolecular distance. The oxygen and nitrogen molecules in a room have (a) (b) (c) (d) (e) (f) equal average translational kinetic energies.1718 Chapter 17 vrms.40 and: In general. but the oxygen molecules have a larger average speed than the nitrogen molecules have. Thus. N= V = wh N= P wh kT Assume atmospheric pressure. One reason for the difference in price is that while nitrogen is the most common constituent of the atmosphere. room temperature and that the room is 15 m square and 5 m high to obtain: 101. only small traces of helium can be found in the atmosphere. From the ideal-gas law. Picture the Problem The number of air molecules in your classroom is given by the ideal-gas law. the number of molecules N in a given volume V at pressure P and temperature T is given by: Assuming that a typical classroom is a rectangular parallelepiped. Use ideas from this chapter to explain why it is that only small traces of helium can be found in the atmosphere. (b) is correct. Estimation and Approximation 22 • Estimate the total number of air molecules in your classroom.Temperature and the Kinetic Theory of Gases 1719 uniform. we can conclude that the oxygen and nitrogen molecules have equal average kinetic energies.4 km/s.2 km/s). Determine the Concept The average molecular speed of He gas at 300 K is about 1. so a significant fraction of He molecules have speeds in excess of earth’s escape velocity (11. Over time. while liquid helium is relatively expensive.381× 10 −23 J/K (293 K ) ( ) . its volume is given by: Substituting for V yields: PV kT where k is Boltzmann’s constant. they must be moving slower than the nitrogen molecules. the He content of the atmosphere decreases to almost nothing.325 kPa ⎞ ⎛ ⎜1 atm × ⎟ (15 m )(15 m )(5 m ) atm ⎝ ⎠ N= ≈ 3 × 10 28 1. Because the oxygen molecules are more massive. 21 •• [SSM] Liquid nitrogen is relatively cheap. they "leak" away into space. 00 mL of water at its bottom. The density of air at sea level is given by: The mass of the air molecules occupying a given volume is the product of the number of moles n and the molar mass M of the molecules. Picture the Problem Assuming the steam to be an ideal gas at a temperature of 373 K. The test tube is held over a flame until the water has completely boiled away. we can use the ideal-gas law to estimate the pressure inside the test tube when the water is completely boiled away. Picture the Problem We can use the definition of mass density. relate the pressure inside the test tube to its volume and the temperature: P= NkT V .0 mL has 1. and the ideal-gas law to estimate the density of air at sea level on a warm day.2 kg/m 3 24 •• A stoppered test tube that has a volume of 10. The water has a temperature of 100ºC and is initially at a pressure of 1. Estimate the final pressure inside the test tube. Using the ideal-gas law. Substitute for m to obtain: From the ideal-gas law.325 kPa ⎞ ⎛ ⎜1 atm × ⎟(29 g/mol) atm ⎝ ⎠ ρ= (8. the definition of the molar mass of a gas. the number of moles in a given volume depends on the pressure and temperature according to: Substitute in the expression for ρ and simplify to obtain: Assuming atmospheric pressure. a temperature of 27°C (93°F).314 J/mol ⋅ K )(300 K ) = 1.1720 23 •• Chapter 17 Estimate the density of dry air at sea level on a warm summer day. substitute numerical values and evaluate ρ: ρ= m V ρ= nM V n= PV RT ρ= M ⎛ PV ⎞ PM ⎜ ⎟= V ⎝ RT ⎠ RT 101.00 atm. and 29 g/mol for the molar mass of air. 346 × 1022 particles Substitute numerical values and evaluate P: P= particles 1. How does this account for the absence of an atmosphere on the moon? Picture the Problem We can find the escape temperatures for the earth and the moon by equating. and M (18 g/mol) and evaluate N: N = (1. virtually all of the molecules of that gas will escape the atmosphere of the planet.00 mL of water to its density: N m M = ⇒N = m A N NA M m = ρV = 1. and Avogadro’s number NA: Relate the mass of 1. We can compare these .15ve and vrms of O2 and H2.00 g ) 6. If the rms speed of a gas is greater than about 15 to 20 percent of the escape speed of a planet. in turn.381× 10 −23 J/K (373 K ) 10 − 6 m 3 10.00 × 103 kg/m3 ( × 1.346 ×10 )( ) = 170 atm 25 •• [SSM] In Chapter 11. 0.0 mL × mL 1atm = 1. (a) (b) (c) (d) At what temperature is vrms for O2 equal to 15 percent of the escape speed for Earth? At what temperature is vrms for H2 equal to 15 percent of the escape speed for Earth? Temperatures in the upper atmosphere reach 1000 K.00 g ( ) ) Substitute for m. its molar mass M.00 × 10− 6 m3 = 1.325 ×10 3 N/m 2 22 (3.022 × 1023 particles/mol 18 g/mol = 3. where g is about one-sixth of its value on Earth and R = 1738 km.723 ×10 7 N/m 2 × 101. NA. where g is the acceleration due to gravity at the surface of the planet.Temperature and the Kinetic Theory of Gases 1721 Relate the number of particles N to the mass of water. we found that the escape speed at the surface of a planet of radius R is ve = 2gR . How does this help account for the low abundance of hydrogen in Earth’s atmosphere? Compute the temperatures for which the rms speeds of O2 and H2 are equal to 15 percent of the escape speed at the surface of the moon. 81m/s 2 6. Because the temperature is high there.81 m/s 2 6.045 g moon Rmoon M 3R 1 0. and M O 2 is the molar mass of oxygen.045( 6 g earth )Rmoon M = 3R 0.0025 g earth Rmoon M = R Substitute numerical values and evaluate T for O2: 0.0025 9.37 ×106 m 2.1722 Chapter 17 temperatures to explain the absence from Earth’s upper atmosphere and from the surface of the moon.37 × 106 m 32.02 ×10 −3 kg/mol = 230 K 3(8. O 2 : 0. See Appendix C for the molar masses of O2 and H2.045 gRearth M 3R (1) Substitute numerical values and evaluate T for O2: T= 0. T is the absolute temperature. (a) Express vrms for O2: vrms. a greater fraction of the molecules reach escape speed.0 × 10−3 kg/mol = 3.314 J/mol ⋅ K ( )( )( ) .81 m/s 2 1.0 × 10−3 kg/mol T= = 160 K 8.15ve and v rms.314 J/mol ⋅ K ) ( )( )( ) (c) Because hydrogen is lighter than air it rises to the top of the atmosphere. (d) Express equation (1) at the surface of the moon: T= 0.60 × 103 K 3(8.15 2 gRearth = 3RT M Solve for T to obtain: T= 0.045 9.738 × 106 m 32.314 J/mol ⋅ K ) ( )( )( ) (b) Substitute numerical values and evaluate T for H2: T= 0. Equate 0.045 9. O 2 = 3RT M O2 where R is the gas constant. but not H2 should be present. a larger percentage of the molecules are moving at escape speed. (b) O2.Temperature and the Kinetic Theory of Gases 1723 Substitute numerical values and evaluate T for H2: T= 0.314 J/mol ⋅ K )(273 K ) 44.314 J/mol ⋅ K )(273 K ) 2.02 × 10 −3 kg/mol = 1.314 J/mol ⋅ K )(273 K ) 32. (d) Are H2. the escape speed is lower.O2 = 3(8. .0 km/s Because v is greater than v rms CO2 and O2 but less than v rms for H2. See Appendix C for molar masses.CO2 = 3(8. Express the rms speed of an atom as a function of the temperature: (a) Substitute numerical values and evaluate vrms for H2: vrms = 3RT M v rms. and CO2 at 273 K and then compare these speeds to 20% of the escape velocity on Mars to decide the likelihood of finding these gases in the atmosphere of Mars.0025 9.84 km/s (b) Evaluate vrms for O2: v rms. 26 •• The escape speed for gas molecules in the atmosphere of Mars is 5.H2 = 3(8.0 × 10 −3 kg/mol = 393 m/s (d) Calculate 20% of vesc for Mars: v = 1 vesc = 5 1 5 (5. Calculate the rms speeds for (a) H2.0 × 10 −3 kg/mol = 461 m/s (c) Evaluate vrms for CO2: v rms.738 × 106 m 2. O2.0 km/s) = 1. and (c) CO2 at this temperature.0 km/s and the surface temperature of Mars is typically 0ºC.314 J/mol ⋅ K ( )( )( ) Because g is less on the moon. Thus. and CO2 likely to be found in the atmosphere of Mars? Picture the Problem We can use vrms = 3RT M to calculate the rms speeds of H2. O2 and CO2.81 m/s 2 1. O2.02 × 10−3 kg/mol = 10 K 8. all three gasses should be found on Jupiter. and the dimensions of the court.H2 = = 1. and (c) CO2 at this temperature.0 ×10 −3 kg/mol = 264 m/s (d) Calculate 20% of vesc for Jupiter: v = 1 vesc = 5 1 5 (60 km/s) = 12 km/s Because ve is greater than vrms for O2. O2.CO2 = 3(8. Express the rms speed of an atom as a function of the temperature: (a) Substitute numerical values and evaluate vrms for H2: vrms = 3RT M 3(8. O2. The average force. O2. (d) Are H2.1724 Chapter 17 27 ••[SSM] The escape speed for gas molecules in the atmosphere of Jupiter is 60 km/s and the surface temperature of Jupiter is typically –150ºC. that the court measures 5 m by 5 m. Is the average pressure from the ball significant compared to that from the air? Picture the Problem The average pressure exerted by the ball on the wall is the ratio of the average force it exerts on the wall to the area of the wall. CO2 and H2. Use any reasonable numbers for the mass of the ball. is the rate at which the momentum of the ball changes during each collision with the wall. See Appendix C for the molar masses of H2. and that the speed of the racquetball is 10 m/s.314 J/mol ⋅ K )(123 K ) 2.23 km/s (b) Evaluate vrms for O2: v rms.0 × 10 −3 kg/mol = 310 m/s (c) Evaluate vrms for CO2: v rms. its typical speed. (b) O2. and CO2 likely to be found in the atmosphere of Jupiter? Picture the Problem We can use vrms = 3RT M to calculate the rms speeds of H2. and CO2.02 × 10 −3 kg/mol v rms. We’ll . Calculate the rms speeds for (a) H2.314 J/mol ⋅ K )(123 K ) 32. and CO2 at 123 K and then compare these speeds to 20% of the escape velocity on Jupiter to decide the likelihood of finding these gases in the atmosphere of Jupiter.314 J/mol ⋅ K )(123 K ) 44. O2.O2 = 3(8. 28 •• Estimate the average pressure on the front wall of a racquetball court. in turn. due to the collisions of the ball with the wall during a game. Assume that the mass of a racquetball is 100 g. (b) Estimate the rms speeds of the protons and the electrons at the center of the Sun.08 N/m 2 (5 m )(5 m )(1 s ) Substitute numerical values and evaluate Pav: ≈ 0. Because the temperature is so high. Then we can apply the ideal gas law to an arbitrary volume of the material. the pressure at the center of the Sun is given by: P= nRT V . To a first approximation. (a) Estimate the pressure at the center of the Sun. (a) To the extent that the plasma acts like an ideal gas. (The masses of these particles can be found in Appendix B.1 Pa Express the ratio of Pav and Patm to obtain: Pav 0. at least to a first approximation. Pav = Assuming head-on collisions with the wall. the protons and electrons are separate particles (rather than being joined together to form hydrogen atoms). the Sun consists of a gas of equal numbers of 29 •• protons and electrons. say one cubic meter. Fav = Substituting for Fav yields: Pav = Pav = 2mv whΔt 2(. respectively.Temperature and the Kinetic Theory of Gases 1725 also assume that the interval Δt between collisions with the wall is 1 s.1 kg )(10 m/s ) = 0.) The temperature at the center of the Sun is about 1 × 107 K. the change in momentum of the ball during each collision is 2mv and the average force exerted by the ball on the wall is: Δp 2mv = Δt Δt where Δt is the elapsed time between collisions of the ball with the wall. Picture the Problem We’ll assume that we can model the plasma as an ideal gas.1 Pa ≈ 5 = 10 − 6 Patm 10 Pa or Pav ≈ 10 −6 Patm and the average pressure from the ball is not significant compared to atmospheric pressure. The average pressure exerted on the wall by the racquetball is given by: Fav Fav = A wh where w and h are the width and height of the wall. and the density of the Sun is about 1 × 105 kg/m3. counting electrons. The sample of metal is 30 cm from the surface to which the metal atoms will adhere. the number of moles of protons would be: Substitute numerical values and evaluate P: (2 ×10 P= mol (8. a small sample of metal will be vaporized so that its atoms travel in straight lines (the effects of gravity are negligible during the brief time of flight) to a surface where they land to form a very thin film.314 J/mol ⋅ K ) 10 7 K 1 m3 1 atm ≈ 2 ×1016 Pa × 101. electrons : vrms.1726 Chapter 17 105 kg nprotons = = −3 = 108 mol M protons 10 kg or.314 J/mol ⋅ K ) 107 K 0. You are designing a vacuum chamber for fabricating reflective 30 •• coatings. electrons = 3RT = M electrons 1 2000 3RT M protons vrms. n = 2nprotons = 2 × 108 mol mprotons The mass of our 1 m3 volume of matter is 105 kg and this mass consists mostly of protons. protons : vrms.325 kPa 8 ) ( ) ≈ 2 ×1011 atm (b) To one significant figure. so in 105 kg. protons = 3RT M protons 3(8. One mole of protons has a mass of 1 g. protons = ( ) = 5 × 105 m/s The rms speed of an electron at the center of the Sun is given by: Substitute numerical values and evaluate v rms. Inside this chamber. 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? . the average speed of the protons is the same as the rms speed: Substitute numerical values and evaluate v rms.314 J/mol ⋅ K ) 10 7 K 1 2000 (0. The very high speed we calculated for the plasma electrons is still an order-of-magnitude smaller than the speed of light.001 kg ) ( ) = 2 × 10 7 m/s Remarks: The huge pressure we calculated in (a) is required to support the tremendous weight of the rest of the sun.001 kg v rms. electrons = 3(8. Let’s estimate that of the 20% of the air that is breathed in as oxygen. Express the difference in mass between an inhaled breath and an exhaled breath: Δm = mO 2 − mCO 2 = f CO 2 M O 2 − M CO 2 nbreath ( ) where f CO 2 is the fraction of the air breathed in that is exchanged for carbon dioxide. we’ll use 32 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. Picture the Problem One breath (one’s lung capacity) is about half a liter. The number of moles per breath is given by: nbreath = Vbreath 22. Because this is an estimation problem. The only thing that occurs in breathing is that oxygen is exchanged for carbon dioxide.Temperature and the Kinetic Theory of Gases 1727 Picture the Problem We can use the expression for the mean free path of a molecule to eliminate the number of molecules per unit volume nV from the idealgas law. approximately 5 percent of each exhaled breath is carbon dioxide. Given this information and neglecting any difference in water-vapor content.381×10 −23 2 (30 cm )π 4 × 10 −10 m ( J/K (300 K ) ) ) 2 ≈ 20 mPa 31 ••• [SSM] In normal breathing conditions. Apply the ideal gas law to express the pressure in terms of the number of molecules per unit volume and the temperature: The mean free path of an air molecule is given by: Substitute for nV to obtain: P= NkT = nV kT V λ= 1 2nV π d kT 2 ⇒ nV = 1 2λπ d 2 P= 2λπ d 2 Substitute numerical values and evaluate P: P= (1. ¼ is exchanged for carbon dioxide. Assume that the air in the chamber is at room temperature (300 K) and that the average diameter of an air molecule is 4 × 10−10 m. estimate the typical difference in mass between an inhaled breath and an exhaled breath.4 L/mol . Then 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. 4 ºF. [SSM] The melting point of gold is 1945.1728 Chapter 17 ⎛ Vbreath ⎞ Δm = f CO 2 M O 2 − M CO 2 ⎜ ⎟ ⎝ 22.4° − 32°) = 1063°C 34 • A weather report indicates that the temperature is expected to drop by 15.0 ºC. Express this 33 • temperature in degrees Celsius.5 L ) 22.4 L/mol ≈ 1× 10 −4 g Temperature Scales 32 • A certain ski wax is rated for use between –12 and –7.4°F to the equivalent Celsius temperature: tC = 5 9 5 (tF − 32°) = 9 (1945. What is this temperature range on the Fahrenheit scale? Picture the Problem We can use the fact that 5 C° = 9 F° to set up a proportion that allows us to make easy interval conversions from either the Celsius or Fahrenheit scale to the other. Convert 1945.0°C over the next four hours. The proportion relating a temperature range on the Fahrenheit scale of a temperature range on the Celsius scale is: Δt F 9 F° ⎛ 9 F° ⎞ = ⇒ Δt F = ⎜ ⎟Δt C Δt C 5 C° ⎝ 5 C° ⎠ Substitute numerical values and evaluate ΔtF: ⎛ 9 F° ⎞ Δt F = ⎜ ⎟(− 7°C − (− 12°C )) = 9 F° ⎝ 5 C° ⎠ Remarks: An equivalent but slightly longer solution involves converting the two temperatures to their Fahrenheit equivalents and then subtracting these temperatures.05)(44 g/mol − 32 g/mol) (0. Picture the Problem We can use the Fahrenheit-Celsius conversion equation to find this temperature on the Celsius scale. By how many degrees on the Fahrenheit scale will the temperature drop? .4 L/mol ⎠ Substituting for nbreath yields: ( ) Substitute numerical values and evaluate Δm: Δm = (0. Temperature and the Kinetic Theory of Gases 1729 Picture the Problem We can use the fact that 5 C° = 9 F° to set up a proportion that allows us to make easy interval conversions from either the Celsius or Fahrenheit scale to the other. The proportion relating a temperature range on the Fahrenheit scale of a temperature range on the Celsius scale is: Substitute numerical values and evaluate ΔtF: Δt F 9 F° ⎛ 9 F° ⎞ = ⇒ Δt F = ⎜ ⎟Δt C Δt C 5 C° ⎝ 5 C° ⎠ ⎛ 9 F° ⎞ Δt F = ⎜ ⎟(15.0 C°) = 27.0 F° ⎝ 5 C° ⎠ The length of the column of mercury in a thermometer is 4.00 cm 35 • when the thermometer is immersed in ice water at 1 atm of pressure, and 24.0 cm when the thermometer is immersed in boiling water at 1 atm of pressure. Assume that the length of the mercury column varies linearly with temperature. (a) Sketch a graph of the length of the mercury column versus temperature (in degrees Celsius). (b) What is the length of the column at room temperature (22.0ºC)? (c) If the mercury column is 25.4 cm long when the thermometer is immersed in a chemical solution, what is the temperature of the solution? Picture the Problem 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 25.4 cm. (a) A graph of the length of the mercury column versus temperature (in degrees Celsius) is shown to the right. The equation of the line is: cm ⎞ ⎛ L = ⎜ 0.200 ⎟ t C + 4.00 cm (1) °C ⎠ ⎝ L, cm 24.0 4.00 0 100 t C , °C (b) Evaluate L(22.0°C ) cm ⎞ ⎛ L = ⎜ 0.200 ⎟ (22.0°C ) + 4.00 cm °C ⎠ ⎝ = 8.40 cm (c) Solve equation (1) for tC to obtain: tC = L − 4.00 cm cm 0.200 °C 1730 Chapter 17 t C (25.4 cm ) = 25.4 cm − 4.00 cm cm 0.200 °C Substitute numerical values and evaluate t C (25.4 cm ) : = 107°C 36 • The temperature of the interior of the Sun is about 1.0 × 107 K. What is this temperature in (a) Celsius degrees, (b) kelvins, and (c) Fahrenheit degrees? Picture the Problem We can use the temperature conversion equations tF = 9 tC + 32° and tC = T − 273.15 K to convert 107 K to the Fahrenheit and 5 Celsius temperatures. Express the kelvin temperature in terms of the Celsius temperature: (a) Solve for and evaluate tC: T = tC + 273.15 K t C = T − 273.15 K = 1.0 × 10 7 K − 273.15 K ≈ 1.0 × 10 7 °C (b) Use the Celsius to Fahrenheit conversion equation to evaluate tF: t F = 9 1.0 × 107°C + 32° 5 ≈ 1.8 × 107°F ( ) 37 • The boiling point of nitrogen, N2, is 77.35 K. Express this temperature in degrees Fahrenheit. Picture the Problem While we could convert 77.35 K to a Celsius temperature and then convert the Celsius temperature to a Fahrenheit temperature, an alternative solution is to use the diagram to the right to set up a proportion for the direct conversion of the kelvin temperature to its Fahrenheit equivalent. F 212 K 373.15 32 273.15 tF 77.35 Use the diagram to set up the proportion: 32°F − t F 273.15 K − 77.35 K = 212°F − 32°F 373.15 K − 273.15 K or 32°F − t F 195.8 = 180 F° 100 Temperature and the Kinetic Theory of Gases 1731 Solving for tF yields: t F = 32°F − 195.8 × 180 F° = − 320 °F 100 38 • The pressure of a constant-volume gas thermometer is 0.400 atm at the ice point and 0.546 atm at the steam point. (a) Sketch a graph of pressure versus Celsius temperature for this thermometer. (b) When the pressure is 0.100 atm, what is the temperature? (c) What is the pressure at 444.6 ºC (the boiling point of sulfur)? Picture the Problem We can use the equation of the graph plotted in (a) to (b) find the temperature when the pressure is 0.100 atm and (c) the pressure when the temperature is 444.6ºC. (a) A graph of pressure (in atm) versus temperature (in degrees Celsius) for this thermometer is shown to the right. The equation of this graph is: atm ⎞ ⎛ P = ⎜1.46 × 10−3 ⎟t C + 0.400 atm °C ⎠ ⎝ (b) Solving for tC yields: P , atm 0.546 0.400 0 100 t C , °C tC = P − 0.400 atm atm 1.46 × 10 −3 °C 0.100 atm − 0.400 atm atm 1.46 × 10−3 °C Substitute numerical values and evaluate t C (0.100 atm ) : t C (0.100 atm ) = = − 205°C Substitute numerical values and evaluate P (444.6°C ) : atm ⎞ ⎛ P (444.6°C ) = ⎜1.46 × 10−3 ⎟ (444.6°C ) + 0.400 atm = 1.05 atm °C ⎠ ⎝ 39 • [SSM] A constant-volume gas thermometer reads 50.0 torr at the triple point of water. (a) Sketch a graph of pressure vs. absolute temperature for this thermometer. (b) What will be the pressure when the thermometer measures a temperature of 300 K? (c) What ideal-gas temperature corresponds to a pressure of 678 torr? 0 torr ⎠ = 3. .0 torr when it reads a temperature of 373 K. (a) Sketch a graph of pressure vs.175 torr. torr 50. (a) A graph of pressure (in torr) versus temperature (in kelvins) for this thermometer is shown to the right.70 × 103 K Evaluate T (678 torr ) : 40 • A constant-volume gas thermometer has a pressure of 30. The equation of this graph is: ⎛ 50.175 torr? Picture the Problem We can use the equation of the graph plotted in (a) to (b) find the triple-point pressure P3 and (c) the temperature when the pressure is 0.0 torr ⎞ P (300 K ) = ⎜ ⎟ (300 K ) ⎝ 273 K ⎠ = 54.1732 Chapter 17 Picture the Problem We can use the equation of the graph plotted in (a) to (b) find the pressure when the temperature is 300 K and (c) the ideal-gas temperature when the pressure is 678 torr.0 torr ⎞ P =⎜ ⎟T ⎝ 273 K ⎠ (1) P . absolute temperature for this thermometer. (b) What is its triple-point pressure P3? (c) What temperature corresponds to a pressure of 0. K (b) Evaluate P when T = 300 K: ⎛ 50.0 torr ⎠ ⎛ 273 K ⎞ T (678 torr ) = ⎜ ⎟ (678 torr ) ⎝ 50.9 torr (c) Solve equation (1) for T to obtain: ⎛ 273 K ⎞ T =⎜ ⎟P ⎝ 50.0 0 0 273 T. 0 0 0 373 T.16 K ) = ⎜ ⎟ (273.0 torr ⎞ P =⎜ ⎟T ⎝ 373 K ⎠ (1) P .0 torr ⎞ P (273. The equation of this graph is: ⎛ 30.175 torr ) ⎝ 30.175 torr ) = ⎜ ⎟ (0. Set tF = tC in tC = 5 9 (tF − 32°) : tF = 5 9 (tF − 32°) Solve for and evaluate tF: t C = t F = − 40.Temperature and the Kinetic Theory of Gases 1733 (a) A graph of pressure versus absolute temperature for this thermometer is shown to the right. Their intersection is at (−40.175 torr ) : ⎛ 373 K ⎞ T =⎜ ⎟P ⎝ 30.0 torr ⎠ = 2.0°C = − 40.0 torr ⎠ ⎛ 373 K ⎞ T (0. −40). torr 30. 42 • Sodium melts at 371 K.16 K) pressure: ⎛ 30.16 K ) ⎝ 373 K ⎠ = 22.0°F Remarks: If you’ve not already thought of doing so. K (b) Solve for and evaluate the thermometer’s triple-point (273.0 torr (c) Solve equation (1) for T to obtain: Evaluate T (0.18 K 41 • At what temperature do the Fahrenheit and Celsius temperature scales give the same reading? Picture the Problem We can find the temperature at which the Fahrenheit and Celsius scales give the same reading by setting tF = tC in the temperatureconversion equation. What is the melting point of sodium on the Celsius and Fahrenheit temperature scales? . you might use your graphing calculator to plot tC versus tF and tF = tC (a straight line at 45°) on the same graph. 2 K.2 K − 273. Picture the Problem We can use the following diagram to set up proportions that will allow us to convert temperatures on the Réaumur scale to Celsius and Fahrenheit temperatures.15 K = 98°C Use the Celsius-to-Fahrenheit conversion equation to find tF: tF = 9 tC + 32° = 5 9 5 (97.00 atm is 90.15 K = − 183°C Use the Celsius-to-Fahrenheit conversion equation to find tF: tF = 9 tC + 32° = 5 9 5 (− 183°) + 32° = − 297°F 44 •• On the Réaumur temperature scale. What is the boiling point of oxygen at 1.2 K. Express the absolute temperature as a function of the Celsius temperature: Solve for and evaluate tC: T = tC + 273. the melting point of ice is 0ºR and the boiling point of water is 80ºR.2 K on the Celsius scale and the Celsius-to-Fahrenheit conversion equation to find the Fahrenheit temperature corresponding to 90.15 K = 371 K − 273.15 K = 90. . Express the absolute temperature as a function of the Celsius temperature: Solve for and evaluate tC: T = tC + 273.15 K t C = T − 273.9°) + 32° = 208°F 43 • The boiling point of oxygen at 1.15 K t C = T − 273.00 atm on the Celsius and Fahrenheit scales? Picture the Problem We can use the Celsius-to-absolute conversion equation to find 90.1734 Chapter 17 Picture the Problem We can use the Celsius-to-absolute conversion equation to find 371 K on the Celsius scale and the Celsius-to-Fahrenheit conversion equation to find the Fahrenheit temperature corresponding to 371 K. Derive expressions for converting temperatures on the Réaumur scale to the Celsius and Fahrenheit scales. and R0 and B are constants that can be determined by measuring R at calibration points such as the ice point and the steam point. set up a proportion to convert temperatures on the Réaumur scale to Fahrenheit temperatures: Simplify to obtain: t F − 32°F t − 0°R = R 212°F − 32°F 80°R − 0°R t F − 32 t R = ⇒ tF = 180 80 9 4 R t + 32 45 ••• [SSM] A thermistor is a solid-state device widely used in a variety of engineering applications.6 ºF? (c) What is the rate of change of the resistance with temperature (dR/dT) at the ice point and the steam point? (d) At which temperature is the thermistor most sensitive? Picture the Problem We can use the temperature dependence of the resistance of the thermistor and the given data to determine R0 and B. where R is in ohms (Ω). (a) If R = 7360 Ω at the ice point and 153 Ω at the steam point. we can use the temperature-dependence equation to find the resistance at any temperature in the calibration range. Once we know these quantities.Temperature and the Kinetic Theory of Gases 1735 C 100 212 F 80 R tC tF tR 0 32 0 Referring to the diagram. (b) What is the resistance of the thermistor at t = 98. set up a proportion to convert temperatures on the Réaumur scale to Celsius temperatures: Simplify to obtain: tC − 0°C t − 0°R = R 100°C − 0°C 80°R − 0°R tC t = R ⇒ tC = 100 80 5 4 R t Referring to the diagram. find R0 and B. Differentiation of R with respect to T will allow us to express the rate of change of resistance with temperature at both the ice point and the steam point temperatures. Its temperature dependence is given approximately by R = R0eB/T. T is in kelvins. Its primary characteristic is that its electrical resistance varies greatly with temperature. . 1 = 3.1736 Chapter 17 7360 Ω = R0e B 273 K (1) (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: 153 Ω = R0e B 373 K (2) Divide equation (1) by equation (2) to obtain: Solve for B by taking the logarithm of both sides of the equation: 7360 Ω = 48.6 °F to kelvins to obtain: Substitute for T to obtain: R = 3.944×10 = 3.94 × 10 3 K Solve equation (1) for R0 and substitute for B: R0 = = (7360 Ω )e −3.944×10 = 1.944 × 10 3 K B= 1 ⎞ −1 ⎛ 1 − ⎜ ⎟K ⎝ 273 373 ⎠ = 3.91×10 −3 Ω 7360 Ω = (7360 Ω )e − B 273 K B 273 K e 3 K 273 K (b) From (a) we have: Convert 98.913 ×10 −3 Ω = 3.1 = B⎜ ⎟K ⎝ 273 373 ⎠ and ln 48.913 × 10 −3 Ω e 3.913 × 10 −3 Ω e 3.944×10 ( ) 3 K T T = 310 K R(310 K ) = 3.31 kΩ ( ) 3 K 310 K (c) Differentiate R with respect to T to obtain: dR d d B = (R0e B T ) = R0e B T dT ⎛ T ⎞ ⎜ ⎟ dT dT ⎝ ⎠ RB −B = 2 R0e B T = − 2 T T .10 = e B 273 K − B 373 K 153 Ω 1 ⎞ −1 ⎛ 1 − ln 48. Temperature and the Kinetic Theory of Gases 1737 Evaluate dR/dT at the ice point: (7360 Ω ) 3. The Ideal-Gas Law 46 • An ideal gas in a cylinder fitted with a piston (Figure 17-20) is held at fixed pressure.0-L vessel contains gas at a temperature of 0.00ºC and a pressure of 4. Apply the ideal-gas law for a fixed amount of gas: P100V100 P50V50 = T100 T50 or. . If the temperature of the gas increases from 50° to 100°C.00 atm. V100 T100 = V50 T50 Substitute numerical values and evaluate V100/V50: V100 (273.15 (273.16 K )2 ⎝ dT ⎠steam point = − 4. 47 • [SSM] A 10.33 Ω / K ( ) (d) The thermistor is more sensitive (has greater sensitivity) at lower temperatures.15 + 50)K V50 or a 15% increase in volume.943 ×103 K ⎛ dR ⎞ =− ⎜ ⎟ (273. We can apply the ideal-gas law for a fixed amount of gas to find the ratio of the final and initial volumes. How many moles of gas are in the vessel? How many molecules? Picture the Problem We can use the ideal-gas law to find the number of moles of gas in the vessel and the definition of Avogadro’s number to find the number of molecules.943 ×103 K ⎛ dR ⎞ =− ⎜ ⎟ (373. by what factor does the volume change? Picture the Problem Let the subscript 50 refer to the gas at 50°C and the subscript 100 to the gas at 100°C.15 + 100 )K = = 1.16 K )2 ⎝ dT ⎠ice point = − 389 Ω / K ( ) Evaluate dR/dT at the steam point: (153 Ω ) 3. because P100 = P50. the same gas now has a pressure of 8.00 atm )(10. At a temperature of 22 glips. the gas has a pressure of 12. Solve the ideal-gas law for the number of molecules in a gas as a function of its pressure. How many molecules are there in 1.00 ×10 −8 ( torr (133.786 mol = 1.35 zaks. its pressure is directly proportional to its temperature. At a temperature of –10 glips. Picture the Problem Because the gas is ideal.7 klads. and temperature: N= PV kT Substitute numerical values and evaluate N: N= (1.00 × 10–8 torr can be achieved using an oil diffusion pump.00 cm3 of a gas at this pressure if its temperature is 300 K? Picture the Problem We can use the ideal-gas law to relate the number of molecules in the gas to its pressure. Hence.00 × 10 −6 m 3 = 3.206 ×10 (4.32 Pa/torr ) 1.″ Determine the temperature of absolute zero in glips.5 klads.22 × 108 1. volume.1738 Chapter 17 PV = nRT ⇒ n = Apply the ideal-gas law to the gas: PV RT Substitute numerical values and evaluate n: n= (8. a graph of P versus T will be linear and the linear equation relating P and T can be solved for the temperature corresponding to zero pressure.08 × 10 24 molecules ( ) 48 •• A pressure as low as 1. and temperature. We’ll assume that the data was taken at constant volume. volume.381× 10−23 J/K (300 K ) ) ) ( ) You copy the following paragraph from a Martian physics textbook: 49 •• ″1 snorf of an ideal gas occupies a volume of 1.786 mol) 6. .0 L ) −2 L ⋅ atm/mol ⋅ K (273 K ) ) = 1.022 × 10 23 molecules/mol = 1.79 mol Relate the number of molecules N in the gas in terms of the number of moles n: N = nN A Substitute numerical values and evaluate N: N = (1. P 1 245 kPa + 101 kPa 180 kPa + 101 kPa Solve for T2: Substitute numerical values to obtain: T2 = (265 K ) = 326.Temperature and the Kinetic Theory of Gases 1739 A graph of P. we can apply the ideal-gas law for a fixed amount of gas to relate the temperatures to the pressures and volumes of the tires.8 klads ⎞ P =⎜ ⎜ 32 glips ⎟T + 9. Then. the tire pressure has increased to 245 kPa. glips When P = 0: ⎛ 3. T2 = T1 P2 because V1 = V2. (a) Apply the ideal-gas law for a fixed amount of gas to the air in the tires: P2V2 PV1 = 1 T2 T1 (1) where the temperatures and pressures are absolute. in glips.9 klads ⎟ ⎝ ⎠ T0 = − 83 glips Solve for T0 to obtain: 50 •• A motorist inflates the tires of her car to a gauge pressure of 180 kPa on a day when the temperature is –8. 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 volume of the enclosed air increases by 7 percent? Picture the Problem Let the subscript 1 refer to the tires when their gauge pressure is 180 kPa and the subscript 2 to conditions when their gauge pressure is 245 kPa. When she arrives at her destination. as function of T.3 K = 53°C . The equation of this graph is: ⎛ 3.7 0 22 T . in klads. Assume that the air in the tires behaves as an ideal gas.5 8.0ºC. klads 12. is shown to the right.8 klads ⎞ 0=⎜ ⎜ 32 glips ⎟T0 + 9.9 klads ⎟ ⎝ ⎠ T0 −10 P. 07 ) ⎜ 2 T1 ⎟ ⎜P ⎟ PV1 1 ⎝ 1 ⎠ T2 = (1. find the number of moles of air in the room. Solve for T2: Substitute numerical values and evaluate T2: ⎛P ⎞ P2V2 T1 = (1.07 V1.0 m by 3. (b) If the temperature increases by 5. volume.0 m )(3.1740 Chapter 17 T2 = (b) Use equation (1) with V2 = 1.325 kPa )(6. how many moles of air leave the room? Picture the Problem We can apply the ideal-gas law to find the number of moles of air in the room as a function of the temperature.7 ×10 3 mol (b) Letting n′ represent the number of moles in the room when the temperature rises by 5 K.314 J/mol ⋅ K )(300 K ) = 3. (a) Use the ideal-gas law to relate the number of moles of air in the room to the pressure.0 m ) (8.0 K and the pressure remains constant.07 )(326. express the number of moles of air that leave the room: Apply the ideal-gas law to obtain: Δn = n − n' n' = PV RT ' (2) Divide equation (2) by equation (1) to obtain: Substitute for n′ to obtain: n' T T and n' = n = n T' T' Δn = n − n T ⎛ T⎞ = n⎜1 − ⎟ T' ⎝ T' ⎠ .3 K ) = 349. (a) If the air pressure in the room is 1. and temperature of the air: Substitute numerical values and evaluate n: n= PV RT (1) n= (101.1 K = 76°C 51 •• A room is 6.0 m.0 atm and the temperature is 300 K.66 ×10 3 mol = 3.0 m by 5.0 m )(5. 0 K ) = 5.Temperature and the Kinetic Theory of Gases 1741 Substitute numerical values and evaluate Δn: ⎛ 300 K ⎞ Δn = 3.0 K and (b) 293 K? Picture the Problem Let the subscript 1 refer to helium gas at 4.00 L holds 10. because P1 = P2.2 K and the subscript 2 to the gas at 293 K.0 K 53 •• A closed container with a volume of 6. evaporate into an empty balloon that is kept at 1. initially at 4. See Appendix C for the molar mass of helium.1L 25. (a) Apply the ideal-gas law to the helium gas to express its volume: m RT nRT1 M 1 mRT1 V1 = = = P1 P1 MP1 Substitute numerical values and evaluate V1: V1 = (10.00 atm) P2V2 PV1 = 1 T2 T1 and.08206 L ⋅ atm/mol ⋅ K )(25.13 L (b) Apply the ideal-gas law for a fixed amount of gas and solve for the volume of the helium gas at 293 K: Substitute numerical values and evaluate V2: V2 = 293 K (5.125 L ) = 60.125 L = (4.00 atm pressure. We can apply the ideal-gas law to find the volume of the gas at 4. What is the final pressure inside the container? Picture the Problem We can apply the law of partial pressures to find the final pressure inside the container.0 K and enough air to fill the rest of its volume at a pressure of 1.66 × 103 mol ⎜1 − ⎜ 305 K ⎟ ⎟ ⎝ ⎠ ( ) = 60 mol 52 •• Image that 10.0 g of liquid helium at 25. The helium then evaporates and the container warms to room temperature (293 K). What is the volume of the balloon at (a) 25.0 g of liquid helium.003 g/mol)(1. See Appendix C for the molar mass of helium and of air. T V2 = 2 V1 T1 5. .0 g )(0.2 K and a fixed amount of gas to find its volume at 293 K.00 atm.20 K. Picture the Problem Let the subscript 1 refer to the tire when its temperature is 20°C and the subscript 2 to conditions when its temperature is 50°C.293 3 ⎟ 10.81 g ⎟ ⎝ ⎛ ⎜ ⎜ 4. find the gauge pressure of the air in the tire.314 J ⎞ (293 K ) =⎜ + ⎟ ⎜ 3 −3 mol ⋅ K ⎠ g ⎞⎛ 10 m ⎞ 28.1742 Chapter 17 The final pressure inside the container is the sum of the partial pressures of helium gas and air: The pressure exerted by the air molecules at room temperature is given by the ideal-gas law: Pfinal = PHe gas + Pair (1) Pair = = nair RT mair RT ρ airVRT = = V M airV M airV ρ air RT M air nHe RT mHe RT = V M HeV The pressure exerted by the helium molecules at room temperature is also given by the ideal-gas law: Substituting in equation (1) yields: PHe gas = Pfinal = mHe RT ρ air RT + M HeV M air ⎞ ⎟ RT ⎟ ⎠ ⎛ m ρ = ⎜ He + air ⎜M V M air ⎝ He Substitute numerical values and evaluate P2: ⎛ kg ⎞ ⎜ 1.0 g ⎜ m ⎟ ⎛ 8. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.325 kPa Pfinal 54 •• An automobile tire is filled to a gauge pressure of 200 kPa when its temperature is 20ºC.00 L × ⎜⎝ mol ⎟ mol ⎠ ⎜ L ⎟ ⎝ ⎠ ⎝ ⎠ 1 atm = 1. the tire temperature increases to 50ºC.003 ⎟ ⎟ ⎟ ⎜ 6.1 atm 101. 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.) After the car has been driven at high speeds.124 × 106 Pa × = 11. (b) Calculate the gauge pressure if the tire expands so the volume of the enclosed air increases by 10 percent. . (a) Assuming that the volume of the tire does not change and that air behaves as an ideal gas. gauge: P2 = 323 K (200 kPa + 101kPa ) (1.gauge = 332 kPa − 101 kPa = 231 kPa (b) Solve equation (1) for P2: P2 = V1T2 P 1 V2T1 T2 V1T2 P1 = P1 1.10V1T1 1.Temperature and the Kinetic Theory of Gases 1743 (a) Apply the ideal-gas law for a fixed amount of gas and solve for pressure at the higher temperature: P2V2 PV1 = 1 T2 T1 (1) and P2 = T2 P 1 T1 because V1 = V2 Substitute numerical values to obtain: P2 = 323 K (200 kPa + 101 kPa ) 293 K = 332 kPa and P2. the ideal-gas law tells us that the total number of molecules per unit volume. H2O. However.10 V1: P2 = Substitute numerical values and evaluate P2 and P2. The denser gas will. N/V. (a) At a given temperature and pressure. between a cubic meter of air with no water vapor molecules.gauge = 302 kPa − 101 kPa = 201 kPa 55 •• [SSM] After nitrogen (N2) and oxygen (O2).10 T1 Because V2 = 1. and a cubic meter of air in which 4 percent of the molecules are water vapor molecules? Picture the Problem (a) At a given temperature and pressure. is constant. the fraction of H2O molecules in a given volume of air varies dramatically. at room temperature and atmospheric pressure. (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 .10)(293 K ) = 302 kPa and P2. from practically zero percent under the driest conditions to as high as 4 percent where it is very humid. would air be denser when its water vapor content is large or small? (b) What is the difference in mass. the most abundant molecule in Earth's atmosphere is water. therefore. be the one in which the average mass per molecule is greater. . at room temperature and atmospheric pressure.04nM humid = 0.325 kPa )(1. and H2O.314 J/mol ⋅ K )(300 K ) 18 g 56 •• A scuba diver is 40 m below the surface of a lake.0152 amu. Assume that the air in the bubble is always in thermal equilibrium with the surrounding water. See Appendix C for the molar masses of N2. (a) The molecular mass of N2 is 28.1744 Chapter 17 mass. (b) Express the difference in mass between a cubic meter of air containing no water vapor. where the temperature is 5.04nM dry − 0. What is the volume of the bubble right before it breaks the surface? Hint: Remember that the pressure also changes.04nM dry and m humid = 0. that of O2 is 31. and a cubic meter of air containing 4% water vapor. between a cubic meter of air containing no water vapor. and assume that there is no exchange of molecules between the bubble and the surrounding water.0ºC. and that of H2O is 18.04)(101.0152 g/mol) ≈ Δm = (8. The bubble rises to the surface.811 g/mol is an 80% nitrogen and 20% oxygen weighted average) and evaluate Δm: (0.811 g/mol − 18. O2. where the temperature is 25ºC.999 amu. He releases an air bubble that has a volume of 15 cm3.04nM humid Δm = 0. a given volume of air will be less when its water vapor content is lower.014 amu (see Appendix C). and a cubic meter of air containing 4% water vapor: The masses of the dry air and humid air are related to their molar masses according to: Substitute for mdry and mhumid and simplify to obtain: From the ideal-gas law we have: Δm = mdry − mhumid mdry = 0.04n(M dry − M humid ) PatmV RT and so Δm can be written as 0.00 m3 ) (28. Because H2O molecules are lighter than the predominant molecules in air.04 PatmV (M dry − M humid ) Δm = RT n= Substitute numerical values (28. . See Appendix C for the molar masses of oxygen and nitrogen. The balloon has a volume of 446 m3 is filled with air with an average temperature of 100°C. Apply the ideal-gas law for a fixed amount of gas: The 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: Substituting for P1 yields: P2V2 PV1 VT P = 1 ⇒ V2 = 1 2 1 T1 P2 T2 T1 P1 = Patm + ρgh V2 = V1T2 (Patm + ρgh ) T1 P2 Substitute numerical values and evaluate V2: V2 = (15 cm )(298 K )[101.00 atm. (Neglect the volume of both the payload and the envelope of the balloon. 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.0°C and a pressure of 1.0 g/mol for the molar mass of air. The contents of the balloon are the air molecules inside it. You’ll need to determine the molar mass of air. How large a payload (including the envelope of the balloon itself) can the balloon lift? Use 29. express the buoyant force on the balloon: Fnet = B − wair inside the balloon B = wdisplaced fluid = mdisplaced fluid g (1) or B = ρoVballoon g where ρo is the density of the air outside the balloon. The air outside the balloon has a temperature of 20.) Picture the Problem Assume that the volume of the balloon is not changing.325 kPa + (1.Temperature and the Kinetic Theory of Gases 1745 Picture the Problem Let the subscript 1 refer to the conditions at the bottom of the lake and the subscript 2 to the surface of the lake and apply the ideal-gas law for a fixed amount of gas.00 atm. Express the net force on the balloon and its contents: Using Archimedes principle. Then the air inside and outside the balloon must be at the same pressure of about 1.325 kPa ) 3 )( ) ] = 78 cm3 57 •• [SSM] A hot-air balloon is open at the bottom.81 m/s 2 (40 m ) (278 K )(101.00 ×10 3 kg/m 3 9. 381 × 10 −23 J/K ⎛ [ ]( ) )( ) = 3.0 m )3 9.0 N. The balloon is inflated with a sufficient amount of helium gas that the net upward force on the balloon and its load is 30.″ what is the maximum altitude attained by the balloon? .325 kPa )⎜ ⎜ ( 1 1 ⎞ 1 ⎟ 6 π (15.0 kN 58 ••• A helium balloon is used to lift a load of 110 N. Fnet = ρ oVballoon g − ρiVballoon g = (ρ o − ρ i )Vballoon g M ⎛N⎞ ⎜ ⎟ NA ⎝ V ⎠ Express the weight of the air inside the balloon: Substitute in equation (1) for B and wair inside the balloon to obtain: Express the densities of the air molecules in terms of their number densities.81 m/s 2 − ⎟ ⎝ 297 K 348 K ⎠ 6.0 m3.1746 Chapter 17 wair inside the balloon = ρiVballoon g where ρi is the density of the air inside the balloon. molecular mass.0 N and the volume of the helium when the balloon is fully inflated is 32. The temperature of the air is 0ºC and the atmospheric pressure is 1.00 atm. relate the number density of air N/V to its temperature and pressure: Substitute for N to obtain: V (2) ρ= PV = NkT and N P = V kT ρ= M ⎛ P ⎞ ⎜ ⎟ N A ⎝ kT ⎠ Substitute in equation (2) and simplify to obtain: Fnet = MP ⎛ 1 1 ⎞ ⎜ − ⎟Vballoon g N A k ⎜ To Ti ⎟ ⎝ ⎠ MP ⎛ 1 1 ⎞ 1 ⎜ − ⎟ 6π d3 g = N A k ⎜ To Ti ⎟ ⎝ ⎠ ( ) Assuming that the average molar mass of air is 28.81 g/mol. (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 ever reach the altitude at which it is fully inflated? (d) If the answer to (c) is ″Yes.81g/mol)(101.022 × 10 23 particles/mol 1. substitute numerical values and evaluate Fnet: Fnet = (28. and Avogadro’s number: Using the ideal-gas law. Neglect any effects due to the changes of temperature as the altitude changes. The weight of the envelope of the balloon is 50. 13 km −1 .0 N + 50.179 kg/m 3 9.206 × 10−2 L ⋅ atm/mol ⋅ K (273 K ) 1. .81 m/s 2 ⎜ (1.0 N V = 30. volume. In Part (c).0 N + 110 N )⎛ ⎜ ( ) ( )( ) = 776 mol (b) Using the result of Problem 13-91.00 atm )(30.293 kg/m 3 − 0. In Part (b).0 N + wskin + wload (ρ air − ρ He )g P(30. 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. we can apply the result of Problem 13-91 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. express the variation in atmospheric pressure with altitude: P(h ) = P0 e −Ch where C = 0.Temperature and the Kinetic Theory of Gases 1747 Picture the Problem (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.0 N + wskin + wload ) RT (ρ air − ρ He )g Substituting for V in equation (1) yields: n= Substitute numerical values and evaluate n: 1L ⎞ ⎟ 3 ⎟ −3 ⎝ 10 m ⎠ n= 8. (a) Apply the ideal-gas law to the helium in the balloon and solve 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 volume of the balloon: Substitute to obtain: Solve for the volume of the helium: n= PV RT (1) FB − wskin − wload − wHe = 30 N FB = wdisplaced air = ρ airVg ρ airVg − wskin − wload − ρ HeVg = 30. and temperature of the helium and Archimedes principle to find the volume of the helium. because wHe = ρ HeVg . express P: P= Substitute for P in equation (2) and simplify to obtain: ⎡ 1 ⎢ P h = ln ⎢ 0 C ⎢ nRT ⎣ V Substitute numerical values and evaluate h: ⎡ ⎤ ⎛ ⎞ (1. hVg (4) P P ρ air. h = ρ air (5) P0 ρ air P0 FB = P ρ airVg P0 . balloon skin.13 km −1 ⎠ ⎢ (776 mol) 8.84 km = 4.h in equation (4) to obtain: Fnet = FB − wtot ≥ 0 (3) w tot = wload + wskin + wHe or.00 atm )⎜ 32. w tot = w load + w skin + ρ HeVg FB = ρ air. h = ⇒ ρair.1748 Chapter 17 h= 1 ⎡ P0 ⎤ ln C ⎢ P(h ) ⎥ ⎣ ⎦ nRT V ⎤ ⎥ 1 ⎡ P0V ⎤ ⎥ = ln ⎢ ⎥ ⎥ C ⎣ nRT ⎦ ⎦ Solving for h yields: (2) Using the ideal-gas law.0 m 3 × 13L 3 ⎟ ⎢ ⎥ ⎜ 10 − m ⎟ 1 ⎛ ⎞ ⎢ ⎝ ⎠ ⎥ h=⎜ ⎟ ln 0.8 km (c) Express the condition that must be satisfied if the balloon is to reach its fully inflated altitude: The total weight is the sum of the weights of the load.206 ×10 − 2 L ⋅ atm/mol ⋅ K (273 K ) ⎥ ⎝ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ( ) = 4. 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 ρair. 293 kg/m ⎤ h=⎜ ln ⎢ ⎥ = 6. because P(h ) = P0 e −Ch .13 km )(4.179 kg/m 3 ) − 110 N − 50 N = 30 N Because Fnet = 30 N > 0. h = ⎡ ⎤ ⎢ ρ ⎥ 1 ⎡Vgρ ⎤ 1 air h = ln ⎢ air ⎥ = ln ⎢ FB ⎥ C ⎣ FB ⎥ C ⎢ ⎦ ⎢ Vg ⎥ ⎣ ⎦ Substitute numerical values and evaluate h: 3 2 3 1 ⎛ ⎞ ⎡ 32.84 km ) (32. Substitute equation (5) in equation (2) to obtain: Using equation (4). h FB Vg ρ air. (d) The balloon will rise until the net force acting on it is zero.0 m 9. Because the buoyant force depends on the density of the air.13 km ⎠ ⎣ ⎦ ( )( )( ) Kinetic Theory of Gases 59 • [SSM] (a) One mole of argon gas is confined to a 1.0-liter container at a pressure of 10 atm.Temperature and the Kinetic Theory of Gases 1749 Substitute for wtot and FB in equation (3) and simplify to obtain: P Vg (ρ air − ρ He ) − wload − wskin P0 Fnet = or.81 m/s 2 )(1. Fnet = e −ChVg (ρ air − ρ He ) − wload − wskin Substitute numerical values and evaluate Fnet: −1 Fnet = e − (0. . What is the rms speed of the argon atoms? (b) Compare your answer to the rms speed for helium atoms under the same conditions.0 m 3 )(9.81 m/s 1. express the density of the air in terms of FB: Substituting for ρ air.293 kg/m 3 − 0.5 N ⎝ 0.0 km −1 ⎟ 190. the balloon will rise higher than the altitude at which it is fully inflated. h and simplifying yields: h= ρ 1 ln air C ρ 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. V.325 kPa/atm ) 1. Picture the Problem We can express the total translational kinetic energy of the oxygen gas by combining K = 3 nRT and the ideal-gas law to obtain an expression 2 for K in terms of the pressure and volume of the gas. Relate the total translational kinetic energy of translation to the temperature of the gas: Using the ideal-gas law. Express the rms speed of an atom as a function of the temperature and its molar mass: From the ideal-gas law we have: vrms = 3RT M RT = PV n 3PV nM Substitute for RT to obtain: vrms = (a) Substitute numerical values and evaluate vrms for argon atoms: v rms.1750 Chapter 17 Picture the Problem We can express the rms speeds of argon and helium atoms by combining PV = nRT and vrms = 3RT M to obtain an expression for vrms in terms of P.87 km/s (1mol) 4.28 km/s (1mol) 39.0ºC and a pressure of 1.0 L of oxygen gas at a temperature of 0.003 ×10−3 kg/mol ( ( ) ) The rms speed of argon atoms is slightly less than one third the rms speed of helium atoms. and M. Ar = 3(10 atm )(101.0 atm.948 ×10−3 kg/mol ( ( ) ) (b) Substitute numerical values and evaluate vrms for helium atoms: v rms.0 × 10 −3 m 3 = 0. See Appendix C for the molar masses of argon and helium. He = 3(10 atm )(101. substitute for nRT to obtain: K = 3 nRT 2 K = 3 PV 2 .325 kPa/atm ) 1.0 × 10−3 m 3 = 0. 60 • Find the total translational kinetic energy of the molecules of 1. . M He See Appendix C for the molar mass of helium.314 J/mol ⋅ K ) 1.Temperature and the Kinetic Theory of Gases 1751 Substitute numerical values and evaluate K: K= 3 2 (101. (At this temperature.1×10 −16 J 62 • Liquid helium has a temperature of only 4.0 × 10 7 K 2 ( )( ) = 2.381×10 −23 J/K 1. and comment on the result. Picture the Problem The rms speed of helium atoms is given by vrms = 3RT .0079 × 10−3 kg/mol ( ) = 5.0 × 107 K 1. hydrogen atoms are ionized and become protons.) Picture the Problem Because we’re given the temperature of the hydrogen atom and know its molar mass.325 kPa )⎜1.0 × 107 K. we can find its rms speed using vrms = 3RT M and its average kinetic energy from K av = 3 kT .0 L × 10 ⎜ ⎝ ⎛ −3 m3 ⎞ ⎟ L ⎟ ⎠ = 0. See Appendix C for the molar mass of 2 hydrogen. which is approximately the temperature in the interior of a star.20 K and is in equilibrium with its vapor at atmospheric pressure. Relate the rms speed of a hydrogen atom to its temperature and molar mass: Substitute numerical values and evaluate vrms: vrms = 3RT MH v rms = 3(8. Calculate the rms speed of a helium atom in the vapor at this temperature.0 × 105 m/s Express the average kinetic energy of the hydrogen atom as a function of its temperature: Substitute numerical values and evaluate Kav: K av = 3 kT 2 K av = 3 1.15 kJ 61 • Estimate the rms speed and the average kinetic energy of a hydrogen atom in a gas at a temperature of 1. 1752 Chapter 17 v rms = The rms speed of helium atoms is given by: Substitute numerical values and evaluate vrms: 3 RT M He v rms = 3(8. Suppose that a chamber contains helium at this pressure and at room temperature (300 K). Express the mean free path of a molecule in an ideal gas: λ= 1 2nvπ d 2 (1) where N nN A nv = = V V Solve the ideal-gas law for the volume of the gas: Substitute for V in the expression for nv to obtain: Substitute for nv in equation (1) to obtain: V = nRT P nN A P P= nRT kT kT 2 Pπ d 2 nv = λ= 64 •• State-of-the-art vacuum equipment can attain pressures as low as –11 7.20 K. Picture the Problem We can combine λ = 1 and PV = nRT to express 2nvπ d 2 the mean free path for a molecule in an ideal gas in terms of the pressure and temperature.0 × 10–10 m.0 × 10 Pa.314 J/mol ⋅ K )(4.20 K ) 4. Assume the diameter of a helium atom is 1. We can use the result of Problem . Estimate the mean free path and the collision time for helium in the chamber. are very large compared to most of the speeds we experience directly. even at temperatures as low as 4. 63 • Show that the mean free path for a molecule in an ideal gas at temperature T and pressure P is given by λ = kT 2 Pπ d 2 . Picture the Problem We can find the collision time from the mean free path and the average (rms) speed of the helium molecules.003 g/mol = 162 m/s Thermal speeds. the mean free path of the gas is given by: Substitute numerical values and evaluate the mean free path λ: τ= λ vav ≈ λ vrms (1) λ= kT 2 Pπ d 2 λ= 2 7. Express the collision time in terms of the mean free path for and the average speed of a helium molecule: From Problem 63. See Appendix C for the molar mass of helium.381×10 −11 −23 Pa π 1. Express the average kinetic energy of a molecule of the gas as a function of its temperature: K av = 3 kT 2 .332 × 109 m ) 4. Compare the average kinetic energy of a molecule of the gas to the change in its gravitational potential energy if it falls 15 cm (the height of the container). See Appendix C for the molar mass of oxygen.332 ×10 9 m = 1.Temperature and the Kinetic Theory of Gases 1753 63 to find the mean free path of the molecules and vrms = 3RT M to find the average speed of the molecules.7 × 10 5 s 3(8. Picture the Problem We can use K = 3 kT and ΔU = mgh = Mgh N A to express 2 the ratio of the average kinetic energy of a molecule of the gas to the change in its gravitational potential energy if it falls from the top of the container to the bottom.3 ×10 9 m Express the rms speed of the helium molecules: Substitute for vrms in equation (1) and simplify to obtain: vrms = 3RT M He τ= λ 3RT M He =λ M He 3RT Substitute numerical values and evaluate τ : τ = (1.0 ×10 ( (1.0 ×10 −10 m ) ( J/K (300 K ) ) ) 2 = 1.007 g/mol = 9.314 J/mol ⋅ K )(300 K ) 65 •• [SSM] Oxygen (O2) is confined to a cube-shaped container 15 cm on an edge at a temperature of 300 K. in fact.15 m ) ( ( )( )( ) ) *The Distribution of Molecular Speeds 66 •• Use calculus to show that f(v). a maximum.022 ×10 23 particles/mol 1. See Figure 17-17 and note that it is concave downward at v = 2kT m . Picture the Problem Equation 17-36 gives the Maxwell-Boltzmann speed distribution.0 × 10 kg/mol 9.1754 Chapter 17 ΔU = mgh = M O 2 gh NA Letting h represent the height of the container. Differentiate Equation 17-36with respect to v: 32 df d ⎡ 4 ⎛ m ⎞ 2 − mv 2 = ⎜ ⎟ ve ⎢ dv dv ⎢ π ⎝ 2kT ⎠ ⎣ 2 kT ⎤ ⎥ ⎥ ⎦ 2 kT 4 ⎛ m ⎞ = ⎜ ⎟ π ⎝ 2kT ⎠ Set df/dv = 0 for extrema and solve for v: 32 ⎛ mv3 ⎞ − mv 2 ⎜ 2v − ⎟e ⎜ kT ⎟ ⎝ ⎠ 2kT m 2v − mv3 =0⇒ v= kT Examination of the graph of f(v) makes it clear that this extreme value is. has its maximum value at a speed v = 2 kT / m . Remarks: An alternative to the examination of f(v) in order to conclude that v = 2kT m maximizes the Maxwell-Boltzmann speed distribution function is to show that d2f/dv2 < 0 at v = 2kT m .81 m/s (0.381×10 −23 J/K (300 K ) = = 7. .9 ×10 4 2 −3 ΔU 2 32. given by Equation 17-36. Setting its derivative with respect to v equal to zero will tell us where the function’s extreme values lie. express the change in the potential energy of a molecule as it falls from the top of the container to the bottom: Express the ratio of Kav to ΔU and simplify to obtain: 3 kT K av 3N A kT = 2 = ΔU M O 2 gh 2M O 2 gh NA Substitute numerical values and evaluate Kav/ΔU: K av 3 6. Because f(v) dv gives the fraction of molecules that have speeds in the range between v and v + dv. Express the integral of Equation 1736: Let a = m 2kT to obtain: ∞ ∫ 0 4 ⎛ m ⎞ f (v )dv = ⎜ ⎟ π ⎝ 2kT ⎠ f (v )dv = 4 ∞ 3 2∞ ∫v e 0 2 2 − mv 2 2 kT dv ∞ ∫ 0 π 4 a 3 2 ∫ v 2e − av dv 0 Use the given integral to obtain: ⎛ π −3 2 ⎞ ⎟ a3 2 ⎜ ⎜ 4 a ⎟= 1 π 0 ⎝ ⎠ That is. f(v) is normalized. Given that the integral ∞ 0 ∫ ∞ 0 v 2 e− av dv = 2 π 4 a−3/ 2 . 68 •• Given that the integral ∫0 v ∞ 3 − av 2 e Picture the Problem In Problem 67 we showed that f(v) is normalized. Picture the Problem We can show that f(v) is normalized by using the given integral to integrate it over all possible speeds. ∞ ∫ f (v )dv = dv = 1 2 . 0 ∞ The average speed of the molecules in the gas is given by: vav = ∫ vf (v )dv 0 ∞ = Substitute a = m 2kT : 4 ⎛ m ⎞ ⎜ ⎟ π ⎝ 2kT ⎠ 4 ∞ 32 3 2∞ ∫v e 0 2 3 − mv 2 2 kT dv vav = π a 3 − av ∫ v e dv 0 . where f(v) is given by Equation 17-36. Hence we can evaluate vav using ∫ vf (v )dv . calculate the average 2a speed vav of molecules in a gas using the Maxwell–Boltzmann distribution function. show that ∫ f (v )dv = 1 .Temperature and the Kinetic Theory of Gases 1755 67 •• [SSM] The fractional distribution function f(v) is defined in Equation 17-36. the integral of f(v) dv over all the possible ranges of speeds must equal 1. (a) The Maxwell-Boltzmann energy distribution function is: ⎛ 2 ⎞⎛ 1 ⎞ f (E ) = ⎜ ⎟⎜ ⎟ ⎟ ⎜ ⎝ π ⎠ ⎝ kT ⎠ 32 E e − E kT Differentiate this expression with respect to E to obtain: d ⎛ 2 ⎞⎛ 1 ⎞ f (E ) = ⎜ ⎟⎜ ⎟ ⎟ ⎜ dE ⎝ π ⎠ ⎝ kT ⎠ 32 ⎛ 1 −1 2 − E kT ⎞ ⎛ −1 ⎞ ⎜2E e + E 1 2 ⎜ ⎟e − E kt ⎟ ⎜ ⎟ ⎝ kT ⎠ ⎝ ⎠ 1 2 Set this derivative equal to zero for extrema values and simplify to obtain: Solving for Epeak yields: The average value of the energy of the gas molecules is given by: Express the ratio of Epeak to Eav: −1 2 Epeak e − E peak kT 1 2 ⎛ −1 ⎞ −E + Epeak ⎜ ⎟e peak ⎝ kT ⎠ kt =0 E peak = 1 2 kT Eav = 3 kT 2 E peak Eav = 1 2 3 2 kT 1 = 3 ⇒ E peak = kT 1 3 Eav (b) The following graph of the energy distribution was plotted using a spreadsheet program.1756 Chapter 17 Use the given integral to obtain: vav = = 4 ⎛ a −2 ⎞ 2 1 a3 2 ⎜ ⎟ ⎜ 2 ⎟= π a π ⎠ ⎝ 2 π 2kT m The translational kinetic energies of the molecules of a gas are 69 •• distributed according to the Maxwell-Boltzmann energy distribution.″ What feature(s) of the graph support her claim? Picture the Problem We can set the derivative of f(E) with respect to E equal to zero in order to determine the most probable value of the kinetic energy of the gas molecules. (b) Sketch a graph of the translational kinetic energy distribution [f(E) versus E] and label the most probable energy and the average energy. (a) Determine the most probable value of the translational kinetic energy (in terms of the temperature T) and compare this value to the average value. the peak value occurs for E = 0.5. Note that kT was set equal to 1 and that.) (c) Your teacher says. Equation 17-38. ″Just looking at the graph f(E) versus E of allows you to see that the average translational kinetic energy is considerably greater than the most probable translational kinetic energy. (Do not worry about calibrating the vertical scale of the graph. . as predicted by our result in (a). 5 1.10 0.35 0.50 0.Temperature and the Kinetic Theory of Gases 1757 0.0 2.25 0.15 0. the volume is fixed at 20 L.45 0. Picture the Problem We can use vrms = 3RT M to relate the temperature of the H2 molecules to their rms speed. When the gas is heated at constant pressure.016 ×10 kg/mol (343 m/s ) 3(8.0 (c) The graph rises from zero to the peak much more rapidly than it falls off to the right of the peak.5 4.0 atm.05 0.20 0.0 4.5 5. Because the distribution is so strongly skewed to the right of the peak. the outlying molecules with relatively high energies pull the average (3kT/2) far to the right of the most probable value (kT/2). See Appendix C for the molar mass of hydrogen.00 0. it expands to a volume of 20 L.314 J/mol ⋅ K ) −3 ) 2 = 9. and the gas’s temperature is increased to 350 K.0 0.40 0. What is the temperature of the gas in kelvins? (c) Next.5 2.5 E 3. what is the temperature of the gas in kelvins? (b) The cylinder is fitted with a piston so that the volume of the gas (Figure 1720) can vary.0 1. General Problems 70 • Find the temperature at which the rms speed of a molecule of hydrogen gas equals 343 m/s.51K 71 •• (a) If 1.0 3.30 f (E ) 0. Relate the rms speed of the hydrogen molecules to its temperature: Substitute numerical values and evaluate T: vrms = 2 M H 2 vrms 3RT ⇒T = M H2 3R T= (2. What is the pressure of the gas now? .0 mol of a gas in a cylindrical container occupies a volume of 10 L at a pressure of 1. and compare it to your answer to Part (b). because V1 = V2. (b) Take the cube root of your answer to Part (a) to obtain a rough estimate of the average distance d between air molecules.325 kPa ) (10 ×10 −3 m 3 ) (1. (a) Find the average volume per molecule for dry air at room temperature and atmospheric pressure. because P1 = P2. (a) Apply the ideal-gas law to express the temperature of the gas: Substitute numerical values and evaluate T: T= PV nR T= (101.4 × 10 2 K (c) Use the ideal-gas law for a fixed amount of gas to relate the temperatures and pressures: PV1 P2V2 1 = T1 T2 or.0 mol)(8.9 K = 1. (d) Sketch the molecules in a cube-shaped volume of air.7 K V1 = 2. (c) Find or estimate the average diameter D of an air molecule.7 K Substitute numerical values and evaluate P2: P2 = 72 •• (a) The volume per molecule of a gas is the reciprocal of the number density (the number of molecules per unit volume).9 K ) = 243.4 atm 243. pressures.2 × 10 2 K (b) Use the ideal-gas law for a fixed amount of gas to relate the temperatures and volumes: PV1 P2V2 1 = T1 T2 or.1758 Chapter 17 Picture the Problem 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 volumes. and temperatures resulting from the given processes. P P2 1 = T1 T2 Solve for T2: P2 = T2 P 1 T1 350 K (1.0 atm) = 1. with the edge length of the cube equal .314 J/mol ⋅ K ) = 121. V1 V2 = T1 T2 Solve for and evaluate T2: T2 = V2 T1 = 2(121. equivalently.99 × 10−26 m 3 ≈ 3. so this distance is also the average distance between neighboring molecules.375 nm or about 1/10 the average distance between molecules. (e) Use your picture to explain why the mean free path of an air molecule is much greater than the average distance between molecules. then this distance would be the width of each cube or. 3d 3d 3d . (a) The ideal-gas law relates the number of molecules N in a gas to the volume V they occupy: Substitute numerical values and evaluate V/N: PV = NkT ⇒ V kT = N P 1.325 kPa = 3. the distance from the center of one cube to the centers of the neighboring cubes.Temperature and the Kinetic Theory of Gases 1759 to 3d. we can imagine that the molecules are at the centers of their respective cubes. Make your figure to scale and place the molecules in what you think is a typical configuration.99 × 10 −26 m 3 ( ) (b) Taking the cube root of V/N yields: (c) Example 17-10 gives the average diameter of an air molecule as: d = 3 3. (d) A sketch of the molecules in a cube-shaped volume of air.42 nm D = 3. with one cube per molecule. The random distribution of the molecules is a typical configuration.75 × 10 −10 m = 0. On average. follows. Picture the Problem (a) We can use the ideal-gas law to find the average volume per molecule. with the edge length of the cube equal to 3d. (b) If one were to divide a container of air into little cubes.381× 10 −23 J/K (293 K ) V = N 101. 1760 Chapter 17 (e) If a particular molecule in the diagram is moving in a random direction. Determine the Concept (a) To escape from the surface of a droplet of water. and is also used to cool buildings in hot.1T1. its chance of colliding with a neighbor is very small because it can miss in either of the two directions perpendicular to its motion. (b) As long as the temperature isn’t too high. (a) Explain in terms of molecular motion why a drop of water becomes cooler as molecules evaporate from the drop’s surface. at the high-energy ″tail″ of the Maxwell-Boltzmann distribution.366 F (T1 ) ⎜ T2 ⎟ ⎝ ⎠ −E +E an increase of almost 37%. The box is sealed so that the enclosed volume remains constant. leaving the slower molecules behind. but also to the molecular motions within liquids.) (b) Use the Maxwell-Boltzmann distribution to explain why even a slight increase in temperature can greatly increase the rate at which a drop of water evaporates. should be many times larger than the distance to the nearest neighbor. The fact that not all molecules have the same speed helps us understand the process of evaporation. At the threshold. Therefore the molecules that escape will be those that are moving faster. the molecules that evaporate from a surface will be only those with the most extreme speeds. The slower molecules have less kinetic energy. and it is heated to a temperature of 400 K. molecules must have enough translational kinetic energy to overcome the attractive forces from their neighbors. Find the force due to the internal air pressure on each wall of the box. 74 •• A cubic metal box that has 20-cm–long edges contains air at a pressure of 1. suppose that we set an initial threshold at E = 5kT1. . (Evaporative cooling is an important mechanism for regulating our body temperatures.0 atm and a temperature of 300 K.1) 2 e 1. which is proportional to the average translational kinetic energy per molecule. increasing the temperature only slightly can greatly increase the percentage of molecules with speeds above a certain threshold. Within this part of the distribution.1e 5 = 1. So the mean free path. decreases. then imagine increasing the temperature by 10% so T2 = 1. For example. or average distance between collisions. 73 •• [SSM] The Maxwell-Boltzmann distribution applies not just to gases. dry locations. the ratio of the new energy distribution to the old one is −3 − 5 F (T2 ) ⎛ T1 ⎞ kT2 kT1 = ⎜ ⎟e e = (1. so the temperature of the droplet. . Because there are two hydrogen atoms in each molecule of water.0 L of water. there must be half as many oxygen molecules in the gas formed by electrolysis as there were molecules of water. express the net force on each wall of the box: Use the ideal-gas law for a fixed amount of gas to relate the initial and final pressures of the gas: F = AΔP = A(Pinside − Patm ) (1) PatmVinitial PinsideVfinal = Tinitial Tfinal or. See Appendix C for the molar masses of hydrogen and oxygen. Patm P = inside Tinitial Tfinal Pinside = Tfinal Patm Tinitial Solve for Pinside to obtain: Substituting in equation (1) and simplifying yields: ⎛T ⎞ F = A⎜ final Patm − Patm ⎟ ⎜T ⎟ ⎝ initial ⎠ ⎛T ⎞ = A⎜ final − 1⎟ Patm ⎜T ⎟ ⎝ initial ⎠ ⎞ 2 ⎛ 400 K F = (0.0 L of water? Picture the Problem We can use the molar mass of water to find the number of moles in 2. one of the proposals is to convert plain old water (H2O) into H2 and O2 gases by electrolysis.4 kN Substitute numerical values and evaluate F: 75 •• [SSM] In attempting to create liquid hydrogen for fuel. there must be as many hydrogen molecules in the gas formed by electrolysis as there were molecules of water and. because Vinitial = Vfinal. because there is one oxygen atom in each molecule of water.325 kPa ) ⎝ 300 K ⎠ = 1. How many moles of each of these gases result from the electrolysis of 2. 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.20 m ) ⎜ − 1⎟ (101.Temperature and the Kinetic Theory of Gases 1761 Picture the Problem 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’ll assume that the gases are at the same temperature.before Express the distance the cylinder will move in terms of the movement of the center of mass when the membrane is removed: Apply the ideal-gas law to both collections of molecules to obtain: PN 2 VN 2 = nN 2 kT and PO 2 VO 2 = nO 2 kT . N2 O2 x . We can find the distance the cylinder moves by finding the location of the CM after the membrane is removed. One section contains nitrogen and the other contains oxygen.after − xcm.1762 Chapter 17 Express the electrolysis of water into H2 and O2: Express the number of moles in 2.0 L × 1000 nH 2 O Because there are two hydrogen atoms for each water molecule: Because there is one oxygen atom for each water molecule: nH 2 = 110 mol nO 2 = 1 n H 2 O = 2 1 2 (110 mol) = 55 mol 76 •• A hollow 40-cm-long cylinder of negligible mass rests on its side on a horizontal frictionless table. The cylinder is divided into two equal sections by a vertical non-porous membrane. See Appendix C for the molar masses of oxygen and nitrogen.02 g/mol 2. The approximate location of the center of mass (CM) is indicated. cm 0 CM 40 Δx = xcm.0 L of water: n H 2 O → n H 2 + 1 nO 2 2 g L = 110 mol = 18. How far will the cylinder move if the membrane breaks? Picture the Problem The diagram shows the cylinder before removal of the membrane. The pressure of the nitrogen is twice that of the oxygen. O 2 2M (N 2 ) + M (O 2 ) 2n(O 2 )M (N 2 ) + n(O 2 )M (O 2 ) Substitute numerical values and evaluate xcm. N 2 + n(O 2 )M (O 2 )xcm.27 cm 2(28.00 g ) = 17.01g ) + 32.after = Substitute to obtain: 2(20 cm )(28.O 2 2 M (N 2 )xcm.O 2 n(N 2 )M (N 2 ) + n(O 2 )M (O 2 ) = = 2n(O 2 )M (N 2 )xcm.01g ) + (20 cm )(32.00 g Locate the center of mass after the membrane is removed: xcm.before = 2(10 cm )(28. N 2 + n(O 2 )M (O 2 )xcm.before ∑x m = ∑m i i i i i = n(N 2 )M (N 2 )xcm. express the center of mass before the membrane is removed: xcm.27 cm = 2.00 g Δx = 20.7 cm .before: xcm.01g ) + (30 cm )(32.00 cm − 17.Temperature and the Kinetic Theory of Gases 1763 Divide the first of these equations by the second to obtain: PN 2 PO 2 nN 2 nO 2 nN 2 nO 2 = or.00 g ) = 20.00 cm 2(28.01g ) + 32. N 2 + M (O 2 )xcm. because PN 2 = 2 PO 2 . 2 PO 2 PO 2 = ⇒ nN 2 = 2nO 2 Express the mass of O2 in terms of its molar mass and the number of moles of oxygen: Express the mass of N2 in terms of its molar mass and the number of moles of nitrogen: mO 2 = nO 2 M O 2 mN 2 = nO 2 M N 2 Using the definition of center of mass. the cylinder moved 2. 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: Divide the second of these equations by the first and simplify to express nH 2 in terms of nN2 : Relate the mN to nN2 : P1V = 2nN 2 + nH 2 RT1 3P1V = 2nN 2 + 2nH 2 2 RT1 nH 2 = 2nN 2 [ ] [ ] (1) mN = nN 2 M N 2 = n N 2 (28.1764 Chapter 17 Because momentum must be conserved during this process and the center of mass moved to the right. In reality. If the mass of the hydrogen in the cylinder is m.016 g/mol) mH 2 2. we can relate the mass of each gas to the number of moles of each gas and their molar masses. what is the mass of the nitrogen in the cylinder? Picture the Problem 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. at pressure P2 = 2P1 the temperature is T2 = 3T1.01g/mol) and nN 2 = Relate the m to nH2 : mN 2 28. If the temperature is doubled to T2 = 2T1.016 g/mol and nH 2 = . 77 •• A cylinder of fixed volume contains a mixture of helium gas (He) and hydrogen gas (H2) at a temperature T1 and pressure P1.01g/mol m = nH 2 M H 2 = nH 2 (2. the pressure would also double. except for the fact that at temperature the H2 is essentially 100 percent dissociated into H1.7 cm to the left. 10 ×10 −8 m ( ( ) ) = 0.10 × 10–8 m. 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. These atoms are trapped and cooled using magnetic .00 atm pressure is 7.605 nm 79 •• [SSM] Current experiments in atomic trapping and cooling can create low-density gases of rubidium and other atoms with temperatures in the nanokelvin (10–9 K) range.01g/mol and mN 2 ≈ 7m 78 •• The mean free path for O2 molecules at a temperature of 300 K and at 1. and hence the radius. We can express the area. Use this data to estimate the size of an O2 molecule. which in cross section look something like two circles.325 kPa ) 7.381×10 − 23 J/K (300 K ) π (101. we can estimate the radius of the molecule from the formula for the area of a circle. Picture the Problem Because the O2 molecule resembles 2 spheres stuck together.016 g/mol 28. Express the area of two circles of diameter d that touch each other: Relate the mean free path of the molecules to their number density and cross-sectional area: Substitute for A in equation (1) to obtain: Use the ideal-gas law to relate the number density of the O2 molecules to their temperature and pressure: Substituting for nv yields: ⎛ πd 2 ⎞ πd 2 A = 2⎜ ⎜ 4 ⎟ = 2 ⇒d = ⎟ ⎝ ⎠ 2A π (1) λ= 1 1 ⇒A= nv A nv λ 2 πnv λ d= PV = NkT or nv = N P = V kT d= 2kT πPλ Substitute numerical values and evaluate d: d= 2 1.Temperature and the Kinetic Theory of Gases 1765 Substitute in equation (1) and solve for mN: 2m N 2 m = 2. We can use a constant-acceleration equation to relate the fall distance to the initial velocity of the molecule.81m/s 2 t 2 2 Use the quadratic formula or your graphing calculator to solve this equation for its positive root: (b) If the atom is initially moving upward: t = 0. substitute in equation (1) to obtain: 0.100 m = 5.14218 s = 142 ms ( ) ( ) v rms = v0 = −5. and the fall time: Relate the rms speed of rubidium atoms to their temperature and mass: Substitute numerical values and evaluate vrms: y = v0t + 1 gt 2 2 (1) vrms = 3kT mRb v rms = 3 1. the acceleration due to gravity. (a) Using a constant-acceleration equation.918 ×10 −3 m/s .381× 10−23 J/K (120 nK ) (85. Consider a gas of rubidium atoms at a temperature of 120 nK. the acceleration due to gravity. 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 given distance. Assume that the atom doesn’t collide with any others along its trajectory. Calculate how long it would take an atom traveling at the rms speed of the gas to fall a distance of 10.6606 ×10−27 kg/u ( ( ) ) = 5.1766 Chapter 17 fields and lasers in ultrahigh vacuum chambers.0 cm if (a) it were initially moving directly downward and (b) if it were initially moving directly upward. the fall time.918 × 10 −3 m/s t + 1 9.918 × 10−3 m/s Letting vrms = v0. relate the fall distance to the initial velocity of a molecule.47 u ) 1. and vrms = 3kT m to find the initial velocity of the rubidium molecules. Picture the Problem Choose a coordinate system in which downward is the positive direction. 100 m = − 5.81 m/s 2 t 2 2 Use the quadratic formula or your graphing calculator to solve this equation for its positive root: t = 0.4 m )APatm nRT = APatm + mg APatm + mg 2. and a 1. We can use the ideal-gas law to find the height of the piston under equilibrium conditions.4-m high.10 mol of an ideal gas at standard temperature and pressure. (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: Solve for h to obtain: Pin = Patm + mg A (1) Pin = nRT nRT = V hA Patm + mg nRT = A hA h= = (2.4-kg piston seals the gas in the cylinder (Figure 17-21) with a frictionless seal. (a) Find the height of the gas column. Picture the Problem (a) Let A be the cross-sectional area of the cylinder. for small displacements from its equilibrium position.1464 s = 146 ms ( ) ( ) 80 ••• A cylinder is filled with 0. (b) Suppose that the piston is pushed down below its equilibrium position by a small amount and then released. we can apply Newton’s 2nd law and the ideal-gas law for a fixed amount of gas to the show that. the piston executes simple harmonic motion. The piston is released from rest and starts to fall.Temperature and the Kinetic Theory of Gases 1767 Substitute in equation (1) to obtain: 0.918 × 10−3 m/s t + 1 9. The trapped column of gas is 2. find the frequency of vibration of the piston. In (b).4 m mg 1+ APatm . The piston and cylinder are surrounded by air. The motion of the piston ceases after the oscillations stop with the piston and the trapped air in thermal equilibrium with the surrounding air. also at standard temperature and pressure. Assuming that the temperature of the gas remains constant. 4 m (1.24 × 10−3 m3 and A = 9.10 mol of gas occupies 2.4 m )A = 2.096 m = 2. relate Pin' to Pin : V V + Ay Substituting for V and simplifying yields: P' in A = Pin A Ah 1 = Pin A y Ah + Ay 1+ h Substitute in equation (3) to obtain: y⎞ ⎛ Pin A⎜1 + ⎟ − Pin A = ma y ⎝ h⎠ or. f = Letting y be the displacement from equilibrium.24 L. y⎞ ⎛ Pin A⎜1 − ⎟ − Pin A ≈ ma y ⎝ h⎠ − Pin A y ≈ ma y h −1 (4) Simplify equation (4) to obtain: .4 kg ) 9.1768 Chapter 17 At STP.1 m (b) Relate the frequency of vibration of the piston to its mass and a ″stiffness″ constant: 1 k (2) 2π m where m is the mass of the piston and k is a constant of proportionality. for y << h. apply ∑ Fy = ma y to the piston in its equilibrium position: For a small displacement y above equilibrium: Pin A − mg − Patm A = 0 P' in A − mg − Patm A = ma y or P' in A − Pin A = ma y P'inV' = PinV or P' in (V + Ay ) = PinV ⇒ P'in = Pin (3) Using the ideal-gas law for a fixed amount of gas and constant temperature.333 ×10 − 4 m 2 (101.81 m/s 2 1+ 9. 0. Therefore: (2.333 × 10 −4 m 2 h= 2.325 kPa ) Substitute numerical values and evaluate h: ( ( ) ) = 2. you will use a spreadsheet to study the distribution of molecular speeds in a gas. M and T. A B C 1 R= 8. Create a graph of f(v) versus v using the data in columns A and B. This formula contains parameters v. what percentage of the molecules has speeds less than 200 m/s? (e) For nitrogen gas at 300 K.028 kg/mol 3 T= 300 K 4 5 f(v) sum f(v)dv v . and T as shown. enter values of speed ranging from 0 to 1200 m/s. (This spreadsheet will be long. Then use the FILL DOWN command to enter the formula in the cells below B7. in the rows above and including the row in question.31 J/mol-K 2 M= 0. (c) Add a third column in which each cell contains the cumulative sum of all f(v) values. R. (b) Explore how the graph changes as you increase and decrease the temperature. and describe the results.4 kg )(2. M. Note that the column for Part (c) is included. (a) Enter the values for constants R.0 Hz 81 ••• During this problem.314 J/mol ⋅ K )(300 K ) = (1. enter the formula for the MaxwellBoltzmann fraction fractional speed distribution.096 m )2 1. Substitute A7 for v. Figure 17-22 should help you get started. multiplied by the interval size dv (which equals 1). Then in column A. the condition for SHM m k nRT where = m mh 2 ay = − Substitute in equation (2) to obtain: f = 1 2π Solving for ay yields: nRT y mh 2 nRT mh 2 Substitute numerical values and evaluate f: f = 1 2π (0.) In cell B7.10 mol)(8. B$2 for M and B$3 for T. in increments of 1 m/s. B$1 or R. what percentage of the molecules has speeds greater than 700 m/s? (a) The first few rows of the spreadsheet are shown below. What is the physical interpretation of the numbers in this column? (d) For nitrogen gas at 300 K.Temperature and the Kinetic Theory of Gases 1769 Substitute in equation (1) to obtain: ⎛ nRT ⎞ y ⎛ nRT ⎞ −⎜ ⎟ A ≈ ma y ⇒ − ⎜ 2 ⎟ y ≈ ma y ⎝ Ah ⎠ h ⎝ h ⎠ ay = − or k y . m/s 800 1000 1200 (b) As the temperature is increased the horizontal position of the peak moves to the right in proportion to the square root of the temperature.0015 0. s/m . (c) Each number in column C of the spreadsheet (shown in (a)) is approximately equal to the integral of f(v) from zero up to the corresponding v value. while the height of the peak drops by the same factor.0010 0.00E+00 3. This integral represents the probability of a molecule having a speed less than or equal to this value of v.00E−08 1.01E−07 1.20E−07 2.51E−07 (unitless) 0.0020 0.50E−07 4.0).1770 Chapter 17 6 7 8 9 10 11 12 (m/s) 0 1 2 3 4 5 (s/m) 0.0000 0 200 400 600 v .70E−07 4.65E−06 A graph of f(v) for nitrogen at 300 K follows: 0. preserving the total area under the graph (which must be 1.0005 0.00E+00 3.0025 0. f (v ).80E−07 7.00E−08 1.20E−07 9. m/s 800 1000 1200 (d) Looking in cell C207. or a little under 14%.2 1.2 0. we find that the probability (at 300 K) of a nitrogen molecule having a speed less than 200 m/s is 0.862 or 0. or about 7%. we find that the probability of a nitrogen molecule having a speed less than 700 m/s is approximately 0. (e) Looking in cell C707.0 0. so the probability of it having a speed greater than this would be 1 – 0.862. .Temperature and the Kinetic Theory of Gases 1771 1.8 f (v ) dv 0.0706 .0 0 200 400 600 v . Note that this value is consistent with the graph of f (v )dv shown immediately above.138 .4 0.6 0. 1772 Chapter 17 .