Chapter 19: Ideal Gases Chapter 19: Ideal Gases In-Class Exercises 19.1. b 19.2. d 19.3. c 19.4. b 19.5. a 19.6. c 19.7. c 19.8. a 19.9. a Multiple Choice 19.1. a 19.2. a 19.3. c 19.4. c 19.5. a 19.6. c 19.7. b 19.8. d 19.9. d 19.10. b Questions 19.11. As the hot air rises, its volume expands due to a decrease in pressure. If we assume that there is no heat exchanged between the hot air and the environment (an adiabatic process) then the temperature of the hot air decreases since the air molecules do work to expand its volume. From the First Law of Thermodynamics, it is known that for an adiabatic process, EWint . Since the work done is positive, the change in the internal energy, and consequently its temperature, is negative. This is known as adiabatic cooling. 19.12. If the gas molecules do not exchange energy with the walls of their container or with each other, then they will never reach equilibrium, unless they are already in equilibrium. In the kinetic theory derivations in the text, it is assumed that the gas is already in equilibrium; that is, the speeds of the gas molecules are already distributed according to the Maxwell speed distribution. Yes, it is true that if all gas molecules had the same speed initially, then due to collisions and interactions between the molecules, the speed would be redistributed according to the Maxwell speed distribution to put the gas in equilibrium. 19.13. The surface of our skin loses heat mostly by evaporative cooling and convection. The rate of heat loss of the surface depends on the speed of the air passing over the surface. This means that as you blow hard on your skin, the evaporation from your skin increases and therefore the rate of heat loss also increases. This is why you feel a cool sensation. When you breathe softly, you feel a warm sensation because the rate of heat loss is reduced and because your breath is also warmer than your skin. 19.14. The velocity of an air molecule has a magnitude and a direction. The auditorium is closed, so there is no net flow of molecules into or out of the room. For a molecule moving with a certain speed in a certain direction in the auditorium, there is another molecule moving with the same speed in the opposite direction. In this way, the average velocity is zero. However, since the root-mean-square speed is a scalar, the direction of the molecules is not considered. Therefore, the average speed is greater than zero since all of the molecules are in motion. 19.15. This is an adiabatic process (0),Q so EWint , where W represents the work done by the system. Since the fuel-air mixture is compressed, there is work done on the system, so the work done by the system is negative: EWint WW . Thus, the internal energy, or temperature, of the mixture increases causing the fuel to ignite. The speed of this compression is irrelevant since no heat flows into or out of the system. 19.16. By the First Law of Thermodynamics, the change in internal energy is EQW, where Q is the heat flow into the gas and W is the work done by the gas. Under condition 1, the piston is blocked to prevent it from moving. Therefore, no work is done on the system and the change in internal energy is due only to the heat added, EQ. As a result the temperature of the gas increases. Under condition 2, some of the heat energy transferred to the gas is used to move the piston (e.g. work is done by the gas on the piston). 759 Bauer/Westfall: University Physics, 1E Therefore, the change in internal energy is EQW. Since the change in temperature is proportional to the change in internal energy, the final temperature of the gas under condition 1 is larger. The only way for the final temperature to be the same is if the space behind the piston head is under vacuum. Then the gas under condition 2 undergoes a free expansion and no work is done by the gas to move the piston. 19.17. The adiabatic bulk modulus is defined as BVdpdV /. For an adiabatic process, pV constant. Taking the full derivative of both sides of this equation gives: dp p V 1 p 1 VdppV dV 0. dVV V Therefore, the adiabatic bulk modulus for an ideal gas is given by: dp p BV V p, dV V as required. 19.18. (a) The monatomic ideal gas undergoes three processes: (1) (,pVT111 , ) to (,pVT221 , ), (2) (,pVT221 , ) to (,pVT122 , ) and (3) (,pVT122 , ) to (,pVT111 , ). The First Law of Thermodynamics states that the change in internal energy is EQW , where Q is the heat flow into the gas and W is the work done by the gas. In process (1), the pressure and volume change but the temperature stays the same. Since this V step is isothermal, E 0, so QW2 pdV. Using the Ideal Gas Law, the integral can be written as 1 11 V1 V2 nRT V WdVnRT12ln . 11 V1 VV1 However, the answer can be simplified further by noting that nRT11122 p V p V . Therefore, the heat flow into the gas during process (1) is V 2 QWpV1111ln . V1 In process (2), the volume is constant so no work is done and the heat flow is 33 QEnRTnRTT . 2222 21 pV pV Since T 22and T 12, the heat flow into the gas during process (2) is 1 nR 2 nR 33pV12 pV 22 QnR2212 Vpp. 22nR nR In process (3) the pressure remains constant so, 33V1 EQW Q EW nRTT pdVnRTT pVV . 333 3 33 12 1 12112 22V2 Substituting nRT111 p V and nRT212 p V gives: 3335 Q pV pV p V V pV pV pV pV pV pV . 32222 11 12 1 1 2 11 12 11 12 11 12 (b) The total heat flow into the gas is VV3355 Q Q Q Q pVln2 pV pV pV pV Q pV1 ln2 pV , total 1 2 3 1 1 1 2 2 2 1 1 1 2 total 1 1 1 2 VV1 2222 1 using pV22 pV 11. 19.19. Consider two atomic gasses reacting to form a diatomic gas that proceeds as AB AB, where A is one of the atomic gases and B is the other. There is 1 mole of each gas and the reaction happens in a thermally 760 Chapter 19: Ideal Gases isolated chamber. Conservation of energy can be applied to solve this problem. Initially the chamber is at temperature Ti , so the sum of the initial energies Ei of the two monatomic gases is 33 EE E RTRTRT 3 (with n 1 for each gas). iAB22 i i i The energy Ef of the diatomic gas at temperature Tf is ERTff (5/2) , using n 1 and CV 5/2 for a diatomic gas. By conservation of energy, 56 EE3. RTRTT T if i25 f f i Therefore, the temperature of the system increases since TTfi . 19.20. The Ideal Gas Law is pV nRT, where p is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature. Rearranging to solve for p gives pnRTV /. The number of moles, n, is equal to the mass of the gas, m, divided by the molar mass, M : nmM /. The mass of the gas can be written as mV , where is the density, so nVM /. Therefore, the equation of state form of the Ideal Gas Law is pRTM /. 19.21. The compression and rarefaction of a sound wave in gas can be treated as an adiabatic process. (a) The speed of sound vS in an ideal gas of molar mass M is vBS /, where B is the bulk modulus and is the density. For an ideal gas, the adiabatic bulk modulus is defined as BVdpdV /. In an adiabatic process, pV constant. Taking the full derivative of this equation on both sides gives: dd dp p pV dp pV dV0 V dp p V1 dV 0 . dp dV dV V Substituting this into the equation for the bulk modulus gives BV pVp/. For an ideal gas with density where the number of moles can be expressed as nVM /, VRT pM pV nRT . MRT Therefore, the speed of sound is pRT RT v . S pM M (b) The speed of sound cannot exceed the speed of light: vcS . Since the speed of sound is directly proportional to the temperature, the maximum temperature occurs at the maximum sound speed, c. Therefore, RT Mc2 vcmax T . S,maxMR max (c) For a monatomic gas, the ratio of the molar specific heats is 5/3. Therefore, the maximum temperature is 2 1.008 1038 kg/mol 2.998 10 m/s T 6.538 1012 K. max 5 8.314 J/ mol K 3 (d) At this maximum temperature, the equations used would not properly describe the situation. As a particle approaches the speed of light, Newtonian mechanics no longer applies and this situation requires quantum mechanics for an accurate description. 19.22. For a monatomic gas, the internal energy is EkTmBm 3/2 , where kB is Boltzmann’s constant. The factor of three corresponds to the translational degrees of freedom the monatomic gas has.