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Chapter 18 - PHYS 132 WEEK 9 NOVEMBER 15-18 18.16 Solve The...

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PHYS 132, WEEK 9 NOVEMBER 15-18 18.16. Solve: The average translational kinetic energy per molecule is e avg = 1 2 mv rms 2 = 3 2 k B T v rms = 3 k B T m Since we want the v rms for H 2 and N 2 to be equal, 3 k B T H 2 m H 2 = 3 k B T m N 2 T H 2 = m H 2 m N 2 T N 2 = 2 u 28 u 373 K ( ) = 26.6 K 18.20. Solve: (a) The average translational kinetic energy per molecule is e avg = 1 2 mv rms 2 = 3 2 k B T This means e avg doubles if the temperature T doubles. (b) The root-mean-square speed v rms increases by a factor of 2 as the temperature doubles. (c) The mean free path is l = 1 4 2 p N V ( ) r 2 Because N / V and r do not depend on T , doubling temperature has no effect on l . 18.25. Solve: The volume of the air is V = 6.0 m ¥ 8.0 m ¥ 3.0 m = 144.0 m 3 , the pressure p = 1 atm = 1.013 ¥ 10 5 Pa, and the temperature T = 20 ° C = 293 K. The number of moles of the gas is n = pV RT = 5991 mols This means the number of molecules is N = nN A = 5991 mols ( ) 6.022 10 23 mol - 1 ( ) = 3.61 10 27 molecules. Since air is a diatomic gas, the room’s thermal energy is E th = N e avg = N 5 2 k B T ( ) = 3.65 10 7 J Assess: The room’s thermal energy can also be obtained as follows: E th = nC V T = (5991 mols)(20.8 J/mol K)(293 K) = 3.65 ¥ 10 7 J 18.29. Visualize: Refer to Figure 18.13. At low temperatures, C V = 3 2 R = 12.5 J / mol K. At room temperature and modestly hot temperatures, C V = 5 2 R = 20.8 J / mol K. At very hot temperatures, C V = 7 2 R = 29.1 J / mol K. Solve: (a) The number of moles of diatomic hydrogen gas in the rigid container is 0.20 g 2 g / mol = 0.1 mol
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