# Sol-Ch20 - CHAPTER 20 E xercises 1 v r m s = 3RT/M 1 2...

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Exercises 1. v r m s = (3RT/M) 1/2 , where the unit of M is kg/mol. (a) 1350 m/s; (b) 604 m/s; (c) 181 m/s 2. v r m s = (3kT/m) 1/2 : (a) 12.0 km/s ; (b) 780 m/s 3. (a) K a v = 3kT/2 = 4.14 × 10 –16 J; (b) 7.04 × 10 5 m/s 4. T = Mv 2 /3R : (a) 1.41 × 10 5 K;(b) 1.61 × 10 5 K; (c) 10,060 K 5. (a) v r m s = (3RT/M) 1/2 , where the unit of M is kg/mol. v r m s = 2740 m/s; (b) 967 m/s 6. Let M 2 = 235 + 6 ×   19 = 349 u; and M 1 = 352 u, then v 2 /v 1 = (M 1 /M 2 ) 1/2 = 1.004. 7. v = (3kT/m) 1/2 = 2.73 km/s 8. T α v 2 : (a) 1200 K; (b) 4800 K 9. (a) 3kT/2 = 2.07 × 10 –14 J; (b) 3.52 × 10 6 m/s 10. T = PV/nR = 180.4 K; K av = 1.5 kT = 3.74 × 10 –21 J 11. (3kT/m) 1/2 = 273 m/s 12. (a) K a v = 1.5kT = 6.28 × 10 –21 J; (b) P = nRT/V = 3.78 × 10 4 Pa 13. (a)v r m s =(3RT/M) 1/2 = 517 m/s. (M is in kg/mol.) (b) P = nRT/V = 1.48 MPa 14. V = NkT/P = 1.87 × 10 –16 m 3 15. Volume for 1 mole = RT/P = 2.25 × 10 –2 m 3 Volume for 1 molecule, d 3 = 2.25 × 10 –2 m 3 /N A = 3.73 × 10 –26 m 3 Thus, d = 3.34 × 10 –9 m. CHAPTER 20

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16. K a v = 1.5 kT = 5.655 ×   10 –21 J, mgΔy = 5.655 × 10 –24 J, thus Δy = 10.9 m. 17. v 2 /v 1 = (T 2 /T 1 ) 1/2 = (373/273) 1/2 = 1.17 18. (a) n 1 = n 2 , m =mM, so m 2 /m 1 = M 2 /M 1 = 28/32 = 7/8. (a) m 1 = m 2 , n = m/M, so n 2 /n 1 = M 1 /M 2 = 32/28 = 8/7 19. E = 3kT/2, so T = 1.93 × 10 5 K 20. (a) 7.45 cal/K = 31.19 J/K. C v = 1.5R, thus n = 2.5 moles
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