# Sol-Ch18 - CHAPTER 18 E x ercises 1(a 21.1 C(b 90.6 C(c...

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1. (a) 21.1 ° C; (b) 90.6 ° C ; (c) 37.0 ° C 2. (a) 621 ° F; (b) –423 ° F; (c) 113 ° F 3. –40 ° F = –40 ° C 4. (a) 33.3 ° C; (b) 20.5 cm 5. 273x9/5 = 491 ° R 6. Find a = 8.62 and b = –207. (a) 31.0 Ω ; (b) 43.0 ° C 7. N/V = P/kT = 2.68 × 10 25 m –3 8. Mass of n moles is m = nM, so ρ = nM/V 9. PV = nRT = (m/M)RT, so P = ρ RT/M, where M is in kg/mol. Thus ρ = 44.5 kg/m 3 . Note 1 g/mol = 10 –3 kg/mol. (a) 1.25 kg/m 3 ; (b) 1.42 kg/m 3 ; (c) 0.089 kg/m 3 10. (a) PV = nRT and n = m/M, thus P α n α 1/M. P 2 (M 1 /M 2 )P 1 = 3.43 atm. (b) m 2 /m 1 = P 2 /P 1 = 2/3.43 = 0.583 kg. 11. (a)V = nRT/P = 2.44 × 10 –2 m 3 = 24.4 L; (b) 37i.2 L 12. (a) n =PV/RT =9122 moles, so m = nM = 265 kg. (b) P 2 = P 1 (T 2 /T 1 ) = 1.017 atm (c) When the pressure dropsby a factor of 1/1.017, the number of moles also drops by the same factor : n 2 /n 1 = 1/1.017, thus Δn = n 1 – n 2 = (1 – 0.9833)n 1 = 0.0167n 1 = 152.5 mol. Thus, Δm = Δn M = 4.42 kg. 13. (a) T

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Sol-Ch18 - CHAPTER 18 E x ercises 1(a 21.1 C(b 90.6 C(c...

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