Đang chuẩn bị liên kết để tải về tài liệu:
Physics exercises_solution: Chapter 19

Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 19 | 19.1 a b pAV nRAT 2.00 mol 8.3145 J mol K 80 Co 1.33 x 103 J. 19.2 a b If the pressure is reduced to 40.0 of its original value the final volume is 5 2 of its original value. From Eq. 19.4 W nRT ln V2 3 8.3145 J mol K 400.15 K ln 5 I 9.15 x 103 J. V1 v 2 J 19.3 pV nRT T constant so whenp increases V decrease 19.4 At constant pressure W pAV nRAT so AT W 75 X103 J_ 35.1K nR 6 mol 8.3145 J mol K and ATk ATc so T2 27.0OC 35.1OC 62.1OC. 19.5 a b At constant volume dV 0 andso W 0. 19.6 b pAV 1.50X105 Pa 0.0600 m3 -0.0900 m3 -4.50X103 J. 19.7 a b In the first process W1 pAV 0 . In the second process W2 pAV 5.00 x 105 Pa -0.080 m3 -4.00 x 104 J. 19.8 a W13 p1 V2 - V1 W32 0 W24 p2 V1 - V2 and W41 0. The total work done by the system is W13 W32 W24 W41 p1 - p2 V2 - V1 which is the area in thep- V plane enclosed by the loop. b For the process in reverse the pressures are the same but the volume changes are all the negatives of those found in part a so the total work is negative of the work found in part a . 19.9 Q 254 J W -73 J work is done on the system and so AU Q - W 327 J. 19.10 a pAV 1.80x 105 Pa 0.210m2 3.78x 104 J. b AU Q - W 1.15 x 105 J - 3.78 x 104 J 7.72 x 104 J. c The relations W pAV and AU Q - W hold for any .