tailieunhanh - Physics exercises_solution: Chapter 19

Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 19 | a b pAV nRAT mol J mol K 80 Co x 103 J. a b If the pressure is reduced to of its original value the final volume is 5 2 of its original value. From Eq. W nRT ln V2 3 J mol K K ln 5 I x 103 J. V1 v 2 J pV nRT T constant so whenp increases V decrease At constant pressure W pAV nRAT so AT W 75 X103 J_ nR 6 mol J mol K and ATk ATc so T2 . a b At constant volume dV 0 andso W 0. b pAV Pa m3 m3 J. a b In the first process W1 pAV 0 . In the second process W2 pAV x 105 Pa m3 x 104 J. 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 . Q 254 J W -73 J work is done on the system and so AU Q - W 327 J. a pAV 105 Pa 104 J. b AU Q - W x 105 J - x 104 J x 104 J. c The relations W pAV and AU Q - W hold for any .

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