1. CV,m = 3/2 R undergoes the transformations = 310.K. and
1 1 1 0 00 0 0 (constant ) 2 (275 177K) 4.073kJ 2 so P 7 5703.4kJ 5 =2Rln 18.2JK 4.073 14.81JK 275 3.39JK 1 and so far you have 0.33 3 TV because vm f f f i sur sur t f f wV q U C K RT V HU V S V q kJ S TK S P bar P T P and T =T V =3V f 0 f 0 f f V P Part(c) -1-1 reversible isothermal process 0 ( 0) 1 2 8.314 275 ln( ) 5.0kJ 3 5.0kJ 11 ln 18 ... ................
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