Solution P2=2 bars Process1→2 State 1 = RT U=C po-RT P1=I bar p=C P RT 600K 300K 2 bars (300K)=600K p At state 3: v3=RT3/p37 RT2 T Mp 2 3T3=T2 kJ 8.31 kmol·K kg 1P3=P1 28.97 ol 3 =1.72m3/kg 上游充通大 March 21,2018 9 SHANGHAI JIAO TONG UNIVERSITYMarch 21, 2018 9 Solution State 1 State 2 1 1 1 p v RT  2 2 2 p v RT  Process 12 At state 3: 2 3 T p 3 1