(1)(T1=250K,P1`=200KPa)→(T1=T2,P1=50KPa)
W=-P2ΔV=-P2(V2-V1)=-(RT2-RT1P2/P1)=-1.559KJ
ΔU=CvΔT=0
Q=ΔU-W=1.559KJ
ΔH=CpΔT=0
(2)绝热 Q=0
(T1=250K,P1`=200KPa)→(T2=?,P1=50KPa)
W=ΔU
ΔU=CvΔT
理想双原子气体 Cv=5/2R Cp=7/2R
-(RT2-RT1P2/P1)=5/2R(T2-T1)
解之 得11/14T1=T2 T2=196.4K
ΔT=53.5K
ΔU=CvΔT=1.113KJ=W
ΔH=CpΔT=1.559KJ
我才刚高中毕业,都是自学的,您在检查下吧