又 A-白Y-26410 P=P2=2.674×10Pa 由理想状态方程得 7=Pk7=1081×10K PoVo (2)I中气体内能的增量为 △=C(☑-Z)=R7-T,) =,(p-P'6)=2.69x10J 5 I中气体对外作的功为 4=△E,=292×103J1.081 10 K 3 0 0 0 1 1 1 = T =  p V pV T ( ) 2.674 10 Pa 5 2 0 2 = 0 =   V V p p 又 2.674 10 Pa 5 p1 = p2 =  由理想状态方程得 (2)Ⅰ中气体内能的增量为 ( ) E1 =CV T1 −T0 ( ) 2 5 = R T1 −T0 ( ) 2.69 10 J 2 5 4 = p1V1 − p0V0 =  2.92 10 J 3 A1 = E2 =  Ⅰ中气体对外作的功为