解(1)QAE=C△T 32 ×8.31×(350-300)=623J A=0 (2)4Es3 ×831×(350-300)=623J 2 A=V=RT=8.31×50=416J QE+A=623+416=1019J 目录结A =0 ( )=623J 2 3 = ×8.31× 350 300 A = pΔV=RΔT = 8.31×50=416J Q =Δ E+A =623+ 416=1019J 解(1) Q =Δ E =CVΔT (2) Δ E ( )=623J 2 3 = ×8.31× 350 300 目录 结束