Question

1-3 2molH2在0℃,压力为100kPa下恒压可逆膨胀至100L,求Q,W,△U,△H.

Answer 1

T=273.15K p=100kPa n=2mol V1=2*8.314*273.15/100=45.4L T2=T1*V2/V1=601.65K

恒压可逆膨胀过程

W=-pdV=-100*(100-45.4)=-5460J

Cp,m=2.5R

dU=Cp,m*n*dT=(601.65-273.15)*2.5*8.314*2=13656J

Q=dU-W=19116J

dH=dU+d(pV)=Q=19116J