I'm going from V2 to V2 dv - what do you think this integral is? Right, W1 so this is easy part, zero here.

这样才保证了过程的可逆性,下面来计算，路径1的功。

麻省理工公开课 - 热力学与动力学课程节选

So if you get these two guys together you get CvdT=-pdV Cv dT is minus p dV.

把它们联系,起来。

麻省理工公开课 - 热力学与动力学课程节选

dU=-pdV So du is just minus p dV.

于是。

麻省理工公开课 - 热力学与动力学课程节选

So it's minus R T1 dV over V, right?

那么这是-RT1，dV/V，对吧？

麻省理工公开课 - 热力学与动力学课程节选

p2 It's minus p external. Well p external is actually p2, so that makes it actually easier.

它将等于-Pext，dV，而外界压强就是，这样简单多了。

麻省理工公开课 - 热力学与动力学课程节选

RT/V this expression becomes Cv dT over T is equal to CvdT/T=-RdV/V minus R dV over V.

这样，or，RT，over，V，bar。，So，now，这个等式就,可以化成。

麻省理工公开课 - 热力学与动力学课程节选