P dV is equal to R dT. pV = RT for 1 mole, pdV=RdT so I just take dV here.

对等压过程，那么。

麻省理工公开课 - 热力学与动力学课程节选

I know I only need 2, so I can relate dV dV to dp through the ideal gas law.

我只需要两个就够了，因此可以用,理想气体状态方程消去。

麻省理工公开课 - 热力学与动力学课程节选

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

Normally I couldn't do that Vdp because this term would have p dV plus V dp, but we've specified the pressure is constant, so the dp part is zero.

一般情况下我不能这么写,因为这一项会包含pdV和，但是我们已经假定压强为常数，所以包含dp的部分等于零。

麻省理工公开课 - 热力学与动力学课程节选

I also want to assume for our present purposes that there's only pressure volume work going on, which is to say I want to put pdV p dV in here minus p dV for dw.

同时假定,对于我们目前的目的而言，只有压强做功这就是说,我要把这里的dw替换为负。

麻省理工公开课 - 热力学与动力学课程节选

And I know pdV what those turn out to be. It's minus S dT minus p dV.

我们知道,这最终就是负SdT减去。

麻省理工公开课 - 热力学与动力学课程节选

So I can instead of the dV here, I can insert this R over p dT.

两个p消掉了，结果。

麻省理工公开课 - 热力学与动力学课程节选