So, the total work that we can get out must be zero, out of the whole sum of them.

因此，我们可以从这两个热机,得到的总功，一定是零。

麻省理工公开课 - 热力学与动力学课程节选

The total work is less than zero.

说明系统对外界做功。

麻省理工公开课 - 热力学与动力学课程节选

So, therefore, we can rewrite our equation in two ways. One is just talking about it in terms only of energy where our kinetic energy here is going to be equal to the total energy going in -- the energy initial minus this energy of the work function here.

所以我们可以把方程,写成两种形式，一个是,只考虑能量，动能等于总的,入射能量-初始能量减去,功函数的能量，我们如果想解决,比方说，我们想知道。

麻省理工公开课 - 化学原理课程节选

V Minus p2, V2. Minus p delta V. So the total work is the work from the left hand side plus the work on the right hand side, p1 V1- p2 V2 which is p1 V1 minus p2 V2.

负p2V2【w=-p2v2】，即负pΔ,所以全部的功就是,左边的功加上右边的功，等于。

麻省理工公开课 - 热力学与动力学课程节选

the total amount of the work that we can get out is just given by the area inside this curve.

能够输出的功就是,我们可以得出来,曲线所包围的面积。

麻省理工公开课 - 热力学与动力学课程节选