And the path that I'm describing then, let's assume that we're raising the temperature up is this path right here.

经过一个，等压过程,路径就是这样。

麻省理工公开课 - 热力学与动力学课程节选

so I'm heating up the system in this path here, and then to connect the 2 endpoints here, a constant temperature path.

需要再用，一个等温过程，这两个状态。

麻省理工公开课 - 热力学与动力学课程节选

Depending on what you call external and internal environment, this path that I'm tracing here, deep within your digestive system, is really directly connected to the outside world through both ends.

基于你们所谓的外环境和内环境,我在这里我所指的消化系统内部通路,它的两端确实与外界环境直接相通

耶鲁公开课 - 生物医学工程探索课程节选

So I'm going to write that immediately here, u=0 because I know the answer to that path here.

等温过程，我们可以得到，Δ

麻省理工公开课 - 热力学与动力学课程节选

So what I'm saying is that du through path 1 plus du going through path 3.

于是1,因为能量。

麻省理工公开课 - 热力学与动力学课程节选

If I look at different points in my container during that path, I'm going to have to use a different value of pressure or different value of temperature That's not an equilibrium state, and that process turns out then to be an irreversible process.

如果我要研究在路径中容器里的,不同的点，我就得在容器里不同的点上使用,不同的压强值,或不同的温度值,实际上这不是个平衡态，这个过程是,不可逆过程。

麻省理工公开课 - 热力学与动力学课程节选

Path number 1 I'm going straight up in the V-T diagram.

路径1在T-V图上,竖直向上。

麻省理工公开课 - 热力学与动力学课程节选