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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。
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All right, so I'm burning some energy, I'm burning some fuel to do this somehow, to get that work to happen.
而这个路径需要先加热再冷却，这样，我需要能量来源，比如燃料，来完成做功。
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Hess' Law states that for any chemical reaction, the energy change is path independent.
盖斯定律表明，对于任意化学反应，能量变化并非是路径依赖的。
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So there's a return branch on every possible path through the code. And that's valuable, it's something you want to think about as your right your own.
所以代码中每条可能的路径,都会有一个返回值分支，这是非常有价值的,当你自己写程序的时候也应该考虑这件事。
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It's not self-renewing, in that division of the zygote results in two daughter cells that are no longer the zygote anymore, they're down some pathway.
它是不会自我更新的,受精卵分裂后成了两个子细胞,而这两个子细胞就不再是受精卵了,它们沿着某种路径进行了分化
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Path number 2 on my diagram it's a reversible, this path number 2, it's a reversible constant pressure path.
路径，首先是一个，等压过程。
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So, this observation is equivalent to saying that there must be something that is path independent.
这个表述相当于说,一定存在某个物理量是与路径无关的。
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All right, so I need to tell you the path, when I go from one state to the other.
好，当我从一个状态过渡到另一个状态时，我需要告诉你们路径。
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Now, suppose I took this two paths, and I took -- couple them together with one the reverse of the other.
其中一条的方向反过来，这是初态，末态,路径1，先冷却。
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so we're going to use this concept of the path to go from the initial point to the end point.
在处理状态函数时,路径的选取。
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And you already saw last time there was this relationship between the temperature and volume changes along an adiabatic path.
是条绝热路径，而上次你已经看到,沿着绝热路径温度和体积,的变化有这个关系。
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dU=CvdT So du is still going to be equal to Cv dT, and we're still going to be able to use the first law, all these things don't matter where the path is.
于是，第一定律也依旧成立，这些关系都与路径无关。
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So let's just compare. Let's compare what C happened in path A to what happened in paths B and C. Yes?
我们比较一下沿,路径A和路径B和,所发生的事有什么不同，什么？
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It's going to be the same temperature V+dV as before but the volume is V plus dV now.
将升温到跟路径1的结果一样,但是现在的体积是。
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All right, so when you do a problem, the path is going to turn out be extremely important.
好，当你解题时，路径会变得,极其重要。
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For instance, you want to know how much energy you're going toget out from doing this expansion.
比如，你想知道在这个膨胀过程中,到另一个平衡态的路径。
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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。
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All right, now suppose that I took these two paths and coupled them together.
现在假设我把这两条路径结合起来,这是路径1做的功。
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So the first path then, the first path, 1 constant volume constant V, so I'm going to, again, let's just worry about energy.
首先，是路径，等压过程。
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Then I can go back, the gas is included at every point of the way.
我就能走回去，气体被包含在路径上的每个点中。
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This depends on the path. It tells you right here the path is constant pressure. These don't depend on the path, right. V doesn't care how you v get there. u doesn't care how you get there.
这由变化的具体路径决定,这个小脚标表明过程是恒压的,这些量都与具体路径无关,即不管是通过什么路径使得体积变化为Δ
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That's going to be number 2 right here.
这就是路径。
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First one, when I write a piece of code, especially code that has branches in it, when I design test cases for that piece of code, I should try and have a specific test case for each possible path through the code.
第一点，当我们写代码的时候，尤其是一些含有分支语句的代码的时候，当我为这些代码,设计测试用例的时候，我应该为每一个可能的程序路径,都设计一个特别的用例。
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the -- no, I'm not going to cool down, 3 in this case here.
一步，路径。
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It would've been easier, but I wanted to go through the whole path right?
简单许多，但是我希望研究整个循环路径？
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So the problem here, roughly, is given, a number of cities, and say the cost of traveling from city to city by airplane, what's the least cost round trip that you can find?
基本上就是一个这样的问题,你被告知了一些城市,以及在城市之间坐飞机旅行的花费,那一条是你们能找到最短往返路径？
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So, the first obvious one is to take V1 to V2 first with p constant. So take this path here.
从初态到末态可以有无数种路径2,甚至像这样，我们只研究两种。
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So then, going from the initial to final states with a red path, ou start with an isobaric processand then you end with an isothermal process.
因此沿着,这条红色路径，你从一个等压过程开始,以一个等温过程结束。
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I could have that path be very slow and steady, so that at every point along the way, my gas is an equilibrium.
我可以很慢，很稳定地经过这条路径，因此在这条路径上的任意一点，气体都处于平衡态。
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And what you'll find is that the maximum work out is obtain for a reversible path.
对于各种可能的中间路径，我们将发现,做功做多的是可逆路径。
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