-
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。
-
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。
-
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。
-
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。
-
So basically he's trying to put some atoms in the way of the alpha particle.
把一些原子,放在alpha粒子的路径上。
-
If I know about the states involved, I just need to know what the volume was in each of them.
我们并不需要知道一条特定过程的路径，如果我们知道所涉及的态,我仅仅需要知道这些态的体积。
-
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。
-
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。
-
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。
-
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？
-
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。
-
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。
-
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。
-
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。
-
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。
-
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
-
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。
-
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？
-
Now these two endpoints here are different.
两条路径的末态不一样。
-
This has a particular value.
体积恒定的路径。
-
The sum of path number 2 and path number 3 get me to the same place, so the energy change by going through this time path, this intermediate point here back all the way to final state should be the same the red path.
而经过路径2和3可以3,到达同样的末态，因此经过路径,2和3带来的能量的变化，与路径1带来的,能量变化相同。
-
OK. Note, by the way, if I chase through each possible path, like there's some IFs in here, if there's some places to go, at least in this piece of code, every possible path through this code ends in a return. And that's a good programming discipline, to make sure that happens.
注意一下,如果我跟进每一条可能的路径，像是这里的，起码在这段代码中就有很多走向，这段代码中的每一条可能路径,在结尾都会返回一个值,这就是一条很好的编程定律，请确保这样做。
-
I take V1 to V2 first, keeping the pressure constant at p1, then I take p1 to p2 keeping the volume constant V2 at V2. Let's call this path 1.
容易计算的路径，第一条路径，是首先保持压强不变,体积从V1压缩到。
-
W2 So in this case, it's the amount of work is the area under that curve. And in this case here, the amount of work is bigger, w2 is bigger and it's the area under this curve.
这是路径2做的功,比W1要大，现在假设,我把这两条路径结合起来。
-
That's the some of delta u in these two paths and if we look at delta u in just path A, it's zero.
这是这两条路径△U的和,如果我们要看路径A的△U，它也是零。
-
So path number 1 went from i, f let's call this path up here. went to f, and this is how much energy change there was.
从i出发，经过路径1到达，能量的变化是这么多。
-
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。
-
So the system will not be described by a single state variable during the path.
因此系统将不能被,路径上的单一状态变量描述。
-
So, this path right here from this top red path is an isobaric process.
所以上面这条红色的路径,就是个等压过程。
-
So we know that in each case the heat is going to be the opposite of the work, but the work isn't the same in these two different ways of getting from here to here, right. So let's just see it explicitly. Here's our qA.
所以我们知道在每种情形下功,与热量相差一个负号,但从这里到这里，在这两条路径,中的功是不同的，对吧？,那么让我们明确地看一下。
应用推荐