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Delta u, delta H, familiar state functions, q w changes in their values, q, w, heat and work.
U，△H，很熟悉的态函数，它们的值在变化；
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Let me give you, I'm going to create, q a polar point, I'm going to call it q, and we'll give it some random values.
让我给大家讲解下，我要去创建，一个极坐标点，然后我会去命名它为，然后我给它赋一些随机的值，好，现在我想知道。
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q1 The heat input is just q1, Q and we'll define that as capital Q.
输入的热是，我们把它定义为大写的。
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And let me use a big Q to be the total quantity produced-- sorry--the total quantity demanded in the market.
用Q代表总产量,对不起,是市场总需求量
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And lower case q sub 2 is the charge on the electron.
小写q加个下标2是指电子所带电量。
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w And that has to equal q plus w, summed up for all the steps.
等于q加，对所有过程相加。
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w Now, we have u, q and w.
现在我们有u，q和。
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w That's u2 minus u1, and it's q plus w.
就是u2减去u1，等于q加。
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And of course, in either case, w delta u is q plus w, so it's q irreversible plus w irreversible, u being a state function it's the same in either case.
当然了，在这两个过程中，Δu都等于q加，因此现在是q不可逆加w不可逆，同时u在这两个过程中都是态函数。
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So qi and q-i or q1 and q2 will be the strategies.
用qi,q-i,q1,q2表示策略
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And because there is an explicit relationship between u, delta u, q and w, you can always find the easy way to derive the change in internal energy or the heat or the work.
因为Δu，q和w之间,有明确的关系，一般来说,内能或热或功的变化,易于计算。
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U It's u, because u is to q plus w right, heat and work, but it's adiabatic. So there's no heat, exchange with the environment, and it's constant volume, so there's no p dV work, right.
什么是零？是U，因为,等于q加w，热量和功，但这是绝热的，所以系统与环境间没有热量交换；,同时它是灯体的，所以也没有pdV形式的功。
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All I'm doing here, nothing particularly exciting, I'm simply plugging this expression in for P and then I multiplied the whole thing out because it was all multiplied by q.
这些计算过程确实很枯燥,我只是把这个算式代入到P中,然后打开整理,因为每一项都乘以了q
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Or I could have a non-adiabatic, I could take the same temperature change, by taking a flame, or a heat source and heating up my substance. So, clearly q is going to depend on the path.
也能改变温度，绝热指的是没有热传递，在非绝热条件下，也同样可以升温,比如用火或者热源加热，这样，q也应当与路径有关。
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