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That's the most outstandinganswer I've ever heard. You must havea goddamn I. Q Of 160.
VOA: standard.other
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Sorry, the question was, what happens if I said p is less than q?
抱歉，问题是，如果我这里写p小于q的话？
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I am going to highlight this by calling this q1, lower case q being the charge here on the nucleus.
我要把它称作q1来进行强调，小写q是指原子核上的电量。
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The state require sa minimum I. Q of 80 to attend public school.
VOA: standard.other
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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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All right, q has to be just the opposite of this because we've already figured out that there's no change in u.
好，q一定与它,相差一个符号，因为我们已经,指出U是不变的。
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So just to emphasize that the strategies are quantities, rather than using S let me use Q today to be the strategies.
为了突出策略就是商品的产量,这次我们用Q来表示策略而不用S了
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H=qp The u plus p V. Delta H is equal to q V.
括号里面的就是H，等于u+pv，Δ
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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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du We can talk about du for the system well, w that's q plus w.
我们可以讨论它的，等于q加。
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Cp And delta T is given by the heat, which has to do with how much of the candle burnt, divided by the constant pressure heat capacity.
T等于热量q除以恒定的等压热容，其中热量与，蜡烛燃烧的多少相关。
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Adiabatic q equal to zero. It's also delta H 0 which is zero. The two didn't necessarily follow because remember, delta H is dq so p is only true for a reversible constant pressure process.
在这个过程中ΔH等于,绝热的所以q等于0，而ΔH也等于,这两个也不一定有因果关系,因为，记住，ΔH等于dq只有在恒压。
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So it's the same as q reversible plus w reversible.
它们是相等的，因此这也等于q可逆加w可逆。
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All right, so that's just solving out the algebra, so we're saying what q solves, a - c over 2b - q2 over 2 = 0.
没错,只需要解出当,/2b-q2/2=0,时q2的值即可
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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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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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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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