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So I could have written right here immediately CvdT equals Cv dT, and that was the end of my derivation.
于是我在这里就可以直接写,来它等于，完成了整个推导。
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R/Cv OK, so that means that this is really instead of -1 R over Cv. it's really gamma minus one.
现在,变成了γ
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So if you get these two guys together you get CvdT=-pdV Cv dT is minus p dV.
把它们联系,起来。
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You're allowed Cv comes out here for this adiabatic expansion, which is not a constant volume only because this is always true for an ideal gas.
绝热过程写下,这个式子是因为它对理想气体都成立，并没有用到等容过程的条件,只用了理想气体的条件。
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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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du, it's an ideal gas. So this is Cv dT and of UB course we can just integrate this straight away.
因此这是CvdT，当然我们可以,直接算出这个积分，那么△
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dU=CvdT pV=RT So I can write du is Cv dT and pV is equal to RT.
于是。
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Cv+R=Cp Cv is equal, oh Cv plus R is equal to Cp it's a relationship that we had up here that we wanted to prove.
我们就得到了,我们一开始,想要证明的。
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So delta u of reaction is approximately equal T to negative Cv for the calorimeter times delta T.
所以反应的ΔU近似等于,负的量热计的Cv乘以Δ
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We take this Cv and put in the exponent here R right. And we put this R up in the exponent here.
我们把这个Cv提出来,放在这里的指数上，好吧？然后把。
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So somehow this is going to allow us to get Cv.
计算出。
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Cp, I forgot to put my little bar on top here because it's per mole Cp dT that's my dq here.
上面的Cv我忘记加上横线了,因为它也是摩尔热容。
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And I know energy is related to Cv through Cv dT etcetera.
我们可以计算,这两个过程中的能量差。
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is always equal to for an ideal gas? Cv dT right.
对吧？
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Cv So this will give us something about C sub v.
从这里可以计算出。
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Cv Cv is going to be related to this path.
这个。
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Because this isn't Cv mathematically speaking.
我们不知道到这是什么。
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but right now you're going to have to take it for granted. So, if the Joule-Thomson coefficient is equal to zero, just like we wrote, du = Cv dT du = Cv dT for an ideal gas, we're going to dH = Cp dT have dH = Cp dT for an ideal gas as well.
但是现在请你们应该把它看成理所当然的,所以，如果焦耳-汤姆逊系数等于零,就像我们写的，对于理想气体,我们也可以得到对于理想气体。
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Cv So, for Cp and Cv, these are often quantities that are measured as a function of temperature, and one could, in fact, calculate this integral.
一般Cp和,都是温度的函数，因此实际上，我们可以将这个积分计算出来。
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T2 So we know that delta u is just Cv times T1 minus T2.
u等于Cv乘以T1减去。
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By definition I'm going to define gamma by Cp over Cv by definition.
把Cp/Cv记作γ,这完全是定义。
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du It's an ideal gas, and that's equal to w1 prime.
等于CvdT，du，is，Cv，dT。,因为是理想气体，所以等于w1一撇。
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So delta u B is Cv times T2 minus T1, right.
是Cv乘以，对吧？
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T1 So delta you is just Cv times T2 minus T1.
因此Δu等于Cv乘以T2减。
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We want a relationship in p-V space, not in T-V space. So we're going to have to do something about that. But first, it turns out that now we have this R over Cv.
我们想要p-V空间中的结果，而不是T-V空间中的,因此需要做一些变换，先来看现在的关系，它跟R/Cv有关。
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And so that means that delta u is always calculable from Cv dT for any ideal gas change.
这意味着对理想气体,Δu只需利用Cv，dT计算。
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Here's heat exchanged in pathway A and in pathway B heat is zero, and in pathway C, Cv here is qC it's Cv T1 minus T2.
这是qA，这是路径A上的热量交换,路径B中的热量交换是零，而在路径C中，这是qC，它是。
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T So we know that T dS/dT at constant volume is Cv over T, T and dS/dT at constant pressure is Cp, over T.
在恒定压强下定压比热容Cp乘以dT除以,所以在恒定体积下dS/dT等于Cv除以，在恒定压强下dS/dT等于Cp除以。
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/T We've got Cv integral from T1 to T2, dT over T is equal to minus R from V1 to V2 dV over V.
左边是Cv乘以，从T1到T2对dT积分。
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So it would imply that CvdT du was equal to only the first term Cv dT.
这意味,这du进等于第一项。
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