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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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So delta u B is Cv times T2 minus T1, right.
是Cv乘以，对吧？
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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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UC So we can immediately write delta u C is Cv times T1 minus T2. Delta Hc C is Cp times T1 minus T2, right?
所以我们可以直接写出△,是Cv乘以,△HC是Cp乘以，对吧？
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If you have a real gas and you write du is Cv dT and your path is not a constant volume path, then you are making a mistake. But for an ideal gas, you can always write this. And this turns out to be very useful to remember.
对于真实气体，如果其变化过程,不是恒容的，du=Cv*dT就不成立，但对于理想气体,这条规则永远成立,这一点非常有用，请记住。
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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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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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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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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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So somehow this is going to allow us to get Cv.
计算出。
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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 Cv is going to be related to this path.
这个。
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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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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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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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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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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 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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RT/V this expression becomes Cv dT over T is equal to CvdT/T=-RdV/V minus R dV over V.
这样，or，RT，over，V，bar。，So，now，这个等式就,可以化成。
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There I have it Cp is equal to Cv plus R, right?
所以Cp=Cv+R，对吗？
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This is only true for an ideal gas. Since it's true for an ideal gas, then we can go ahead and replace this with Cv, and then we have Cp=Cv+R Cp with Cv plus R, which is what we were after.
常犯的一个错误，这只对理想气体成立,因为对理想气体成立,所以我们可以继续，用Cv代替,这项，最后得到。
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du But here you've got pressure constant. du, T this is du, not H here. du/dT is only equal Cv to Cv when the volume is constant, not when the pressure is constant.
这里是压强横笛，du，这是，不是H，偏U偏,只在体积恒定时等于，而不是压强恒定时。
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Cv The only difference is it'll be Cp instead of Cv, B but there it is for pathway B. There it is for C a pathway C. So the state functions that we're familiar with are doing what we expect they ought to be doing, right? If you go around in a cycle, starting and ending at the same place the state functions have to stay the same.
是Cp而不是，这是路径,这是路径,所以我们熟悉的态函数的行为,正与我们预期的相同，对吧？,如果你沿着循环走一圈,开始和结束于同一个位置。
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