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So, these two are equal to each other as well which tells me that this derivative, Cp dH/dT constant pressure is Cp.
所以这两者也相等,这告诉我们在恒压下微分，等于。
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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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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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dT This is equal to Cp minus R dT.
等于。
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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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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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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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There I have it Cp is equal to Cv plus R, right?
所以Cp=Cv+R，对吗？
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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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