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pV=RT dT here because the pressure is constant, dV=RdT/p so dV is equal to R over p dT.
因为对1摩尔气体有，于是。
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Because what we've done is we forced p, pressure here, to be equal to the external pressure.
因为这里我们让内部的压强，等于外部的压强。
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It's not constant pressure, because we have a delta p going on. It's not constant volume either.
也不是恒容，这个限制,是这个实验的限制。
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It's reversible, that means that p external, equals p. I'm doing it very slowly so that I'm always in equilibrium between the external pressure and the internal pressure so I can go back and forth.
我都可以不损失能量，而把过程逆向进行,整个过程中的,每一步都保持平衡,现在我们来分别。
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Pressure has dropped out of the picture completely here. So there is no p dependence here.
理想气体的H只是温度的函数,这对于真实气体来说是。
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It's tabulated in books, and this we can measure p in the experiment. Delta p here is the change in pressure from the left side to the right side, and we can put a thermometer, measure the temperature before the experiment and measure the temperature after the experiment.
这列在书上，这个量我们在,实验中也可以测量，在这里Δ,是从左边到右边的压强变化，我们可以放一个温度计，去测量实验前的温度,再去测量试验后的温度。
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p So dV/dT at constant pressure is just nR over p.
所以恒定压强下dV/dT等于nR除以。
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Normally I couldn't do that Vdp because this term would have p dV plus V dp, but we've specified the pressure is constant, so the dp part is zero.
一般情况下我不能这么写,因为这一项会包含pdV和，但是我们已经假定压强为常数，所以包含dp的部分等于零。
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I also want to assume for our present purposes that there's only pressure volume work going on, which is to say I want to put pdV p dV in here minus p dV for dw.
同时假定,对于我们目前的目的而言，只有压强做功这就是说,我要把这里的dw替换为负。
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We're going to change the pressure above, Pext right now there's a p external, which is equal to p on the inside.
来改变外界条件，可以改变外界的压强，它将与气体压强p相等。
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I'm pressing on the gas. So I expect that to be a positive number. The pressure is constant 0 p. The V goes from V1 to zero.
我们对气体加压,所以这应该是一个正数,压强是常数，p，V从V1变成。
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The property is the limit as p goes to zero of pressure times molar volume.
与摩尔体积的乘积,在气体压强p趋于0时的极限。
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So we have dH/dT keeping pressure constant, is du/dT keeping pressure constant.
等于偏U偏T，p恒定加上，偏pv偏T，p恒定。
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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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T Remember, we're trying to get delta H, p we're trying to get dH/dT constant pressure and dH/dp constant temperature. OK, these are the two things were trying to get here.
想要得到在恒压状态下的偏H偏，和在恒温状态下的偏H偏,好的，这是两个我们，在这里想要得到的东西。
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