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Now for this experiment, this is a constant enthalpy experiment for the Joule-Thomson experiment, this is equal to zero.
对于这个实验，焦耳-汤姆逊实验,是一个焓不变的实验,焓变化等于0，所以我可以。
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This is just an equality. I have a constant pressure dH process. This term here is equal to zero.
这是一个等式，这是个恒压过程,这项等于零，这意味着。
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And we wrote something that looks, the energy is equal to minus the Madelung constant times Avogadro's number, 0R0 q1 q2 over 4 pi epsilon zero R zero.
我们写下了，晶格能等于负的马德隆常数,乘以阿伏伽德罗常数，乘以q1q2除以4πε
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We can also do a similar thing, and I'll keep my distance from the board, but we can instead be holding x constant, for example, putting x to be equal to zero, and then all we're doing is considering the electric field as a function of t.
我们也可以做类似的事情,把x固定为一个常数,例如令x等于零，然后,考虑电场作为时间的函数,这种情况下，我们划掉。
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This piston is being brought out, so we expect 0 the work to be negative, negative. And we start o V2 ut with zero volume. We end up with a volume p2 of V2, and the external pressure is constant to p2.
所以我们可以想象功是负的,开始的时候体积是,最终的容积是,外界的压力恒为。
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dw=0 Well for constant volume, dw is equal to zero.
约束是恒定体积，此时。
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This is equal to zero. So this irreversible process this Joule-Thomson process, is a constant enthalpy process. Delta h for this process is equal to zero.
等于0。所以这个不可逆过程,也就是焦耳-汤姆逊过程,是一个等焓过程。
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SdT This has minus T dS minus S dT, but the dT part is zero because we're at constant temperature.
这一项包含负的Tds和，但是dT的部分等于零,因为温度为常数。
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And that implies that since the quantity we want is given by this quantity, which is zero times a constant, the quantity we want is also zero.
因为我们需要的量,是由这个量乘以一个常数，因为这个量是零，因此我们需要的量也是零。
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OK, for constant volume, this is zero.
对于恒定体积过程，第二项等于零。
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u=0 Constant temperature isothermal delta u is zero.
对等温过程，Δ
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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'm going to call it r. It's going to be the gas constant, and now I have r times t is equal to the limit, p goes to zero of p r.
如果去掉p趋于0的条件,在有限压强下都,保持RT=pV的关系。
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So we immediately get du at constant S and V is less than zero.
这样我们马上就得到以下结论：,在等熵，等容条件下dU必须小于零。
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And so now we have this quantity, p times v bar, and the limit of p goes to zero is equal to a constant times the temperature.
不仅仅对氢气或氮气适用，在p趋于0的极限下,它适用于任何气体。
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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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du/dV So now our du/dV, dp/dT at constant T is just T times dp/dT which is just p over T minus p, it's zero.
现在我们的恒定温度下的，等于T乘以dp/dT，在这里,等于p除以T，最后再减去p，结果是0。
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Now, for an ideal gas, du/dV under =0 constant temperature is equal to zero.
对于理想气体，温度一定,时偏U偏V等于零。
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Because temperature is constant H only 0 cares about temperature. and that's equal to zero.
因为温度是常数，而H只和温度有关,所以这等于0，如果这等于。
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So if you had a high temperature, this a small compared to b. If you're negative which means that dT/dp at constant H is less than zero.
高于反转温度，这一项相比于b很小,意味着H恒定时,偏T偏p小于零。
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