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We start at p1, V1. and p external is equal to zero.
从状态开始，外部压强为零。
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They don't care that there are other atoms and molecules around. So that's basically what you do when you take p goes to zero.
这正是当压强无限小时,气体的行为,气体的体积无限大。
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So if we draw the 2 p orbital, what we just figured out was there should be zero radial nodes, so that's what we see here.
如果我们画一个２ｐ轨道，我们刚才知道了是没有径向节点的，我们在这可以看到。
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That's different when you have continuous values-- you don't have P because it's always zero.
和离散型随机变量的分布不同的是,连续型随机变量的分布中,某一点的概率值始终是零
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So for an ideal gas then, dH/dp under 0 constant temperature, that has to be equal to zero.
所以对于理想气体,偏H偏p在恒温下，等于。
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pV=RT p plus a over v bar squared times v bar minus b equals r t. All right if you take a equal to zero, these are the two parameters, a and b. If you take those two equal to zero you have p v is equal to r t.
我们就回到,也就是理想气体,状态方程,下面我们来看看,这个方程。
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Yup, zero radial nodes. So, for a 2 p orbital, all the nodes actually turn out to be angular nodes.
没有，对于2p轨道，所有的节点都是角向节点。
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As p goes to zero of p times v bar.
适用于任何气体。
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So, if we say that in this entire plane we have zero probability of finding a p electron anywhere in the plane, the plane goes directly through the nucleus in every case but a p orbital, so what we can also say is that there is zero probability of finding a p electron at the nucleus.
而只要是ｐ轨道,这个平面都直接，穿过原子核，那么我们,可以说在原子核上,找到一个ｐ电子的概率为零。
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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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du external dV minus T surroundings dS is less than zero.
加上压强p，du，plus，p,乘以dV减去环境温度T乘以dS小于零。
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So, du plus p dV is less than zero.
因此，du加上pdV需要小于零。
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So you should know that 0 any time m is equal to zero that it's the p z.
所以当我们讨论p轨道时,只要m等于,它就是pz轨道。
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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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And so they defined them, p after many experiments, the limit of this 0 delta T delta p and the limit of delta p goes to zero as the Joule-Thomson coefficient.
他们定义了这些量，以及它们的范围，ΔT比Δ,Δp的极限趋近于,叫做焦耳-汤姆逊系数。
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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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H We also saw that dS for constant H and p was greater than zero.
我们同样可以看到如果保持自由焓,和压强不变熵的变化dS也是大于零的。
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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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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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And it's p external and p external is zero, so there's no work.
外部的p是零，所以结果是零，没有做功。
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We can look at the 2 p, which should have one radial node, and we just figured it out for the, excuse me, 3d for the 3 p has one radial node, and for the 3 d here, we should have zero radial nodes, we just calculated that.
我们看2p，它有一个节点，我们已经知道对于，不好意思，是3p有一个节点，对于，它应该没有径向节点，我们刚刚算过这个。
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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 in this experiment here, delta p is less than zero. You need to have this whole thing greater than zero. So delta T is less than zero as well. So if you're below the inversion temperature and you do the Joule-Thomson experiment, you're going to end up with something that's colder on this side than that side.
所以在这个实验中，Δp小于零,这全部都大于零,因此ΔT也小于零,所以如果在低于转变,温度的情况下做焦耳-汤姆孙实验,最后的结果是,这边的温度比这边低。
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