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"I was appalled. When you read through it it's sort of like it's a memo out of some science fiction movie - I mean, the temperature of the air can't be any lower than thus and so, you can't do it for any few minutes.
VOA: standard.2009.04.17
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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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Enzymes work best, enzymes are proteins that catalyze chemical reactions and our bodies operate by elaborate networks of chemical reactions, When we're off from that temperature then they don't work properly.
举酶这个例子再好不过,酶是一种具有催化作用的蛋白质,人体依靠精密复杂的化学反应网络来运转,如果我们体温与之相差较大,这些酶就不能正常工作
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In other words, the order of taking the derivatives with respect to pressure and temperature doesn't matter And what this will show is that dS/dp dS/dp at constant temperature, here we saw how entropy varies with volume, this is going to show us how it varies with pressure.
换句话说,对温度和压强的求导顺序无关紧要,结果会表明,恒定温度下的，对应我们上面看到的，熵如何随着体积变化，这个式子告诉我们，熵如何随着压强变化。
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The constraint isn't constant temperature because the temperature is going to be changing.
是在不停变化的，不是恒压,因为我们已经有Δp了。
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So we're going to start with a mole of gas, V at some pressure, some volume, T temperature and some mole so V, doing it per mole, and we're going to do two paths here.
假设有1摩尔气体，具有一点的压强p，体积，温度，我们将让它，经过两条不同的路径。
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V2 And there's going to be some volume V2 and some volume V1, but are not necessarily the same. Especially since the pressures are different. we don't know yet about temperature so I don't know what to say about these volumes because I don't know what the temperatures' are going to do.
这里的体积会变成2,这里的容积是V1，它们不必相等,尤其是当压力不相等的时候,我们还不知道温度,所以我不能说这里的容积,是多少因为我不知道,温度会怎样变化。
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OK, for most gases, T inversion is much 300K greater than 300 degrees Kelvin. Much greater than room temperature.
好，对大多数气体,转变温度都高于,比室温高很多。
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The only function it is -- it doesn't care where the gas is. It only cares where the temperature is.
是温度的函数,它只是温度的函数,不管是什么气体。
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You know, we've done a bunch of thermochemistry, and we've kind of seen that the energy of mixing, your energetics don't change too much as a function of temperature.
我们做过很多的化学热力学习题，某种程度上我们发现混合的能量，并不随着能量,发生大的改变。
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So we don't really need to put in a certain amount of heat and change the temperature of the products and the calorimeter and so on.
所以我们实际上并不需要输入,一定的热量，改变生成物,和量热计的温度之类。
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And so an experiment said the gas didn't increase its temperature when it expanded the vacuum.
这个实验告诉你,气体在向真空膨胀的过程,中温度没有升高。
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And this volume, temperature and pressure doesn't care how you got there. It is what it is.
另一个状态,也有一组确定的体积。
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Because he thought that would be big enough that in Denmark, the temperature wouldn't go below zero.
因为他觉得这个数字,对于丹麦来说够大了，气温不会低到零度以下。
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In other words, T surrounding dS has to be greater than zero, and of course temperature is always positive.
换句话说这就意味着，环境温度T乘以dS必须大于零，当然环境温度T是正的。
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He didn't want to use negative numbers To measure temperature in Denmark outside.
他可不想用负数,来测量丹麦室外的温度。
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So the concept of an absolute zero, a temperature below which you just can't go, that's directly out of the scheme here, this linear interpolation scheme with these two reference points.
这就是绝对零度，这样，从线性插值的图像出发，我们得到了绝对零度的概念，你永远无法达到,低于绝对零度的状态。
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I won't end up at the same temperature.
但不结束在相同的温度。
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We use a liquid helium. And so in order to make a liquid helium, you can't take helium at room temperature and do this, because if you did, you would just heat it up, because the room temperature is above the inversion temperature, so Joule-Thomson would heat up the helium.
为了得到液氦,不能再常温,做这个实验,否则就是加热氦气，因为室温高于它的转变温度,所以焦耳-汤姆孙，实验会加热氦气。
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That is, in real life, the variables that you'd normally control aren't some combination of entropy and these variables, but really their temperature, volume and pressure, any couple of those, might be what you'd really have under experimental control.
在生活中，我们所能控制的,不是熵和其他变量的组合，而是温度，体积，压强，以及其中的两两组合,这些才是试验中所能控制的。
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In this case, V = /P. Have two quantities and the number of moles gives you another property. You don't need to know the volume. All you need to know is the pressure and temperature and the number of moles to get the volume.
以及气体的摩尔数,就可以得到第三个量,知道压强，温度和气体的,摩尔数就可以推导出气体的体积,这称为状态方程,它建立了状态函数之间的联系。
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When you go from here to going outside, to going to a much hotter room, your temperature stays the same and your body is able to control this on your own, you don't have to think about it.
当你从这里走出去,去一个更热的房间的时候,你的体温不会变化,你的身体能自行控制体温,你无需要刻意想它
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The ideal gas constant doesn't change, temperature doesn't change, and so v we just have minus nRT integral V1, V2, dV over V.
理想气体常数不变，温度也不变，因此,是负的nRT，积分从v1到v2，dv除以。
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What is dH/dT as a function, keeping pressure constant, what is dH/dp, keeping temperature constant?
恒定时偏H偏T是什么,温度恒定时的偏H偏p又是什么呢？,好的，让我们解决第一个问题？
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And so redefining then the temperature scale to the Kelvin scale, where t in degrees Kelvin is equal to t in degree Celsius, plus 273.15.
以K为单位的温度,是以℃为单位的温度,加上273。15度。
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du/dT constant pressure is the direct derivative with respect to temperature here, which is sitting by itself under constant volume keeping this constant but there is temperature sitting right here too.
偏U偏T，p恒定是对,温度的直接微分，而它本身对体积不变,保持它不变，但是这里也有一个温度,这就是偏U偏V，T恒定。
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And that generally is just given by the heat capacity the calorimeter times delta T, right.. Because the heat capacity of the calorimeter just like this thing is not strongly temperature dependent, OK.
一般它就由量热计的热容,乘以ΔT给出，因为，量热计的热容,就像这张桌子一样,并不强烈依赖于温度。
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Last time we reach the third law which is telling us that we can't quite get to zero degrees Kelvin .but that as the temperature approaches zero degrees Kelvin, the absolute entropy of a pure substance in perfect crystalline form is zero.
上次课我们得到了热力学第三定律,这个定律告诉我们我们无法,达到0K的温度,但是在我们接近绝对零度的过程中，以完美晶体形式存在的纯物质的绝对熵,也趋向于零。
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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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