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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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So if you get these two guys together you get CvdT=-pdV Cv dT is minus p dV.
把它们联系,起来。
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Now you have to know from elementary calculus that v times dv/dt is really d by dt of v square over 2.
根据初等微积分知识你们得知道,v乘以dv/dt其实就是d/2)/dt
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I know I only need 2, so I can relate dV dV to dp through the ideal gas law.
我只需要两个就够了，因此可以用,理想气体状态方程消去。
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It's constant pressure. OK, so now, last time you looked at the Joule expansion to teach you how to relate derivatives like du/dV.
这是恒压的，好，上节课你们,学习了焦耳定律,以及怎样进行导数间的变换。
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dS/dV And that, now, we know must equal dS/dV, with a positive sign. At constant temperature.
我们知道这个等于恒定温度下的，符号为正。
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And so just like here, w2 now q2 is minus w2, that's integral going from three to four p dV.
就跟这儿一样，现在是q2等于负，等于从第三点到第四点过程的pdv的积分。
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dV we're still going to write that it's an ideal gas.
理想气体的条件,也依然成立。
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V+dV And this is going to be here V plus dV.
这里是。
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TdS It comes from the fact that dq reversible is T dS, pdV and dw reversible is minus p dV.
这个结论来自于：可逆过程下dq等于，做功dw等负的。
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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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dV OK, so we've got dT here dV here.
等式的左边是dT，右边是。
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And I know pdV what those turn out to be. It's minus S dT minus p dV.
我们知道,这最终就是负SdT减去。
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pdv It's an isothermal expansion, so dw is just negative p dV.
因此dw等于负，这就是。
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v dv We have minus V1, V2, nRT over V dV.
负的v1，v2，nRT除以。
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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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pdV Minus S dT, that's the p dV term that's left, minus p dV.
应该是负SdT，留下的应该是pdV项负的。
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pdV So, du is T dS minus p dV.
即du等于TdS减去。
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So there's p v work and it's given by the integral minus p external dv or just the integral from one to two of dw.
我们知道对于pv系统来说,功w可以写成黑板上的两种不同形式。
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p dA/dV, at constant T, must be negative p.
在恒定温度下，dA/dV等于。
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That's where that term comes from, du/dV dV/dT.
乘以偏V偏T，p恒定，这项的来源。
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OK, so for a constant volume process, du we can write du, partial derivative of dT u with respect to T at constant V, dT, dv plus partial derivative of u at constant V, dV.
好，对于一个恒定体积的过程，我们可以写出,等于偏u偏T，V不变，加上偏u偏V，T不变。
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So, from the calculus we know dv/dt=a.
由微积分我们得知dv/dt=a
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dS/dV There's some variation, dS/dV, at constant temperature.
这里有一点变化，即恒定温度下的。
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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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So it's minus R T1 dV over V, right?
那么这是-RT1，dV/V，对吧？
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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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Now, we can immediately see what du/dV at constant entropy is.
我们能够很容易看出,恒定熵的条件下du/dV的数值。
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/T We've got Cv integral from T1 to T2, dT over T is equal to minus R from V1 to V2 dV over V.
左边是Cv乘以，从T1到T2对dT积分。
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