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And they're both Ds,so . "Stephanie Rafaella.
VOA: special.2009.07.06
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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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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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Some of you, unfortunately, won't do very good jobs at the beginning... and my TAs, I'll encourage them to be prepared to give Ds.
其中有些人，很遗憾，在开始时表现不会很好。,我会鼓励我的助教们做好给D的准备。
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p Well, it's not just p dS/dV because there's some dS/dV at constant T.
它不是简单的,因为式子中还包含，恒定温度下的。
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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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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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One thing that makes it pretty clear is that certainly delta S or the sign of it alone is not dictating the outcome here.
这个例子非常清楚的说明,熵增dS或者它的符号,不能单独决定最后的结果。
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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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So dS for u and V fixed is greater than zero.
所以当内能u和体积V固定时，dS大于零。
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It's nR log of p2 over p1 for the process where there's a pressure change.
结果是dS等于nR乘以p2除以p1的对数,这是对压强变化的结果。
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if it's a reversible process then the equality holds, but if it's irreversible, which means it happens spontaneously, T then dS is greater than this.
对于可逆过程，等号成立，但是如果是不可逆过程，这些不可逆过程是自发进行的，那么dS大于dq除以。
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This one, which we already had seen, which is dS, is greater than zero.
这个是我们已经看过的，即熵的增加dS大于零。
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So, the dS term is zero, but the other two are not.
这时dS等于零，但是其他两个不是。
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T So we know that T dS/dT at constant volume is Cv over T, T and dS/dT at constant pressure is Cp, over T.
在恒定压强下定压比热容Cp乘以dT除以,所以在恒定体积下dS/dT等于Cv除以，在恒定压强下dS/dT等于Cp除以。
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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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We saw that in general dS is greater than T or equal to dq over T.
我们发现一般来说dS大于,或者等于dq除以。
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SdT So dG is dH minus T dS minus S dT.
所以dG等于dH减去TdS再减去。
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So then, just like we saw, analogous to what saw just before, dS/dp it's T dS/dp at constant T.
就像我们看到的,就像我们刚才看到的一样，结果是T乘以恒定温度下的。
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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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SdT So we have dA is minus S dT minus T dS.
我们得到dA等于负TdS减去。
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pdV So, du is T dS minus p dV.
即du等于TdS减去。
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T Namely dS is greater than dq over T.
换句话说就是dS大于dq除以。
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dA/dT dS/dV So this is negative dS/dV.
是负S，It’s，negative，S。,这个二阶偏导数是负的恒定温度下的。
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So all that's left is negative T dS is less than zero.
只剩下TdS小于零。
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We'll substitute that in, and the T dS terms are going to cancel.
将其代入，消去TdS项。
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T That dS is greater than dq over T.
对吗，熵的变化dS大于热量dq除以温度。
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This time, the T dS terms are going to cancel.
这次TdS会被消去。
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