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If I look for, say, minus 1, you might go, gee, wait a minute, if I was just doing linear search, I would've known right away that minus one wasn't in this list, because it's sorted and it's smaller than the first elements.
如果我要查找-1，你可能要怒了，呵呵，等一等，如果我用的是线性查找，我不会知道-1不在这个列表中,但是列表是排好序的,1又比第一个元素小。
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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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So this is 3, plus 2 times 3, plus 4 t minus 2.
也就是3+2*3+4t，好。
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Now, I don't have to fill my head 6 with 2.18 times 10 to the minus 18. 13.6, that's a cool number. I can remember that.
现在，我不需要记住,2。18乘10到负18，13。,那数字太给力了，我可以记住它。
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This has a minus and that doesn't have a minus.
这里有一个负号而那里没有
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- Like, it doesn't matter, like, you have minus 32 even though -- >> David: F minus 32?
如果你只是减去32，这是没有关系的-,>>，大卫：F减去32？
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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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So instead of v bar, we write p v bar minus b, equal r t.
现在考虑，这些气体分子之间。
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A It tells me that the partial of A with respect to T at constant V is minus S. Right?
他告诉我们,在恒定体积下对温度的微分等于负S，对吗？
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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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If I reduce that it would be 3 plus t of b minus 3, so in general 3*k+t this is 3 k plus t of b minus k. OK.
把b减去一个,在外面加个3就可以了,因此也就是。
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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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I'm using the fact that when you take a cosine and change the angle inside the cosine, it doesn't care. whereas, if you go to the sine and change the angle inside the sine, it becomes minus sine.
我用了一个结论,当你使用余弦函数时,改变角度的正负,函数值不变,而对于正弦函数,改变角度的符号,则函数值也会变号
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V So it's minus T dV/dT at constant p, plus V.
负的T乘以恒定压强下dV/dT，再加上。
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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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*t t of n minus 1 is 3 plus 2 t of n minus 2.
加上。
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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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How do I rea-- replace the expression FOR t of n minus 1? Substitute it in again.
我们怎么来代替t这个表达式呢？,再来替换掉它，t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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/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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In B it isn't, it's Cv times T2 minus T1, right.
在B中不是，而是Cv乘以。
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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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RT/V this expression becomes Cv dT over T is equal to CvdT/T=-RdV/V minus R dV over V.
这样，or，RT，over，V，bar。，So，now，这个等式就,可以化成。
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In particular, how would I write an expression for t of b minus 1?
尤其是,我是怎么写出来t的表达式的？
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pdV So, du is T dS minus p dV.
即du等于TdS减去。
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This is 3 plus 3 plus t of b minus 2. Right?
+3+t对不对？
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On the next step though, this, we get substituted by that. Right, on the next step, I'm back in the even case, it's going to take six more steps, plus t of b minus 1. Oops, sorry about that, over 2.
这一步就是偶数了，这一步会让我们得到，6+t这样的结果，因为b-1现在是偶数了,别忽略这里的细节。
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r+ And let's say that sodium has a radius, r plus, r- and chlorine has a radius, r minus, when r is very large in comparison to the radii of the ions, I don't need to draw them this way.
让我们假设钠有半径，是，氯也有半径，是，当r比离子半径大很多的时候，我不需要这样来描述。
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