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P dV is equal to R dT. pV = RT for 1 mole, pdV=RdT so I just take dV here.
对等压过程，那么。
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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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Then the present value of $1-- The present PDV or PV of $1 = $1/.
一美元的现值...,一美元的现期贴现值或者现值就等于1/
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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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dq=0 dw=-pdV I wrote dq equals zero I wrote what dw was.
这样的式子。
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So it's, this is just the integral pdv And it's an ideal gas, isothermal, right.
从一点到二点的,的积分。，from，one，to，two，of，p，dV。,这是理想气体，恒温过程，好的。
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dU=-pdV So du is just minus p dV.
于是。
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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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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, du plus p dV is less than zero.
因此，du加上pdV需要小于零。
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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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V2 p1 times V2 minus V1. What that turns out to be, 0 this area right here. It's V1 minus V2 times p1.
第二步做的功是-pdV，积分从V2到，这部分积分出来是多少？应该是。
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So, the work for this process is the integral, pdv or minus the integral, V1, V2, p dV.
此过程中的做功等于负的积分，从v1到v2，外部。
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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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Vdp So dH is just du plus p dV plus V dp.
所以dH等于du加上pdV再加上。
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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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V1 This first integral is zero V1 to V1, then I get minus p2 times V2 minus V1 or p2 times V1 minus V2. Again, 0 a positive number.
先保持体积为，对pdV积分1,然后对p2dV从V1到V2积分，第一部分积分为。
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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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That's equivalent to doing the integral, and so, what we end up getting is that the reversible work v2 pdv is equal to minus integral V1, V2, p dV.
这与刚才的积分过程效果相同，最后，我们得到的结论是可逆过程的功,是负的积分，从v1到。
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U It's u, because u is to q plus w right, heat and work, but it's adiabatic. So there's no heat, exchange with the environment, and it's constant volume, so there's no p dV work, right.
什么是零？是U，因为,等于q加w，热量和功，但这是绝热的，所以系统与环境间没有热量交换；,同时它是灯体的，所以也没有pdV形式的功。
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Then pressure is constant.
代入pdV项中。
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