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And for a couple of reasons, one this is the first P set where we're actually gonna give you code to work from.
由于一些原因，这里是第一个习题集,在此我们提供了一些代码供大家参考。
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Well, presumably because whatever cognitive structures it takes in your brain to underwrite the ability to P-function, those cognitive structures have been broken, so they no longer work.
大概是因为,无论你大脑中的什么认知结构,使得人格功能性成为可能，这种认知结构都被破坏了，它们不再继续工作。
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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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So, it's very easy to calculate, however, the number of radial nodes, and this works not just for s orbitals, but also for p orbitals, or d orbitals, or whatever kind of work of orbitals you want to discuss.
径向节点，的数量，这不仅对s轨道适用，对p轨道，d轨道，或者任何你们想讨论的轨道，都是适用的,它就等于。
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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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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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And it's p external and p external is zero, so there's no work.
外部的p是零，所以结果是零，没有做功。
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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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V Minus p2, V2. Minus p delta V. So the total work is the work from the left hand side plus the work on the right hand side, p1 V1- p2 V2 which is p1 V1 minus p2 V2.
负p2V2【w=-p2v2】，即负pΔ,所以全部的功就是,左边的功加上右边的功，等于。
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