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So we can just write out what those are, minus w prime over q1 prime q1 is greater than minus w over q1.
因此可以写出，负w一撇除以q1一撇,大于负w除以。
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Wirr Minus w irreversible, this is the work which is done to the environment by the system, -Wrev minus w irreversible is always smaller than minus w reversible.
一点提示，不可逆过程中系统,对外界做的功，总是小于,之前我们。
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So w over q1 prime is greater than minus w over q1.
因此w除以q1一撇大于负w除以q1，等于。
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Q So it's just capital W over capital Q, q1 which is to say it's minus all that stuff over q1.
所以效率就是W除以，等于负的所有这些之和除以。
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Wrev So minus w reversible, 1 p2 is less than p1, so p2 over p1 is less than one, log of something less than 1 is negative times negative.
那么，而p2小于p1，p2/p1小于，取对数后是负数,负负得正。
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The efficiency for the one on the left q1 is minus w over q1, The efficiency prime for the one on the right is minus w prime over q1 prime.
左边的效率是,负w除以，右边的效率是,负w一撇除以q1一撇。
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w So that means that minus w prime must equal w And w is greater than zero.
也就是说负w一撇等于,在这之中。
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w That's u2 minus u1, and it's q plus w.
就是u2减去u1，等于q加。
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But we know that minus w prime w is the same thing as w.
我们知道,负w一撇等于。
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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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