And so, we can rewrite this as the work nRTln is equal to minus nRT log p1 over p2, nRTln or nRT log p2 over p1.

因此我们可以把这个式子改写一下,功等于负的，或者。

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If you go and work that out, log2 log of two and you have two log two, by math you can show that for the line r series m 2*log2 is equal to two natural log of two, which is 1 1.386 which is greater than one.

如果你把它计算出来，这二者的对数，你会得到2个，通过数学，你会得到第r列，第m个值,等于,也就是1。386，这大于。

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