Delta v is positive in this direction here, the work is negative. So work on the system is positive. Work done by the system is negative.
功是负的,总而言之,外界对系统做功的符号为正。
This piston is being brought out, so we expect 0 the work to be negative, negative. And we start o V2 ut with zero volume. We end up with a volume p2 of V2, and the external pressure is constant to p2.
所以我们可以想象功是负的,开始的时候体积是,最终的容积是,外界的压力恒为。
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。
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。
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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