Work done in cyclic process [closed]








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I have been told that when $\Delta V=0$, work done will be zero $(W=0)$. I know that, in cyclic process $\Delta V$ will be zero, since it's a path function and $\mid W \mid$ i.e. magnitude of work done will be the area enclosed by the curve. But my doubt is when $\Delta V=0$, shouldn't $W$ also be zero in such processes?




The reversible work is $W=\int{PdV}$ over the path of the process. This integral is not zero for the cyclic process (involving V changing) even through $\Delta V$ from beginning to end of the process is zero. So, whoever told you that $W = 0$ when $\Delta V=0$ is wrong.


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I'm not sure about my answer but I think this is how it goes.


$\Delta G=-W$ (useful work);


$= \Delta H-T \Delta S$


$= P + \Delta V -T \Delta S$
Therefore even though $\Delta V$ is zero it does not imply that all the other terms are zero too. Therefore $W$ is not zero.


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