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Philip Wood
Nov14-12, 11:45 AM
PF Gold
P: 956
Work done for irreversible process

I can't see the logic of your assertions. In the cases I quoted the external work done is zero. This is because there is no external pressure and no external force, as the gas is expanding into a vacuum (empty space). You can calculate pΔV in which p is the gas pressure, if the gas expands into a very small (initially empty) volume, ΔV. The result is non-zero. So pΔV ≠ work in this case.

It is an irreversible change, though I'd hesitate to say that dissipation is involved. That's the point I was trying to make.