Work done for irreversible process
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Nov14-12, 06:29 PM
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
is the gas pressure, if the gas expands into a very small (initially empty) volume, Δ
. The result is non-zero. So
≠ 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.
I know in free expansion , there will be no external pressure so no work done, I don't understand why do you want to use gas pressure times ΔV. So do you mean that PΔV is the dissipative work?