Work done for irreversible process
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Nov14-12, 10:32 AM
Yet the expanding gas does no external work, as no force resists its expansion. [If the expansion is into a large extra evacuated volume, we can't even
, as the change is non-quasi-static and pressure does not change with volume in a defined way.]
Since we can calculate the work done ,that means there must be external pressure, so how you say no force?
So do you mean we can't calculate the work done for irreversible process?