[Physics] Work done in adiabatic process irreversible and reversible

Question

Work done for adiabatic reversible process is $(P1V1 – P2V2)/(γ-1)$ but this is also the work done for adiabatic irreversible process. How?

Answer 1

You quote the case of reversible adiabatic expansion, where the gas (in an insulated cylinder of low heat capacity) expands slowly, pushing a piston and doing the amount of work that you quoted.

An extreme example of an irreversible adiabatic process is called Joule expansion. An insulated vessel is partitioned into two chambers. One has a gas in it, the other a vacuum. The partition breaks, so that the gas expands into a vacuum. We have irreversible adiabatic expansion, and the gas does no work at all!

A less extreme example of irreversible adiabatic expansion would be when the gas expands very rapidly. If the piston's speed is not negligible compared with the rms speed of the gas molecules, the gas pressure next to the piston will be less than that in the bulk of the gas, so the work done will be less than in the reversible case.