Work done for adiabatic reversible process is $$(P_1V_1 - P_2V_2)/(γ-1)$$ but this is also the work done for adiabatic irreversible process. How?
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1$\begingroup$ Your question is not very clear. Why do you say that the work done in an adiabatic process is irreversible? $\endgroup$– By SymmetryCommented Feb 7, 2018 at 13:15
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$\begingroup$ This is the work done for a process that is both adiabatic and reversible. $\endgroup$– Chet MillerCommented Feb 7, 2018 at 13:48
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$\begingroup$ I changed the question. Can you please answer it now? $\endgroup$– Anuraag ReddyCommented Feb 10, 2018 at 6:06
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1$\begingroup$ As Chestermiller and I have both pointed out, the quantity of work you quote is for a REVERSIBLE adiabatic expansion of an ideal gas. What is your problem with this? $\endgroup$– Philip WoodCommented Feb 10, 2018 at 20:01
2 Answers
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.
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$\begingroup$ Please change the answer as i changed the question $\endgroup$ Commented Feb 10, 2018 at 17:33
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$\begingroup$ My answer above still stands. I don't want to seem rude, but you don't seem to have read it. I re-iterate: $\frac{p_1 V_1-p_2 V_2}{\gamma -1}$ is the work done by an ideal gas expanding reversibly. The work done when it expands $irreversibly$ between the same two states is less. $\endgroup$ Commented Feb 10, 2018 at 20:42
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$\begingroup$ I should have said "between the same two volumes" rather than "between the same two states". $\endgroup$ Commented Feb 11, 2018 at 10:45
As long as the initial and final states are equilibrium states, the work done between them in an ADIABATIC process (dQ = 0) is simply dW = -dU, where U is the internal energy of the system. Since dU depends only on the initial and final states, not on the path or process connecting them, miraculously the work dW for adiabatic processes depends only on the initial and final states and can be calculated from the change of the internal energy.
For ideal gas, -dU = nCv(T1-T2) = (P1V1 - P2V2)/(gamma - 1), so you can use either formula because they are the same formula.
Of course, the tricky part is to get the final state in the irreversible adiabatic process, since there are no nice formulas for those. If we start from the same initial state (P1, V1), the final state (P2, V2) will be different for a reversible and an irreversible processes. That will lead to different work dW = - dU. However, if you have info about the final state in an irreversible process, no matter how complicated it was, you can still calculate dW = - dU if you know a formula for the internal energy of the system.