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Andrew Mason
#3
Oct4-11, 12:44 AM
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Quote Quote by Confusus View Post
Now for the irreversible compression, with same initial and final system states.
If they had the same initial and final states there would be no difference in the entropy calculation. So there has to be a difference between the final state of the irreversible process and the final state of the reversible process.

As an example, consider a quasi-static reversible adiabatic expansion of an ideal gas from Vi to Vf and an irreversible adiabatic expansion (let's say a free expansion - no work done - to make it simple) from Vi to Vf. The free expansion results in no change in internal energy/temperature, since no work is done. The reversible expansion does work so the internal energy/temperature decreases.

To calculate the change in entropy of the irreversible free expansion, you have to find the integral of dQ/T over the reversible path between the initial and final states. That would be an isothermal reversible expansion in which there is heat flow into the gas (dQ>0), so the integral of dQ/T over that path is > 0.

AM