Reversible vs irreversible work for adiabatic process

AI Thread Summary
In an adiabatic process involving a gas transitioning from state A to B and back to A, the net work done is questioned due to differing values of dE for reversible and irreversible steps. The key issue is that irreversibly compressing at constant pressure results in a final volume greater than the initial, preventing a return to the original state. The temperature increase during this irreversible compression is attributed to friction and viscous dissipation. The discussion highlights the necessity of reversible processes to achieve the initial state adiabatically. The problem posed is recognized as an interesting challenge in thermodynamics.
Andrew U
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I have a gas transitioning adiabatically between A (P1, V1) and B (P2, V2) where P1>P2 and V2>V1. The question is to determine the net work done on the gas if the gas is first expanded reversibly from A to B (w = dE = Cv(T2-T1)), and then compressed irreversibly from B to A (w = -Pext(V1-V2)) at a constant external pressure defined by A. In this scenario, simply looking at the areas under the graphs the net work should be positive.I am trying to reconcile this with dE for the gas. For the roundtrip transition (A to B to A), dE = 0. And if we take each step as adiabatic, then dE = w for each step, but as I have described above you would end up with two different values for dE for each step, thus dE not equal to zero. My logic is flawed somewhere, can someone help?
 
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Is this a problem you invented yourself, or is it out of a book?

Chet
 
The flaw in the logic is this: If you try to recompress irreversibly at constant P1, you will re-equilibrate at a volume greater than V1 before ever reaching V1. So, you won't be able to reach the initial state in this adiabatic irreversible recompression. You can calculate what final volume and final temperature the system reaches by irreversibly recompressing at P1. The only way you can back to the initial state adiabatically is if you do it reversibly.
 
Chestermiller said:
The flaw in the logic is this: If you try to recompress irreversibly at constant P1, you will re-equilibrate at a volume greater than V1 before ever reaching V1. So, you won't be able to reach the initial state in this adiabatic irreversible recompression. You can calculate what final volume and final temperature the system reaches by irreversibly recompressing at P1. The only way you can back to the initial state adiabatically is if you do it reversibly.
Hi Chet, thanks! I invented this one (and obviously missed an important point). So if I end up at a larger volume, then my temperature must also be higher, correct? Is this because of friction during the irreversible compression?
 
Andrew U said:
Hi Chet, thanks! I invented this one (and obviously missed an important point). So if I end up at a larger volume, then my temperature must also be higher, correct? Is this because of friction during the irreversible compression?
Clarification: When I ask about the increase in temperature, I am referring to an additional increase that is larger than one would normally expect for an adiabatic compression, thus leading to a larger final volume.
 
Andrew U said:
Hi Chet, thanks! I invented this one (and obviously missed an important point). So if I end up at a larger volume, then my temperature must also be higher, correct?
Yes.
Is this because of friction during the irreversible compression?
Yes. Viscous dissipation.

Don't despair. You invented a very interesting problem. I've seen other versions of this before.

For more details in irreversible vs reversible processes, see my Physics Forums Insights article: https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/
 

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