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Entropy change of a system is a state function... But if we consider reversible adiabatic process, the entropy change of the system is zero, whereas the entropy of the system is greater than zero in irreversible adiabatic process. Shouldn't the entropy of the system in both cases must be same?

This answer: How would one calculate the entropy change for an adiabatic irreversible process? deals with how to find the the entropy of the system but it doesn't clear how can we say entropy is state function in case of irreversible adiabatic process.

Natasha J
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  • When a system goes from state A to B once reversibly, the other time irreversibly, the system entropy change is the same. What is not the same is the state change and entropy change of the neighborhood. – Poutnik Dec 16 '23 at 11:51

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It is imposible to design s reversible Adiabatic Process between the same two end states as an irreversible Adiabatic Process. The reversible path cannot be adiabatic.

To learn how to determine the change in entropy for an irreversible Adiabatic Process (or any other irreversible process), see the following link: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Chet Miller
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  • By other words, endpoints of both adiabatic process paths have to be connected by another process, typically isobaric or isochoric one. – Poutnik Dec 16 '23 at 12:05
  • @Poutnik Those would be two possible options for the reversible path.. – Chet Miller Dec 16 '23 at 12:19
  • That would be a different case. I mean irreversible adiabatic compression would lead to higher T and p at the final volume. Then followed by isochoric reversible cooling would lead to the final state of reversible adiabatic compression. As the result, extra work of neighborhood would be converted to thermal energy of neighborhood. – Poutnik Dec 16 '23 at 12:44
  • @ChetMiller, so entropy change of system in case of irreversible system is non-zero. But what about entropy change in reversible adiabatic process? – Natasha J Dec 16 '23 at 12:57
  • @Natasha There is no reversible adibatic path between the same two end states as for an irreversible adibatic process. – Chet Miller Dec 16 '23 at 13:10
  • @Poutnik You can always inset reversible segments into a reversible path, including even Carnot cycles. There are an infinite number of reversible paths between two end states; all have the same entropy change. – Chet Miller Dec 16 '23 at 13:15
  • That is clear, but not my case. I mean joining both endpoints of reversible path and partially irreversible path. Both with the same entropy change for the system, but not the same for neighborhood. – Poutnik Dec 16 '23 at 14:07
  • @Poutnik I don’t follow. – Chet Miller Dec 16 '23 at 14:36
  • Path1: Reversible adiabatic compresion p1,T1,V1->p2,T2,V2. // Path2: Irreversible adiabatic compression(p_ext>p_int or friction or..) p1,T1,V1->p3,T3,V2 where p3,T3 > p2, V2 + reversible isochoric cooling p3,T3,V2->p2,T2,V2. Entropy change for the system is the same for both paths. Entropy change for the neighborhood is the same for both paths. – Poutnik Dec 16 '23 at 15:27
  • I agree with this, but don’t understand what “entropy change for neighborhood" means. – Chet Miller Dec 16 '23 at 16:03
  • Rather "of neighborhood". E.g. that isochoric cooling passes heat to neighbourhood. // $\Delta S_\text{irrev,tot} = \Delta S_\text{irrev,sys} + \Delta S_\text{irrev,nbhd} \gt \Delta S_\text{rev,sys} + \Delta S_\text{rev,nbhd} = \Delta S_\text{rev,tot}$, while $\Delta S_\text{rev,sys} = \Delta S_\text{irrev,sys}$ and $\Delta S_\text{irrev,nbhd} \gt \Delta S_\text{rev,nbhd}$ – Poutnik Dec 16 '23 at 16:37
  • Let us continue this discussion in chat. – Poutnik Dec 16 '23 at 16:40