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Last week I had my thermodynamics exam, and one question asked me to say whether the following is correct, or incorrect:

"The change in entropy of a closed system is the same for any process between two specified states."

My answer was that this need not always be correct, and my reasoning was as follows:

When we observe the entropy balance for a closed system:

$$s_2-s_1=\int_1^2 \frac{\delta Q}{T} + \sigma$$

Even if there were no heat exchange with the environment of the system, the entropy production need not be equal to $0$, implying that it is therefore not always the case that the entropy between two specified states needs to be equal. The statement would be true if there were no internal reversibilities, but nothing was said about that in the question.

This is not approved, and the only argument they give is that entropy is a state quantity, and so this is valid. Can someone explain this to me in more detail?

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    $\begingroup$ I read my answer again, and I think I gave the right reasoning but the wrong conclusion. The change in entropy is indeed "constant.". Is this correct? $\endgroup$
    – Masterrun80
    Commented Jul 20, 2023 at 13:57
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    $\begingroup$ A state variable depends only on the state and not how the process took the system to that state. $\endgroup$
    – hyportnex
    Commented Jul 20, 2023 at 14:05
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    $\begingroup$ Your reasoning using the Clausius equation makes no sense to me. You are aware that the generation term varies with path between the specified end states,right? $\endgroup$
    – Chet Miller
    Commented Jul 20, 2023 at 14:38
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    $\begingroup$ Your reasoning is actually good, but you missed out the fact that the question specifically specified "between two specified states". This means that the question is asking if you understand that entropy is a good state variable, as opposed to asking about spontaneous entropy generation. In particular, yes, you may have irreversible generation of entropy, but in the situation the question specified, the excess entropy generated will be pushed out of the system before reaching the final state. $\endgroup$
    – naturallyInconsistent
    Commented Jul 20, 2023 at 14:45
  • $\begingroup$ Are you aware that the T in your equation is the temperature at the boundary between the system and surroundings through which the heat dQ flows? $\endgroup$
    – Chet Miller
    Commented Jul 20, 2023 at 15:04

1 Answer 1

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"The change in entropy of a closed system is the same for any process between two specified states.".

The statement is correct. The reason is entropy is a state function. That means a change in entropy of the system between the same two equilibrium states is independent of the path between the states. Note emphasis on “system” since entropy generated may be passed to the surroundings in the form of heat.

However, if there is no heat transfer with the surrounding the process is adiabatic and any entropy generated due to irreversible work will be retained in the system. In this special case a reversible adiabatic process and irreversible adiabatic process cannot connect the same two equilibrium states.

To calculate the entropy generated for an irreversible adiabatic process you need to assume any convenient reversible path between the two states, which will necessarily not be adiabatic, and apply the entropy definition. You can do this because the entropy change of the system is independent of the path.

Hope this helps

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