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From the definition of entropy change,

$$S_2-S_1=\left ( \int_{1}^{2} \frac{\delta Q}{T}\right )_{int.rev}$$

From the closed system entropy balance, we have

$$S_2-S_1=\left ( \int_{1}^{2} \frac{\delta Q}{T}\right )_{b}+\sigma $$ where $\sigma$ is the entropy produced within the system, vanishing to zero in the absence of irreversibilities. I don't quite understand how the entropy change between two states is the same for all processes. Is it the case that the entropy transfer in the case of internal irreversibilities present is lower than that of the entropy transfer in an internally reversible process between these same two states, and the entropy production makes up for this difference?

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    $\begingroup$ Yes, that is exactly correct. In both cases, the heat transfer takes place at the boundary temperature. What you have stated is the essence of the Clausius inequality. $\endgroup$ Feb 15 at 4:22
  • $\begingroup$ @ChetMiller Thanks! $\endgroup$ Feb 15 at 6:31
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    $\begingroup$ Seems the post and commentaries could be restructured and gain a lot from a more elaborate answer. Otherwise this will just "hang" in perpetuity as an unanswered "yes/no" question. $\endgroup$
    – Buck Thorn
    Feb 15 at 7:40
  • $\begingroup$ @BuckThorn How should I reconstruct it? I would love someone to expand on it! $\endgroup$ Feb 15 at 19:34
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    $\begingroup$ Well, I meant that it would be nice to have an open-shut post with a clean answer.@ChetMiller evidently knows his stuff but aimed for conciseness and answered your "yes/no" question with a "yes, you're right" comment.. I think if he could expand on the answer it would be nice, but ultimately the fate of the post is up to him, you and the wisdom of the crowds. $\endgroup$
    – Buck Thorn
    Feb 15 at 19:58
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When a system changes from state 1 to state 2, the change in entropy of the system is the same for all processes, because entropy is a state function.

The difference between a reversible and an irreversible process is that, in the latter case, there is net entropy production for the universe, i.e., for the system+surroundings.

Since the entropy change for the system is independent of the process, the difference in entropy production between rev. and irrev. processes is seen in the entropy change of the surroundings.

More broadly, since path differences (which in turn means differences in heat flow and/or work flow) have no effect on the final state of the system, they instead always manifest themselves as differences in the final state of the surroundings.

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  • $\begingroup$ Makes perfect sense, thank you. I also have another question. Does entropy vary within different areas of a system? Is there like an entropy gradient of some sort? $\endgroup$ Feb 15 at 20:15
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    $\begingroup$ @CalebWilliamsUIC You'll want to post that as a separate question. $\endgroup$
    – theorist
    Feb 15 at 20:20
  • $\begingroup$ Alright! I will make another question at some point today. $\endgroup$ Feb 15 at 21:02
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The way that you describe it is exactly correct. This is the essence of the Clausius Inequality.

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