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If we have a system that goes from equilibrium state A to B, since entropy is a state function then for the entropy change we would have:

$\Delta S = S_B - S_A$

Since entropy is a state function it is important the end state and the initial state. It doesn't matter whether the process to bring the system from A to B is rev. or irrev.

But if the process is rev. Then for rev. processes we know that $\Delta S = 0$.

And for irrev. processes $\Delta S > 0$.

So how exactly can entropy change be zero since the process is reversible but at the same time because the system goes to another state, whose multiplicity is different then the one of the previous state, the entropy changes. But if the same process from A to B is irreversible now the entropy is bigger?

In the following link: Why is the work done in reversible process greater than work done in irreversible process?

Bob D says:

"To elaborate, since entropy is a property of the system, the difference in entropy between two equilibrium states is the same, regardless of the process (reversible or irreversible) that connects the two states. Consequently, in an irreversible process the additional entropy generated in the system must be transferred to the surroundings in order for the change in entropy of the system to be the same."

His statement indicates that regardless of the type of process the entropy change is the same (as it should). But as I explained above, for each case entropy behaves differently.

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In a reversible process, the change in entropy of the system plus surroundings is zero, but not necessarily for each of them individually. The entropy change of the system can be positive, negative, or zero. But, for a reversible path, it will be equal to the entropy change for any irreversible path between the same two end states.

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