Does Entropy Change Depend on the Process? The entropy is a state variable, so the entropy change should not dependent on the path between the initial and final state. That's why we can calculate the entropy change of free expansion with an isothermal process. In chapter 20 of Halliday's Fundamental of Physics, it says that:

If a process occurs in a closed system, the entropy of the system increases for irreversible processes and remains constant for reversible processes. It never decreases.

It seems like the entropy change depends on whether the process is reversible or not. Why doesn't the statement violate the assumption of "entropy is a state variable"?
 A: Entropy $S$ is a state function.
However, the way its change is defined indeed depends on the process.
$$S(\mathrm B) - S(\mathrm A)\geqq \int_\mathrm A^\mathrm B \frac{đQ}{T},\tag {I}^\S$$ where the equality only arises for reversible processes i.e., \begin{align}S(\mathrm B)-S(\mathrm A) &=\int_{\mathrm A}^{\mathrm B} \frac{đQ_\textrm{rev}}{T}\tag{I.i}\\ S(\mathrm B)-S(\mathrm A) &\gt \int_{\mathrm A}^{\mathrm B} \frac{đQ_\textrm{irrev}}{T}\tag{I.ii}\end{align}
$\Delta S =S(\mathrm B)-S(\mathrm A) $ doesn't depend on the path and only on the two states $\mathrm A$ and $\mathrm B$ but it only equates with $\displaystyle \int\frac{đQ}{T}$ when the transformation is reversible.

It seems like the entropy change depends on whether the process is reversible or not.

Entropy is a state function and so it doesn't depend on the process connecting the two states. However, whether it equates with $\displaystyle \int\frac{đQ}{T}$ depends on the process.

Why doesn't the statement violate the assumption of "entropy is a state variable"?

For an isolated system, $đQ= 0$ which implies $\mathrm dS\geqq 0$, the equality arising for reversible transformation; it doesn't contradict the fact that $S$ is a state variable in any way.

$^\S$ It is to be noted that $T$ is actually $T_\textrm{reservoir/environment}$ which need not be equal to the temperature $T_\textrm{system}$ of the system or part of the system  which exchanges the thermal energy but when the transformation is reversible, then only$T_\textrm{reservoir/environment}= T_\textrm{system}$.
A: I think there is a hidden assumption in your reasoning that there are two paths (reversible and irreversible) from a state A to a state B. From state A and with reversible path, you can reach state B. But this doesn't mean that, with irreversible path, you still can reach state B. It would be some other state.The entropy change doesn't depend on path selected but depends on state. 
A: State functions, like entropy, depend only on the initial and final states. Mathematically, they are said to have an "exact" differential. Both these are equivalent to saying state functions are path-independent. So entropy change (not absolute entropy itself) is independent of the path taken, and the entropy change will be the same between two states no matter what process is used to transition between them.
However, as I understand it, path-independence is a separate issue from reversible/irreversible. Going from state A to state B, we can take any reversible path and a state function will yield the same value. But irreversible paths are fundamentally different from reversible paths, and should not be regarded as simply another path from A to B. Irreversible processes are a different class of processes than reversible/ideal ones. Taking an irreversible path from A to B, I would generally expect the state function to have a different value than even the "same" path done reversibly.
I am not an expert and I welcome any revisions.
