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"?

  • $\begingroup$ Note that in thermodynamics a closed system means no matter can enter or leave but heat is allowed to enter or leave. $\endgroup$ – John Rennie Dec 15 '16 at 16:32
  • $\begingroup$ I'm not sure @JohnRennie, whether Halliday actually meant closed system what we generally mean by a closed system; I assume he is actually talking about isolated system. $\endgroup$ – user36790 Dec 15 '16 at 17:30
  • $\begingroup$ Yes. I agree with MAFIA36790. Some physicists use the term closed system to describe system which, in thermodynamics, is called an isolated system. In thermodynamics, a closed system allows heat to enter and leave the system, but not mass, while an isolated system allows neither heat nor mass to enter of leave the system. $\endgroup$ – Chet Miller Dec 16 '16 at 4:19

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$T_\textrm{reservoir/environment}= T_\textrm{system}$.

  • $\begingroup$ I have a minor correction. In equation I, the subscript of T should be boundary, not system:$$S(B)-S(A) \geq \int_A^B{\frac{dQ}{T_{boundary}}}$$where $T_{boundary}$ is the temperature on the portion of the system boundary where the heat transfer dQ occurs. For a reversible process path, where the equality sign applies, the system temperature T is uniform, and equal to the boundary temperature T_{boundary}. $\endgroup$ – Chet Miller Dec 16 '16 at 4:26
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    $\begingroup$ Yes, I know @ChesterMiller, it follows from the Clausius' Inequality that $T$ is of the reservoir/environment and $T_\textrm{rev}=T_\textrm{sys}$ only when the transformation is reversible. I'd correct it. Poor typo due to copy-pasting; sorry for that. $\endgroup$ – user36790 Dec 16 '16 at 4:32

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.

  • $\begingroup$ +1 In fact if entropy change when going from state A to state B is zero, then in a isolated system (which seems to be the same as closed system, OP must clarify) there cannot be an irreversible path that goes from A to B. So for an isolated system, set of initial and final states is different for reversible and irreversible processes. $\endgroup$ – Deep Dec 16 '16 at 5:23

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.

  • $\begingroup$ If irreversible processes are different from reversible process, how can we calculate the irreversible free expansion process with reversible isothermal process? $\endgroup$ – LY3000 Dec 15 '16 at 17:25
  • $\begingroup$ I'll bet you that you can't, you are only calculating an approximation of the irreversible process. Experimentally, we go the other direction. We perform an irreversible process (like any real-world process), and if is done fast enough, etc, then it is a good approximation of the reversible process. $\endgroup$ – electronpusher Dec 15 '16 at 17:35

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