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I seem to be able to contradict the second law of thermodynamics. What am I missing?

  1. $dS = \frac{\delta q}{T}$ along reversible paths
  2. $S$ is a potential: $\Delta S_{A \to B} = \int_\gamma\frac{\delta q}{T}$ for every reversible path $\gamma$ from $A$ to $B$ (if any)
  3. In an isolated system $\Delta S_{A \to B} \ge 0$, $= 0$ for a reversible path, $\gt 0$ for an irreversible path (by the second law of thermodynamics)
  4. But, if for a reversible path $\Delta S_{A \to B} = 0$ (by 3). then, by 2., $\Delta S_{A \to B} = 0$ for all paths (including irreversible ones), contradicting the second law.

The only week point that I can spot is the "if any" in 2. Are irreversible paths the ones between states which aren't connected by reversible paths? But then how do we define entropy change between such states?

There is a related question here: How can the entropy of the universe increase when undergoing an irreversible process?

However, I don't fully understand/accept the accepted answers as I'm wondering what happens if I just look at the universe as a whole and not split it into a system and surrounding.

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  • $\begingroup$ I have a suspicion that the answer to my question will tend towards: either the universe is in a thermal equlibrium at which point its entropy will be at its maximum and only reversible processes will occur or it is not in which case it doesn't make sense to talk of "the entropy of the universe" as entropy can only be defined for a system at equilibrium. $\endgroup$
    – Damian Birchler
    Commented Sep 3, 2021 at 22:07
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    $\begingroup$ I think the idea is that in an isolated system, there is no reversible path between two states that are joined by an irreversible path. The only way to reversibly join the two states is to allow the system to interact with an environment of some sort, in which case the entropy of the system and environment both change (in opposite directions). The way to define the entropy change between the two states joined by the irreversible path is exactly to compute the change in entropy along the reversible path traversed while interacting with the environment. $\endgroup$
    – march
    Commented Sep 3, 2021 at 22:11
  • $\begingroup$ In response to my own comment: The second law does talk of the "entropy of an isolated system". However, there it is simply defined as the sum of entropies of the subsystems, which must themselves be in thermal equilibrium. Correct? $\endgroup$
    – Damian Birchler
    Commented Sep 3, 2021 at 22:14
  • $\begingroup$ Each subsystem must be in a state of thermal equilibrium to have a well-defined thermodynamic entropy, but the subsystems don't need to be in thermal equilibrium with each other. They can then exchange energy, and if this exchange is, for instance, heat across a finite temperature difference, the net change in entropy of the two subsystems will be positive. $\endgroup$
    – march
    Commented Sep 3, 2021 at 22:17
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    $\begingroup$ To expand on what @march said, to get the entropy change of an isolated system that has experienced an irreversible process between two thermodynamic end states, the alternative reversible path you choose does not have to treat the system as isolated. You just determine the entropy change for the system when it experiences this alternative non-isolated path. This is what we mean when we say $\Delta S=\int{\frac{dQ_{rev}}{T}}$. And it give you the entropy change for the irreversible isolated path. $\endgroup$
    – Chet Miller
    Commented Sep 3, 2021 at 23:03

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