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By the definition of an isolated system, $dE = dN = dV = 0$ (energy, no. of particles, and volume).

If we assume that a macrostate has a unique ($E$, $V$, $N$) then we can write $$dS = \frac1T\cdot dE + \frac PT\cdot dV - \frac{\mu}T\cdot dN.$$

Doesn't this imply that $dS = 0$ for all isolated systems? if so, how could the total entropy possibly change if a system is truly isolated?

Also, I understand that we should additionally require that the partial derivatives of $S$ with respect to $E$, $N$, $V$ also equal 0 for $S$ to be maximized (and hence in the equilibrium macrostate), but doesn't this mean that we have things like $P/T = 0$?

I'm getting quite confused following this line of reasoning. I'd be very grateful for any clarification.

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    $\begingroup$ If. you have two parts of a isolated system, each under different thermodynamic equilibrium conditions, but separated by a constraint, and you revoke the constraint, allowing them to equilibrate with one another. $\endgroup$ Aug 29 '21 at 12:04
  • $\begingroup$ Hello! I have edited your question using MathJax (LaTeX) math typesetting. For future questions, you can refer to MathJax basic tutorial and quick reference. Thanks! $\endgroup$
    – Jonas
    Aug 29 '21 at 12:30
  • $\begingroup$ Chet Miller - this was actually the problem that lead me to asking this question. If we have 2 systems that are isolated then allow them to exchange particles, then dS1 and dS2 (entropies of the 2 systems) will be non zero since dN1 and dN2 are non zero. However, the overall entropy change, dS would still be zero since the overall dN = 0. I know that this is wrong because there is an entropy increase due to mixing. My question is, how can I fix the line of reasoning that lead to me getting dS = 0. $\endgroup$
    – a h
    Aug 29 '21 at 15:26
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If we assume the whole system is isolated and in equilibrium, then it is in its maximum entropy state. Remember that isolated systems evolve until they reach equilibrium, and irreversible processes, like equilibration, cause the entropy of an isolated system to increase. Once the system reaches equilibrium, it stops evolving, and its entropy remains constant.

If the system is not in an equilibrium state, then it can evolve. An example of this could be a system partitioned into two chambers. Each chamber could have a different temperature or pressure or chemical potential compared to the other. Each part of the total system is locally in equilibrium, but the system as a whole is not. If we allow the two chambers to mix, exchanging energy, changing volume, and mixing particles, the system will evolve until a new equilibrium is reached, increasing its entropy.

Lets say the two subsystems have different temperatures, $T_A > T_B$. In order to equilibrate, energy $dE$ must flow from subsystem $A$ to subsystem $B$. $A$ loses energy ($-dE$), and $B$ gains energy ($+dE$). The total change in energy of the system is $$dE_\mathrm{tot} = -dE + dE = 0.$$ To find the total change in entropy of the system we must integrate $$dS_\mathrm{tot} = dS_A + dS_B = -\frac{1}{T_A}dE + \frac{1}{T_B}dE > 0.$$ To do so we might need to know about how the rate of energy exchange depends on temperature, the specific heats of the subsystems, etc. But the entropy change is not zero.

The entropy of $A$ decreases, while the entropy of $B$ increases by more, because $T_A > T_B$. So the entropy of the whole system increases.

Any isolated, nonequilibrium system will increase its entropy as it equilibrates.

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