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At thermodynamic equilibrium, there is thermal, mechanical, and diffusive equilibrium. Does this imply:

$$d\mu = dT = dV = 0$$

$$dU = TdS - PdV \implies dU = TdS$$

Here, I know entropy is maximum, so perhaps $dS=0$ and hence $dU = 0$? I also don't think I can write $d\mu = 0$ as this may be sort of abusive treating of the chemical potential.

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    $\begingroup$ If you are just sitting at equillibrium it would be surprising for anything to be changing, so $dX = 0$, for all $X$. None zero differentials impy change, which implies some sort of process is taking place. $\endgroup$
    – By Symmetry
    Commented May 2, 2019 at 12:04

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A system is in thermodynamic equilibrium when there are no changes in the macroscopic properties (internal energy, entropy, temperature, etc.) of the system.

So yes, at thermodynamic equilibrium $\Delta U=0$.

Hope this helps.

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