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In thermodynamics of gases one usually writes $dE = \delta Q+\delta W$ and then say that along a reversible path $\delta Q_R=TdS$ and $\delta W_R=-PdV$. One also has $\delta Q\leq \delta Q_R=TdS$ (due to Clausius theorem) and $\delta W\leq \delta W_R=-pdV $ (due to dissipative forces).

However in a system allowing the exchange of particles, one modifies the differential of energy to $dE=TdS -pdV + \mu dN$. My question is

Is it correct to call $\mu dN=\delta Y_R$ where $Y$ is, say, chemical work? If so, can one have an irreversible chemical work? Can you give me an example with irreversible chemical work? Also is it also true that $\delta Y\leq \mu dN$ with equality only when the process is reversible?

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  • $\begingroup$ To add to the question, reversible chemistry sounds like a kinetic equation in a reversible reaction where steady-state is useful. Loss of particles through a hole might be considered irreversible? $\endgroup$ – xxyzzy Sep 19 '17 at 19:18
  • $\begingroup$ The $\mu dN$ term is part of the reversible work term, i.e. $\delta W_R = -pdV+\mu dN$ $\endgroup$ – By Symmetry Sep 19 '17 at 20:49

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