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There's a reaction occurring at constant $T$ and $P$, and it is in chemical equilibrium. Now by adding infinitesimal quantities of reactants, it is taken to another equilibrium state in a reversible way. During this process, if entropy change is $\Delta S$ and volume change $\Delta V$, $$\Delta U = T \Delta S - P \Delta V.$$ If the same process is carried out irreversibly, by suddenly adding a large quantity of the reactants and disturbing the chemical equilibrium, then the work still remains $P \Delta V$ ($ \Delta V$, a state function, has the same value). In that case, the heat $q=T \Delta S$, which is only true for reversible processes. Even though the process is irreversible, there has been no entropy change in the universe? Where am I going wrong? It can only be that the work is not $P \Delta V$, but since the process is at constant pressure, the work should have this value.

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    $\begingroup$ Adding more reactants, reversibly or not, is changing the total number of moles in the system. That's something that your equation doesn't account for. $dU = T dS - P dV + \sum_i \mu_i dNi$ $\endgroup$ – Curt F. Jul 18 '15 at 3:33

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