# Clarification on the Clausius inequality

In Atkins' Physical Chemistry the criteria for spontaneity using Gibbs energy was calculated using the Clausius inequality:

$$\mathrm dS ≥ \frac{\mathrm dq}{T},$$

and at constant pressure, $$\mathrm dq = \mathrm dH.$$ Thus, $$\mathrm dH - T\,\mathrm dS$$ is negative.

So, if we define a function $$G = H - TS$$ at constant pressure and temperature, we get

$$\mathrm dG = \mathrm dH - T\,\mathrm dS ≤ 0,$$

which we use as a criteria for spontaneity.

My issue with this argument is that it implies that the Clausius inequality is for spontaneous processes only, otherwise we would end up with $$\mathrm dG ≤ 0$$ for all processes.

However, the derivation of the inequality simply used the fact that

$$|\mathrm dW_\mathrm{reversible}| > |\mathrm dW_\mathrm{irreversible}|$$

when the system does work.

So, where is it implied that the derivation is for spontaneous processes only?

• It's not the case that $\mathrm dS ≥ \frac{\mathrm dq}{T}$ is the criterion for spontaneity. Furthermore, $\mathrm dS ≥ \frac{\mathrm dq}{T}$ is not a statement of the Clausius inequality. The Clausius inequality states that $\mathrm dS > \frac{\mathrm dq}{T}$ for spontaneous processes. I.e., in the expression $\mathrm dS ≥ \frac{\mathrm dq}{T}$, the inequality sign applies to spontaneous processes, and the equals sign applies only for reversible processes. Oct 19 '19 at 18:39

For this situation, from the first law, we have $$\Delta U=q-w$$ so that, from the definition of Gibbs free energy, it follows that: $$\Delta G=q-w+p\Delta V-T\Delta S$$ Applying the Clausius inequality to this situation yields: $$\Delta S=\frac{q}{T}+\sigma$$where $$\sigma\geq0$$ is the entropy generated, with the = sign applying if the process is reversible and the > sign applying if the process is irreversible or spontaneous. If we combine the previous two equations, we obtain:$$w=-\Delta G+p\Delta V-T\sigma$$ So, for any process path between two specified end states of a system under these pressure and temperature constraints, the maximum work over and above p-V work is $$-\Delta G$$, and all other irreversible cases will involve less work (over and above $$p\Delta V$$).
Your question is how does the inequality $$|\mathrm dW_\mathrm{reversible}| > |\mathrm dW_\mathrm{irreversible}|$$ follow from more general statements of the Second Law (a "corollary" being the Clausius inequality). Or, how is that inequality a definition of spontaneity?
There are various established formulations of the Second Law. Clausius' statement is that a pump cannot transfer heat from a lower T into a higher T reservoir without external work being supplied. Such a process, in other words, is not spontaneous. While Clausius' statement seems obvious, it can be shown to be consistent with other statements of the Second Law, including Clausius' inequality: $$|\mathrm dW_\mathrm{reversible}| > |\mathrm dW_\mathrm{irreversible}|$$. The inequality means that the amount of work that can be performed by a heat engine from a heat input $$dq_{rev}$$ never exceeds the reversible work $$|\mathrm dW_\mathrm{reversible}|$$. A violation of this condition would mean that an irreversible engine requiring only a fraction of the reversible heat could be used to power the reversible engine as a heat pump, driving more heat from the cold into the warm reservoir than was sent in the opposite direction by the irreversible engine, and this would violate the Clausius statement, and thereby the Second Law.