# Helmholtz Free Energy Minimization Proof

In the proof that Helmholtz free energy is minimized at equilibrium, we use the fact that $$\mathrm dS_\text{total}=\mathrm dS_\text{system}+\mathrm dS_\text{bath}\geq 0$$
But we also use the fact that $$\mathrm dS_\text{bath}=\frac{\mathrm dU_\text{bath}}T=-\frac{\mathrm dU_\text{system}}T$$ Why does this not imply that $$\mathrm dS_\text{system}=\frac{\mathrm dU_\text{system}}T$$ and therefore $$\mathrm dS_\text{total}=0$$?
Is this because only the bath is always at temperature $$T$$, so we can imagine a reversible path, whereas the system may have some temperature fluctuations during the process of heat exchange with the bath?

• I went through this for dG here: chemistry.stackexchange.com/questions/124412/… The equiv. argument applies for dA. But the essence of it is we treat the bath as infinite, so all heat flow into it is rev., and thus dS for the bath can be calculated directly from the actual heat flow. But that's not the case for the sys. There, to calculate dS, you need to find a rev. path (where both T and q can be different from the actual T and q). Hence, for a spontaneous process, dS_sys + dS_bath > 0. Oct 17 '20 at 6:13

Who says that $$dS_{bath}=\frac{dU_{bath}}{T}=-\frac{dU_{system}}{T}$$? The equation should correctly read $$\Delta S_{bath}=\frac{\Delta U_{bath}}{T_{bath}}=\frac{Q_{bath}}{T_{bath}}=-\frac{Q_{syst}}{T_{bath}}$$, where the Q's are for the irreversible path. The Q of the system for the reversible path is different from that for the irreversible path because of both temperature gradients and viscous dissipation of mechanical energy to internal energy during the irreversible path (both of which generate entropy). For the reversible path of the system, these entropy generation effects are not present.