# Is the statement $dF=(\delta Q-TdS)-pdV-SdT$ more general than $dF=-SdT-pdV$?

So the Helmholtz free energy is defined as $$F=U-TS$$, which means $$dF=\delta U-TdS-SdT$$ $$=(\delta Q-pdV)-TdS-SdT$$ $$=(\delta Q-TdS)-pdV-SdT.$$ since $$\delta Q \neq TdS$$ ($$dS \geq \delta Q/T$$) in general, it means tha $$dF=-pdV-SdT$$ only holds when the process is reversible?

• No. The definition of F is U-TS. so dF = -SdT-PdV, where these equations refer to thermodynamic equilibrium states. Apr 7, 2022 at 14:56
• @Chet Miller how is the $\delta Q- TdS$ part of the problem explained, is it not to do with equilibrium states? Apr 7, 2022 at 15:01
• Does this answer your question? Fundamental thermodynamic relation and irreversible processes Apr 7, 2022 at 15:46

Consider any irreversible process in which the initial and final thermodynamic equilibrium states are not close together, but the system is in contact with a constant temperature reservoir at temperature T, which happens to be the same as the initial temperature of the system. Under these circumstances, the final equilibrium temperature of the system will also be T. For this situation, it follows from the first and 2nd laws of thermodynamics that $$\Delta U=Q-\int{p_{ext}dV}$$and $$Q=T\Delta S-T\sigma$$where $$\sigma$$ is the entropy generated as a result of irreversibility (always positive). So, if we combine these two equations, we obtain: $$\Delta U=T\Delta S-T\sigma-\int{p_{ext}dV}$$or, equivalently, $$W=\int{p_{ext}dV}=-\Delta F-T\sigma$$From this it follows that the maximum work that can be obtained by all processes between the specified initial and final thermodynamics equilibrium states of the system is $$-\Delta F$$.