For an isotherm, $dF=-pdV$, so $\Delta F=-\int^{V_1}_{V_2}pdV$. Thus, a change in the free energy tells us how much work has been done by or on the system. However, at this point the textbook I've been reading and my lecturer both stated that this is true for reversible isotherms, implying that this might not be true for irreversible ones. Does this interpretation also hold for irreversible isotherms (I don't see why not), and if not then why not?


Partial answer: This certainly does only hold true for quasi-static processes (the reversiblity in your textboox is an even stronger condition), as only for these processes a pressure is defined throughout the process.

  • $\begingroup$ Although it is possible to have an irreversible quasi-static process. Would you say the textbook's condition is stronger than necessary? $\endgroup$ – Pancake_Senpai Apr 6 '20 at 9:34
  • $\begingroup$ @Pancake_Senpai This is a very good question. In Landaus statistical physics he also uses the term "reversible", but I feel like it should hold true for all quasi-static processes. $\endgroup$ – user224659 Apr 6 '20 at 14:05
  • $\begingroup$ If the process is quasi-static but not reversible then it is not guaranteed that the system is in equilibrium with its environment or that all parts of the system are in equilibrium with each other. This means that different parts of the system may have, for example, different temperatures or pressures so the single pressure in the expression above may not be well defined (or at least require some care to interpret) $\endgroup$ – By Symmetry Apr 6 '20 at 14:16
  • $\begingroup$ Aren't quasi-static processes defined such that internal equilibrium is always maintained, though? On a P-V diagram only a quasi-static process can be drawn precisely because all parameters are well defined throughout the process. $\endgroup$ – Pancake_Senpai Apr 7 '20 at 11:02

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