For a single component closed system, my textbook says $$dU=\left(\frac{\partial U}{\partial S}\right)_VdS+\left(\frac{\partial U}{\partial V}\right)_SdV=TdS-PdV$$ is valid for both reversible and irreversible process. But if the process is irreversible and non-quasi-static, (for example, if it is just a set of points in the coordinate system far apart from each other), how can I talk about the total differential of the state function?

Furthermore, the Maxwell's relation $\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V=\frac{\partial^2 U}{\partial S \partial V}$ seems to suggest the smoothness of the state function.

What is it that I am getting wrong here?


1 Answer 1


The thermodynamic functions in your equations apply only to thermodynamic equilibrium states. A reversible process consists of a continuous sequence of thermodynamic equilibrium states, so you can evaluate the changes in these properties using these equations. For an irreversible process between two arbitrary thermodynamic equilibrium end states, you need to identify a reversible path between the exact same two end states and then integrate these equations along the reversible path. This applies even if you have an irreversible process executing a highly tortuous path between two thermodynamic end states which are, in the end, very closely neighboring (differentially separated). This is what they mean when they say that the differential equations are applicable to an irreversible change between two differentially separated end states.

  • $\begingroup$ Does it also imply that the differential expression and Maxwell's relations can be used at the discontinuous regions such as phase change boundaries? I think the total differential still works but the Maxwell's relations are not? $\endgroup$
    – P'bD_KU7B2
    Mar 28, 2022 at 11:54
  • 1
    $\begingroup$ The equations only apply to a single phase. $\endgroup$ Mar 28, 2022 at 14:14

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