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following the question: Why dW=pdV is an inexact differential?

Usually the pressure p is given by the equation: $$p=-\left.\frac{\partial U}{\partial V}\right|_{S},$$ where $U=U(S,V)$ is internal energy of the system (let us not complicate things with number of particles) and S is its enthropy.

For the term $-pdV$ to be exact, the extrerior derivative must be zero. That leads to the condition: $$\frac{\partial^2 U}{\partial V\partial S}=0,$$ that is, the most general form of $U$ for which $-pdV$ is exact is: $$U=f(S)+g(V),$$ where f and g are some functions.

In such system, the work would be function of state and would not depend on history. Obviously such system is pretty wierd and i guess unrealistic, but my question is: is there some fundamental reason in thermodynamics alone that forbids such systems to exist? I.e. can i see that work must depend on the history for any system from thermodynamics alone?

Thanks:)

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Starting with $dU=T(S,V)dS-p(S,V)dV$, the work differential $pdV$ is exact when the pressure $p$ depends only on $V$, $p=p(V)$, in which case the system is purely mechanical and has no thermal properties. The same conclusion holds for more than 2 variables.

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  • $\begingroup$ why doesnt it have thermal properties? The energy can still depend on enthropy, thermal and mechanical properties just need to be decoupled. And thermal properties might still be coupled with other variables, like number of particles in the system (at least mathematically) $\endgroup$
    – Umaxo
    Commented May 10, 2019 at 9:27
  • $\begingroup$ If $p=p(V)$ then $dU+pdV=TdS$ is a total differential, and therefore $T$ can only depend on $S$ in the two-variable case that was your question. If there are more variables, say, particle numbers and chemical potential, etc., then with $p=p(V)$ the mechanical variation part is still separate from the thermal process and is not influenced by temperature or entropy. $\endgroup$
    – hyportnex
    Commented May 10, 2019 at 13:29

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