# Can thermodynamical work be a function of state?

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:)

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.
• 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. May 10, 2019 at 13:29