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Thermodynamic potentials without unconstrained variables?

Physics Asked on December 20, 2020

Principle of minimum energy in thermodynamic states

${left(frac{partial U}{partial V}right)_S} = 0$

at equilibrium $U(S,V)$ at fixed entropy is extremum with respect to the unconstrained variable $V.$

But what if $V$ is fixed as well? Situations like this occurred also when I was dealing with other thermodynamics potentials such as $F(T,V)$ or $H(S,p).$ $F(T,V)$ is obviously 2-dimensional(for closed system), since $S$ and $p$ can be calculated from partial derivatives of $F$ with respect to the independent variables $T$ and $V.$ It seems to me that, a fixed $T,V$, system does not allow any unconstrained variables, and does not evolve at all. How can I understand the change of free energy along constant $T$ and $V$ processes?

I have another question. Since thermodynamic state variables can be properly defined only at equilibrium according to my textbook, so do potentials which are functions of $T,S,V,p,$ I thought. But while studying 3-d $F-T,V$ space, I found even at points apart from minimum-$F$ point, every thermodynamic variable can be properly defined. Do those points indicate equilibrium states for given thermodynamic variables? At first I thought the minimum-$F$ point indicates the unique equilibrium, but now I know this is nonsense since $T = p = 0$ at that point.

Last question: If thermodynamic potentials such as $F$ can be defined only at equilibrium, I think we need some $F$-like quantity for non-equilibrium states to describe the state evolving from non-equilibrium to its equilibrium. Sometimes I find an obvious non-equilibrium state that should go to the corresponding obvious equilibrium state. For example, dividing a box into two parts, filling one part with ideal gas, and removing the wall(system connected to heat reservoir, and total box volume fixed). Quantities like pressure for the whole box is poorly defined since the box is half filled at first, but I could calculate free energy for the initial system just regarding the gas-filled system only, and this led to $Delta F<0$ for the whole process(I know this somehow contradicts my first question). Is this calculation valid? If this is valid, I think it’s possible to imagine one (although not physically relevant) additional dimension (related to non-equilibrium property of the system) can be added to the $F-T,V$ space, and $F$ along this new dimension (with other variables-$T,V$-fixed) is minimized when the non-equilibrium property vanishes. In my opinion this idea resolves my first question, but I’m not sure whether this is a sound argument.

Sorry for my poor English.

One Answer

"But what if V is fixed as well?" Since you have already fixed $S$ in a simple system, ie. one that has only two independent variables nothing will happen and the constraint on the system will not allow it to evolve. To allow the system evolve while two of its independent variables are constrained you must have at least another physical/chemical parameter that is allowed to vary, eg., magnetic or dielectric in an external field. It is not true that a thermodynamic potential is definable only in equilibrium, rather in equilibrium one can calculate its value easier.

Answered by hyportnex on December 20, 2020

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