Unconstrained Equilibrium

Consider a an isolated system, in this case:

$dU = 0$ and $dV = 0$

If we consider a system that can only perform $PV$-work on its surroundings, we can write for any quasi-static (constantly in equilibrium).

$$\input{equations/11-3A}$$ (17-1)

Therefore, at equilibrium the entropy must be a maximum or a minimum ( $dS = 0$) (This is the equation of a tangent plane)

Figure 17-2: The surface of $S$ as a function of $U$ and $V$

Since entropy must always increase as a system approaches equilibrium, it must be a maximum at equilibrium. (In other words, it must be at the summit of a hill with $U$ given as north and $V$ given as west.)

How to express this maximal relationship?

Let $\ensuremath{\delta}S$ represent any possible variation (change internal temperature, distribution, heat flow, etc.) to the entropy. Then, if the supposed variation also has $\ensuremath{\delta}U = \ensuremath{\delta}V = 0$,

$$\input{equations/gibbs-eq-1}$$ (17-2)

implies $S$ is a maximum and is thus the equilibrium state.

This introduces the concept of a virtual change:

Equation 17-2 is equivalent to

$$\input{equations/gibbs-eq-2}$$ (17-3)

(See if you can figure out why--hint: $({\partial U}/{\partial S})_V = T >0$

Implies $U$ is a minimum at constant $S$ and $V$--and that is what we expect from mechanics.

Figure 17-3: Ball rolling on a surface.