# Phase transitions and Helmholtz vs Gibbs Free Energy

Gibbs phase rule for Gibbs free energy says that phases during a phase transition must be in mechanical, chemical, and thermal equilibrium, e.g. $$T_1=T_2$$,$$G_1=G_2$$, and $$P_1=P_2$$, where the subscripts represent different phases (phase 1 and phase 2), T is temperature, P is pressure, and G is Gibbs free energy.

My understanding of these rules is that the constant temperature and pressure come from the fact that the decrease of Gibbs Free energy only becomes synonymous to the second law of thermodynamics when pressure and temperature are constant.

For systems where pressure isn't constant, but where for example, volume is constant, and temperature remains constant, we use Helmholtz free energy. However I was wondering if there is an equivalent Gibbs phase rule for phase transitions in equilibrium, where $$T_1=T_2$$,$$F_1=F_2$$, and $$V_1=V_2$$, where F is Helmholtz free energy and V is volume. I was just wondering if this is right.

• It's not $G_1 = G_2$, but rather $\mu_1 = \mu_2$, where $\mu = G / N$ is the chemical potential (and $N$ is the number of particles). These relations hold for intensive quantities only. – lr1985 Oct 11 '18 at 9:40
• Phase do not have to be in equilibrium during a phase transition, they only have to be in equilibrium if the transition is to be thermodynamically reversible process, or if no transition is occuring - when the phases coexist and their size does not change. – Ján Lalinský May 4 '19 at 22:01

## 2 Answers

The "phase rule" named after Gibbs is an equilibrium coexistence relationship among the possible degrees of freedom $$v$$, the number of chemical components $$c$$, the number of intensive parameters $$k$$ and the number of phases $$f$$: $$v=k+c-f$$ The coexistence is the result of the maximization of the total entropy of the system under the constraint of intensive variables and extensive variables, from which one can derive that the maximum $$S$$ is equivalent to the minimization of the corresponding potential under Legendre transformation, see 1, 2. The totality of the variable intensive and extensive parameters determine the state. In a chemically closed system one assumes that the total mass is constant and that results in the equality of the chemical potentials for all phases, etc.

When using the Gibbs potential one assumes that the intensive parameters is $$k=2$$ representing temperature and pressure as externally controlled and fixed parameters that also prevail within the system itself. So for a single chemical species $$c=1$$ you have $$v=3-f$$ implying for example that for a single phase $$v=2$$ and you can change both $$T$$ and $$p$$ and the system can be in equilibrium for arbitrary $$T, p$$ but when $$f=3$$ then $$v=0$$ and thus the "triple point", that is the coexistence temperature and pressure of the solid-liquid-gas are not variable. Notice that the total volume is arbitrary and will be be determined by whatever external values $$T, p$$ may have.

Now if you set $$k=1$$ and allow only the temperature $$T$$ be controllable externally then the Helmholtz potential is minimum in equilibrium and then the phase rule says that $$v=1+c-f$$ but then one must also assume that the total volume is fixed as external constraint and let the internal system pressure settle as it may.

For two phases to coexist, they must be in equilibrium. To be more specific, they must be in:

• mechanical equilibrium (same $$T$$)
• thermal equilibrium (same $$P$$)
• phase equilibrium (same $$\mu = g$$)

If any of these didn't hold then the system's state wouldn't be steady

• if one were hotter it would transfer energy to the colder one by heat
• if one were at higher pressure it would expand and push the other out of the way (transferring energy by work)
• if one had a higher chemical potential then there would be a net transformation of material from one phase to the other.

This "internal" equilibrium must hold regardless of which external constraints are placed on the system. In the example you gave, a two-phase system at constant $$V$$ would still need to have phases with the same $$T$$, $$P$$, and $$\mu = g$$.