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As far as I can tell, the Gibbs Phase Rule gives as a "law" the number of parameters which can be independently/a priori fixed when in an equilibrium with $r$ components and $M$ phases as $$f = r - M +2.$$ My question is about what it means when $f$ is negative, and why Callen (who I am reading right now) seems to rule out the equilibrium corresponding to $f$ negative.

As a concrete example and for simplicity, take the equilibrium in one component of four phases. The Gibbs Phase Rule says $f = 1 - 4 + 2 = -1$, and so one conventionally concludes that such an equilibrium is impossible. But let us recall the derivation of the phase rule. Strictly speaking, if four phases are in equilibrium then we must have the following three equations (where the subscripts denote phases): $$\mu_1(T,P) = \mu_2(T,P)$$ $$\mu_1(T,P) = \mu_3(T,P)$$ $$\mu_1(T,P) = \mu_4(T,P).$$ The crux of the Gibbs Phase Rule is to note that we have two unknowns in three equations, and thus that the system cannot be satisfied. My question is why is this so? I understand that in general this system may not be satisfied, but why can't things conspire (i.e. the structure of certain, particular $\mu_i$ in a given system) be such that the system is satisfied? Perhaps this has something to do with the nature of nonlinear equations?

EE18
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2 Answers2

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It seems, that you have the wrong equations. For example, for three phases (liquid, vapor, solid) you have

\begin{align} T^s &= T^l =T^v \equiv T^{eq} , \label{eq:thermal_equilibrium_multi} \\ p^s (T^{eq},\rho^s)&= p^l (T^{eq},\rho^l) =p^v (T^{eq},\rho^v) \equiv p^{eq} , \label{eq:mechanical_equilibrum_multi} \\ \mu^s(T^{eq},\rho^s)&= \mu^l(T^{eq},\rho^l) =\mu^v(T^{eq},\rho^v) \equiv \mu^{eq} . \\ \label{eq:chemical_equilibrium_1} \end{align}

which has exactly $N$ equations for $N$ unknowns. This would not change if you consider more or less phases

ConvexHull
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The crux of the Gibbs Phase Rule is to note that we have two unknowns in three equations, and thus that the system cannot be satisfied. My question is why is this so?

To solve for $n$ unknowns we need $n$ independent equations. If we are given $n+1$ equations and $n$ unknowns, then either (i) there is no solution, or (ii) if there is a solution these equations are not independent.

Case (i) means there is no extra phase.

Case (ii) means that what appears as an extra phase is some combination of the other three phases, i.e., not a new phase.

In either case we cannot have more equations than unknowns, hence the number of degrees of freedom cannot be negative.

Themis
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