Inconsistency between Helmholtz and Gibbs Free Energies

Question

I know that if the Helmholtz free energy, $A$, is expressed as a function $A\sim A(N,V,T)$, then this function contains all thermodynamic information about the system. For instance, the pressure of the system is given by:

$$P(N,V,T) = -\frac{\partial A}{\partial V}\Bigg|_{n,T}$$

Indeed, if we imagine $N$ and $T$ are fixed, then this in effect gives us pressure as a function of volume:

$$P\sim f(V).$$

My question relates to the inverse of this equation. If we are given $P$ rather than $V$ (still with $T$ and $N$ fixed and known) then either numerically or analytically we can solve this to find $V$. But if the equation is complicated enough, there maybe multiple values of $V$ for a given $P$, $T$ and $N$.

This didn't seem too unreasonable to me (perhaps physically this represents a gas and a liquid co-existing at the same $T$, $P$ and $N$ at equilibrium), until I realised that, if we worked with the Gibbs free energy instead, and found values of $G(T,P,N)$, then in that case

$$V(T,P,N) = \frac{\partial G}{\partial P} \Bigg|_{N,T}$$

and this gives us an explicit, single-valued equation for $V$ as a function of $P$, $T$ and $N$ (though perhaps the inverse $P(V)$ is now multi-valued?).

What's going on here? One formulation of thermodynamics gives me an EOS in which there can exist multiple values of $V$ for a given $P$, and the other says I can have multiple values of $P$ for a given $V$!

Answer 1

Note that the problem is really a purely mathematical (apparent) paradox you may phrase in terms of functions and their derivatives etc., regardless of their physical interpretations.

But the argument seems to suggest that if $p,A$ are single-valued functions of $V$ so that several values of $V$ may produce the same $p$, then you may prove just the opposite, namely that $V,G$ are single-valued functions of $p$ so that several values of $p$ may yield the same $V$.

This would indeed be a paradox in mathematics so there must be a mistake.

The mistake is the following. The function $A(V,N,T)$ may be a single-valued function of $T$ and $p$ is then a single-valued function of $V$ etc., too. Does it imply that $G(p,N,T)$ is a single-valued function of $p$, as you suggest? Your reasoning is that $$ G = A - pV$$ (fix the sign if it is wrong, please) must also be single-valued because $A$ as well as $V$ as well as $p$ are single-valued. However, the omission is that one should say "single-valued functions of what". Here, $G,A,p,V$ are, by your assumptions, indeed a single-valued functions but of $V$.

Are they – and especially $G$ – single-valued functions of $p$? The answer is, of course, No. To switch from $A$ to $G$, you have to rewrite all functions of $V$ to functions of $p$ which means that you effectively have to replace $$ V \to V(p) $$ everywhere. However, the assumptions only said that $p(V)$ is single-valued. On the contrary, $V(p)$ was assumed not to be uniquely determined. So because $V(p)$ isn't really single-valued by your assumptions, $G(p)$ won't be a single-valued function of $p$, either!

Incidentally, when phases co-exist, $V$ isn't a single-valued function of $p,T$ – just realize that ice and water of different densities may co-exist at the melting point $T$. etc. However, I think that the usual situation is that $p$ often fails to be a single-valued function of $V$, too. This is no inconsistency and it doesn't mean that that $A,G$ are ill-defined. It just means that the state must be described not just by $p$ or $V$ but also by some discrete information about the "phase" (or about the percentage of different phases if they co-exist).

Answer 2

The Gibbs free energy $G(P)$ is defined from $A(V)$ through a Legendre transformation, viz., \begin{equation} G(P) \equiv A(V(P)) + P V(P). \end{equation} (I suppressed the dependence on $T$ and $N$ because they are irrelevant for this discussion.) We have changed the independent variable from $V$ to $P = -dA/dV$, and $V(P)$ should be single-valued for $G(P)$ to be consistently defined.

There is indeed a principle that guarantees this, at least when there is no long-range interaction (e.g., gravity) involved so that the internal energy, entropy, and various free energies are extensive quantities. We have the requirement that $A(V)$ must be a convex function, i.e., \begin{equation} \tag{1} \frac{d^{2}A}{dV^{2}} = - \frac{dP}{dV} \ge 0. \end{equation} (See sections on thermodynamic stability in, e.g., Callen or Kardar.) After all, the Legendre transformation is well-defined only for convex/concave functions. Suppose that the equality in the above condition holds at most for isolated values of $V$ so that $P(V)$ is a monotonic function. Then, $V(P)$ is single-valued, and $G(P)$ is well-defined.

There is subtlety when $dP/dV=0$ for some finite interval of $V$, i.e., $P$ doesn't change in this region. (It corresponds to the case that two phases coexist.) Then, $V(P)$ develops a discrete jump at this value of $P$ but otherwise is single-valued. However, even with the jump in $V$, $G$ remains continuous. To see this, notice that in this interval, $P$ is constant and $A$ is a linear function of $V$ with the slope of $-P$. That is, \begin{equation} \Delta G = \frac{dA}{dV} \Delta V + P\Delta V = -P\Delta V + P \Delta V =0. \end{equation} Thus, there is no discontinuity in $G$ associated with the jump in $V$. We see that as long as the convexity condition, i.e., Eq. (1), is satisfied, we have a consistent definition of $G(P)$.

Note: One may have an equation of state that violates the convexity condition (e.g., the Van der Waals equation). However, the region where the convexity condition doesn't hold is thermodynamically unstable, and one has to supplement it by the Maxwell construction. With this the convexity condition is restored.