Question about Van der Waals EoS and Spinodal Line

[dRic2]

Van der Waals EoS is a cubic EoS that can represent the transition between gas and liquid phase. If you plot a P-V graph you end up with something like this

The part I underlined in red are said to be metastable equilibria (i.e. superheated liquid and supersaturated vapor). I don't understand why. Why should I have a superheated liquid for example ? Why doesn't it boil?

So the first explanation regards nucleation. If you don't have bubbles, the liquid can't boil. That's fine. Then why if I cross the spinodal line I always have separation of phases ?

To answer this question one might notice that the spinodal line is the point where $\frac {\partial P} {\partial V} = 0$, but since $P = -\frac {\partial F}{\partial V}$ ($F$ is the Helmholtz's free energy), then $\frac {\partial^2 F}{\partial V^2} = 0$ so, the spinodal point is a point where Helmholtz's free energy reverses its curvature. Yet I'm missing the final step here. I can see that when $F$ becomes a concave function then I must have phase separation, but it still don't answer my question: "why is the region enclosed bu the spinodal and the binodal line a metastable equilibrium?"

 

[Valerie Park]

The answer to this question is related to the fact that the Helmholtz free energy is a convex function inside the metastable region and has a minimum at the binodal line. This means that, if you have a superheated liquid or a supersaturated vapor, then the system will try to reach the binodal line, where the Helmholtz free energy is lower. However, in order to do that, it needs to overcome an energy barrier, which is the difference between the Helmholtz free energy at the binodal and at the metastable point. This means that these metastable states are stable, but not equilibrium states: they can be maintained for some time, but they will eventually decay into the binodal equilibrium state.