# Stability, unstability (and metastability) in liquid-gas phase transition: unstable in regards to what?

I have a question about the stability, unstability (and extra questin about metastability, between the spinodal lines if you have time), when we are having a liquid gas phase transition.

Here, a curve including the spinodal lines. The main thing to look at is in fact the red line obtained by Maxwell construction. My question is mainly: What do we precisely mean by "unstability" in the phase transition, and how will the system move when I prepare the system in given states.

The volumes $$V=a$$ and $$V=b$$ represent the spinodal lines. Between those two curves the system is said unstable.

Between $$V_1$$ and $$a$$, $$b$$ and $$V_2$$, the state is metastable.

But I want to make things really clear: what do we precisely mean by "stable" and "unstable" zones of the diagram?

Does it mean that if I take an initial state such that:

• the temperature corresponds to this curve $$P(V)$$ (because it is isothermal curves here)

• the volume is such that $$a \leq V\leq b$$

And then from this initial state, I fix $$(T,P)$$ and I let the volume change.

The system will be unstable in this $$(T,P)$$ ensemble (the volume will change until reaching the equilibrium). And in practice this zone also corresponds to the phase transition liquid-gas. Thus when the talk about stability, unstability here, they implicitly assume we are in $$(T,P)$$.

Am I right?

I advise to take a look at the beginning of page 64 of this document for further info if you want a more detailed context https://www.uam.es/personal_pdi/ciencias/evelasco/master/tema_III.pdf

Extra question: If I am right with my previous explanation, what happens if I initialize my system in the metastable zone : $$V_1 \leq V \leq a$$? How do I know if the system stays here, go to do a phase transition to reach the stable state in the "other direction", go to the stable state just at $$V=V_1$$?

• For example, point $V=a$ has two states. One is on the black curve and the other is on the red curve. The free energy of the state on the red curve is lower. Thus that is more stable and likely. – user115350 Jan 4 '18 at 18:23
• Ok so basically you answered my first question, thanks. Then we indeed implicitly talk about the Gibbs free energy, and thus that we are in $(N,P,T)$ ensemble to talk about stability here. Do you have an idea for the extra question ? Thanks a lot ! – StarBucK Jan 5 '18 at 0:02