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I am trying to understand the liq-gas transition with van der Waals model. And I am very confused by everything. Here is what I understood and I hope you will correct me.

I consider the Free energy $F(V,T,N)$ and the Gibbs energy $G(P,T,N)$.

Those functions are thermodynamic potential. It means that their values should be minimum when the system is at thermodynamic equilibrium.

Also:

To have a stable system, thus a system that can have thermodynamic equilibrium, the free energy must be concav and the gibbs must be convex. If it is not possible then no equilibrium is possible in the region

Question 1:

Do you agree with this paragraph? Am I right by saying that if these thermodynamic quantities don't have the nice convexity properties then it physically mean that the system is not stable and thus doing a phase transition. In fact I am a little confused by this, I will explain with an example just below.

Here is the curve showing $P(V)$ for the van der Waals model for $T<T_c$.

enter image description here

And now the associated free energy :

enter image description here

Question 2: In this curve of the Free energy in function of $V$ we then understand that it is not possible to have thermodynamic equilibrium for the red part (because the curve is not concav here). But then, what happens if I fix $V$ to such a value? The system won't be in equilibrium but as $T,V,N$ are fixed what changes then (bc everything is fixed)? Does this mean by contradiction that, thus, I can't fix all those values in this area?

Last question:

When we do maxwell construction, we try to take the concav envelop of $F$ to make it "respect" the convexity property.

And on the graph of $P(V)$ it then become the "area" property that will give the final value of the pressure in function of volume:

enter image description here

But I would like to really understand the point of maxwell construction. What does the curve mean if we don't do it ? Like the pressure in function of volume. Are they physically "wrong" if we don't to this trick ? I am not sure how I should interpret the maxwell construction. Should I understand this as:

(1): I have a model of $P(V)$ that works very well excepted in a specific zone. Then I locally modify my model by using the maxwell construction (that has physical justification of course). At the end I have a good result.

Or:

(2): The maxwell construction is more a "change of point of view". Like the Pressure-Volume curve without doing it are not physically wrong but as we have a phase transition occuring here the pressure when two phase are in contact doesn't have the same meaning. Then we have to change the pressure curve.

My last question is probably not very clear but I hope you will understand what I mean.

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  • $\begingroup$ P(V) convexity is not what the statements refer to. It is the convexity of function G(para) or A(para), where parameter can be volume or pressure etc. $\endgroup$
    – user115350
    Jan 3, 2018 at 0:10
  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/378123/2451 $\endgroup$
    – Qmechanic ♦
    May 18, 2022 at 11:56

2 Answers 2

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Question 2: In your example, there is no equilibrium if the volume is allowed to change, because F has no minimum w.r.t. V. If started in a given state and released, the system will expand to lower F, doing work on its environment. If, however, you hold the volume fixed, then the fact that F has no minimum w.r.t. V does not matter. Since dV is constrained to be 0, dF has no contribution from the PdV term in any state change. In that case, if T, V, and N are all fixed, the system can still exchange heat with its surroundings, changing its internal energy and entropy (example: ice at $0^\circ$C absorbing heat from a $0^\circ$C water bath - OK, its volume isn't quite fixed as it melts, but almost!). Then for any small change from equilibrium, dF = dU - TdS = 0.

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  • $\begingroup$ What I wanted to say is that because of the convexity "problem" no equilibrium is possible here. But in general for any value of $(N,V,T)$ I can indeed have an equilibrium (the free energy will decrease in respect to all the other variables such as $P,U,N$). But the pdf I read said that because of the convexity behavior we have an unstable state for this value of $V$ and then I wondered what would happen to the pressure (for example) because in a general case the pressure just adapt to minimize F to reach the equilibrium value. But here equilibrium seems to be impossible because of the $\endgroup$
    – StarBucK
    Jan 2, 2018 at 20:09
  • $\begingroup$ convexity that shows an unstable behavior. $\endgroup$
    – StarBucK
    Jan 2, 2018 at 20:09
  • $\begingroup$ (I'm trying to find again the pdf on where I read the fact that the convexity behavior of $F$ shows that it is an unstable region, I edit when I find it) $\endgroup$
    – StarBucK
    Jan 2, 2018 at 20:23
  • $\begingroup$ I'm afraid I can't help there. I don't know what concave/convex mean in this context. AFAIK, equilibrium corresponds to a minimum in whichever free energy applies given the constraints on the system. Can you post the document that's causing the confusion? $\endgroup$
    – pwf
    Jan 2, 2018 at 20:34
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Let's consider the point $V_1$ given in your free energy picture. The system knows that there are two preferred (stable) states: one at $V_g$ and one at $V_l$ that are at the same pressure.

Volume is conserved in this case, at $V_1$, so the system chooses to partition itself so that part of the system sits at $V_g$ and part of the system sits at $V_l$ precisely such that the fraction of the system in $V_g$ and the fraction of the system $V_l$ sum to $V_1$.

Because the system is now in two partitions with the same pressure, it is in happy equilibrium and we have conserved volume by determining how much of our system is in the $V_l$ partition and the $V_g$ partition. This is the Maxwell construction.

We're able to do this because we don't have control over the microscopic degrees of freedom, namely the particle positions and velocities. Because pressure is the same in both partitions we can say the system is equilibrated. The two volumes then correspond to the volume of the system in liquid state versus gas state.

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