# Thermodynamic PV-diagram: Why does a critical point =inflection point?

1. Mar 10, 2014

### Nikitin

Why is the first and 2nd order derivatives of pressure with regards to volume equal zero at the critical point for an isotherm phase curve? Why/how is it possible to expand a fluid at its critical point without any change in pressure?

thanks for all help :)

2. Mar 10, 2014

### Nikitin

Also, can somebody explain to me the premise for the maxwell construction? Ie why does dG = 0 lead to $\int_{p_{liquid}}^{p_{gas}} Vdp =0$? I mean, it's fair enough that $\int_{{liquid}}^{{gas}} dG=0$, but $Vdp$ is just one of the many terms in the total differential of dG...

Maxwell construction: http://en.wikipedia.org/wiki/Maxwell_construction

Last edited: Mar 10, 2014
3. Mar 10, 2014

### Khashishi

On the wiki page, the PV diagram shows it pretty well. Above the critical point, you have PV curves where P increases with decreasing V throughout. You take a cylinder of fluid, and you compress it, you expect the pressure to increase. Pressure "tries" to minimize itself over the whole system, and in this case, it does so by distributing the density evenly over the entire container.

Below the critical point, you have a portion of the PV curve with an upward slope (dotted line part). This portion is unstable because if you have a cylinder of the fluid, it will compress itself, allowing the external gas to expand. Therefore, if you have two cylinders pushing against each other, initially placed in the unstable region, one of them will spontaneously contract, and the other will expand. One becomes the liquid and the other the gas. This will reduce the overall pressure and is free-energetically favorable. This partitioning between liquid and gas will always happen as long as you have the unstable region in the PV curve, so that means that the partitioning will stop when the positive slope stops.

Since the curves vary smoothly with parameter changes, this means the positive slope stops when it is replaced by an inflection.

You can't expand a critical fluid without changing the pressure except infinitesimally. Any finite volume change will change the pressure and shift you out of critical.

4. Mar 10, 2014

### Khashishi

I think any terms in the free energy are "included" in the pressure; that is, the partial derivative against volume already includes anything that could factor into the gas liquid transition.

5. Mar 11, 2014

### Nikitin

Ah yes, I forgot that by using the thermodynamic identity you will get rid of almost all the terms in the total differential of Gibbs free energy. But I can't say I understand your first post....

6. Mar 11, 2014

### Khashishi

Consider a Van der Wal's potential between molecules. The free energy is low when the molecules are far apart. But there's also a local minimum when they are close together. So, molecules that are far apart will exert an outward pressure. But if you get some molecules close together, they will be drawn together and exert some kind of collective inward pressure. Now in a container with a fixed number of molecules, the average density can't change, but the local density can change because clumps and voids can form. Below the critical point, the state with clumps and voids is energetically favorable to the state where density is even throughout, because of this Van der Wal's potential. The PVT diagram tells you where exactly this distribution ends up.

At zero temperature, all the molecules could be stuck together in a local minimum in one clump. But at a finite temperature, hotter molecules constantly evaporate off the surface of the clump into the gas. In equilibrium, molecules are also sticking to the clump.