# Critical density of water - phase transitions

1. Aug 24, 2009

### TheDestroyer

Hello people,

I'm studying now phase transitions. I saw that the order parameter of water gas-liquid transition is the d[g]-d[c] or d[l]-d[c].

where
d[g] is gas density
d[l] is liquid density
d[c] is the critical density

what is it?
how can it be a factor between both transitions from liquid-gas and gas-liquid, I mean is it the same for both transitions?
if yes, then how can both become zero in the phase transition on each side of the critical temperature? and is there some theoretical way to calculate the critical density?

Thank you :)

2. Aug 24, 2009

### olgranpappy

I think this d[c] must refer to the density at the critical point (the point where the liquid/gas coexistence curve terminates). At this point there is no longer a difference between the liquid and the gas so that the density at this critical point is the same for both.

3. Aug 24, 2009

### TheDestroyer

But then how do we say that d[g]-d[c]=0 in liquid phase, and d[l]-d[c]=0 in gas phase? this means either that d[c] doesn't have a constant value for a certain liquid, or there is something else I can't find out. Because you have 2 equations for 1 variable (unknown), which is d[c]! does it make sense?

Am I thinking in a wrong way? please let me know :)

4. Aug 26, 2009

### PhaseShifter

At the critical point d[g]=d[l].

The order parameter is actually d[fluid]-d[c], but the fluid can be a gas, a liquid, or a supercritical fluid depending on temperature and pressure.

The distinction "gas" or "liquid" is only meaningful below the critical temperature, where the fluid is "gas" on one side of the coexistence curve and "liquid" on the other side.

As for theoretical calculation, if you have an equation of state it can be calculated--but normally the parameters for the equation of state are calculated from critical temperature and pressure, rather than the other way around.

If you plot isotherms for a fluid, you will see this at the critical point:

$${{\partial P}\over \partial d}=0$$

$${{\partial^{2} P}\over \partial d^{2}}=0$$

Above the critical temperature the slope will never be zero, and below the critical point the slope will have a relative maximum and relative minimum, but the inflection point between them will not be horizontal.

Last edited: Aug 26, 2009