Simple estimation of the critical temperature of water

Question

I'm trying to develop fermi estimation skills and I came up with a question for which I don't even know where to start from. Here goes:

Is it possible to estimate the critical temperature (say in Kelvin degrees) of water in a simple way using fermi estimation?

By critical temperature I mean the temperature of the point at the end of the coexistence line of water and vapour. See this plot.

Answer 1

Estimation: I want the two densities of vater and vapour to become approximately equal.

• the density of water is nearly constant
• the vapour pressure (you can derive this from the above mentioned Clausius-Clapeiron-equation) is approximately exponential in $1/T$. This means, that if you increase pressure by a factor, the inverse of the evaporation temperature $1/T$ will increase by some shift.
• the pressure is 100 kPa at 100°C and 611 Pa (if I remember correctly - which is the point of this question here) at the triple point at 0°C, so you have a factor of 150.
• now the corresponding shift in $1/T$: the temperature changed from 275 to 375. I will now calculate in units of 25K. So 1/T changed from 1/11 to 1/15, i.e. from 15/165 to 11/165. It dropped by 4/165.
• the density of air is 1.3 kg*m^-3, so water vapour might be near 1, i.e. 1000 times less than water.
• 1000/150 is nearly 7, so 1000 might be near $150^{1.4}$. Remember, an exponent of 1 corresponds to 4/165 drop in $1/T$.
• let's say the $1/T$ will drop by 5.5/165 more - i.e. the temperature is doubled.
• this means 470°C, which is a bit too much (it should be below 380°C...)

Probably the temperature-dependence of the evaporation heat (and therefore the deviation from the exponential vapour pressure) is the main error; it's hard to say...

Answer 2

I haven't done a calculation yet, but I would use an extrapolation based on the Clausius-Clapeyron formula:

$$\frac{dP}{dT} = \frac{L}{T\Delta V}$$

You then take any two known thermodynamic quantities of water and water vapor and linearly extrapolate to that point where the difference is zero. A good choice could be the entropy, the entropy of water vapor is easily estimated by treating it as an ideal gas and taking into account the internal degrees of freedom of the H2O molecule. The entropy of water then follows from the latent heat, the way this depends on temperature follows from the well known heat capacity of water.

So, I think it's not that difficult to come up with an estimate with only a blank sheet of paper and a calculator.

Answer 3

I guess you can use the Van der Waals equation and some estimates of the molecule volume and intermolecular attractive forces. The parameters of the critical point depend on these characteristics, so you need to assess them.