Simple estimation of the critical temperature of water

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...