# Methods to solve a vapor pressure equilibrium question with water on the moon? [closed]

"Imagine a tall column of water on the Moon, maintained at a temperature of 50°C and left to the open vacuum at the top. Up to what depth will the water continue to boil? The Moon's radius is 0.273 that of the Earth's and its mass is 1.23% that of the Earth's. "

This question is on a physics prize exam in our area and is the only question on the exam that I have no knowledge on how to complete. I understand that if the vapor pressure of a liquid is less than the pressure of the liquid then the liquid will vaporize, but I still do not understand how I would use that knowledge to solve this problem. I am wondering if there are any resources that will further explain the vapor pressure equilibrium that could potentially lead to a way to solve this question? Or any explanation on how to approach this question to solve it?

Thank you.

NOTE

During this prize exam, anyone who takes it is not allowed any formula sheet or a programmable calculator. We re given a data sheet that does not include a value for the mass of earth. The rules in the exam do state to use the given values as they have changed some values to prevent memorization. i.e. latent heat of vaporization of water is $$2300\frac{kJ}{kg}$$ instead of $$2264\frac{kJ}{kg}$$ so I doubt that they would want us to calculate gravitational force of the moon compared to the force of the earth. They do give us the conversion of Torr to atmospheric pressure of "760 Torr = 1 atm" which could be useful to create a ratio 1/7.6 to continue the calculations, but without creating some type of function where the pressure slowly increases, I do not see how one could find the final depth of water effected.

CREDIT thank you Farcher for your help on this

• A start? The data about the Moon is probably for you to find the value of the gravitational field strength g on the Moon. Then it might be that you have to do a pressure=hρg calculation? Boiling rather than evaporation occurs when the vapour pressure of the liquid is equal to the external pressure. So looking at the graph because you are told that the water is maintained at 50 C the local external pressure just above the liquid surface must be 100 torr - approximately 100 mm of mercury which is about 1/7.6 of an atmosphere. Atmospheric pressure on the Earth is approximately 10 metres of water. Feb 6 '16 at 20:22

The vapor pressure of your water column at 50 degrees C is approximately 100 Torr, which is 1/7.6 of one atmosphere. The pressure in the water column is given by the equation $P = \rho g h$ , where $\rho$ is the density of the water, g is the local acceleration due to gravity, and h is the height of the water column above a stated point. The water will continue boiling in the water column, starting at the top end of the column (the point that is exposed to vacuum), and continuing down the water column until the pressure in the water column exceeds the vapor pressure of the water.
Estimates of the variable values for this problem are: P = 13,300 Pa, density of water = 1000 kg/m^3, and g = 1/6 that of earth's gravitational acceleration, or 1.635 m/s^2. Separation of the unknown variable yields $h = P/ \rho g$ = 8.13 m.