# Homework Help: Vapour pressure and volatility.

1. Sep 21, 2010

### Referos

1. The problem statement, all variables and given/known data

(I)Explain how each of the following factors affect the vapour pressure of a liquid: (a)surface area, (b)volume, (c)attractive intermolecular forces, (d)temperature, (e)density of liquid.

(II)Order the following substances in terms of volatility: CH4, CBr4, CH2Cl2, CH3Cl, CHBr3, CH2Br2.

3. The attempt at a solution

(I)

(a)I guess that if the surface area increases, the rate of molecules that pass from liquid to vapour increases, but so does the rate of molecules from vapour to liquid, so it has no effect?

(b)Well, the liquid's volume can increase by either having a greater surface area or a greater height/depth. Surface area doesn't affect vapour pressure, so that's out. And increasing the liquid's depth doesn't increase the number of molecules close to the surface that can escape, so, again, no effect on vapour pressure?

(c) and (d) are kinda obvious, so I'm skipping these

(e)No idea here. I suppose that with higher density, the mass of each individual molecule is greater and so (assuming constant temperature) their momentum is higher. So the impulse the liquid's surface molecules exert on the vapour is higher, so we get a higher average force and thus higher pressure. But the exact same thing can be said regarding the vapour molecules, so, once again, there's no net effect on the vapour-liquid equilibrium, and the vapour pressure must remain unchanged, I guess.

(II)
Also no idea here. I know that volatility decreases with intermolecular forces. If all these were apolar molecules, then the only intermolecular forces would be the dispersion forces. Since this increases with polarizability which increases with molecular weight, I could compare their intermolecular forces only with their molecular formulae. But some of these are polar molecules, and there's no way I can compare the relative effects of the dispersion forces and the dipole-dipole forces. So I cry :( .

2. Sep 21, 2010

### Staff: Mentor

(I) seems to be right to me, even if your analysis of e is not completely correct, higher density doesn't mean mass of molecules is greater.

(II) after some googling looks to me like dipole interactions can be ignored, but I agree with you that question requires making comparisons that are not obvious.