# What affects the vapor pressure of a pure liquid? [closed]

So pretend we have two samples of pure benezene in equilibrium with its vapor. One sample containing 0.5 mol and one containing 1 mol. They're also both at the same temperature. Let's pretend they're both at, say, 298 K (room temperature). Would the two samples have the same vapor pressure? I was thinking since the 1 mol sample has more molecules that could be vaporized, it would produce a greater pressure. But I was told this is wrong.

• True, 1 mol of liquid has more molecules that could be vaporized, but then again, 1 mol of vapor has more molecules that could be condensed. It is a two-way street, see? – Ivan Neretin Dec 30 '20 at 5:48
• The rate of evaporation is proportional to the area of the liquid/gas boundary, not to liquid mass. And so does the condensation. – Poutnik Dec 30 '20 at 7:35
• The macroscopic evaporation rate will be higher for a larger surface, as by Poutnik comment. And, as you are thinking on number of molecules, more benzene can simply fill of vapour a bigger container. – Alchimista Dec 30 '20 at 9:00

## 2 Answers

According to Raoult's Law, vapor pressure of a component in a solution is directly proportional to its mole fraction. In your case mole fraction is equal to 1(since benzene is the only component), which means that both produce the same vapor pressure.

• One doesn't need ideal stuff to state that a pure liquid has a characteristic vapor pressure at given conditions. Not that the answer is wrong, but it took an unnecessary path. – Alchimista Dec 30 '20 at 8:55
• Vapor pressure of pure solvent is a characteristic of the solvent which depends on the conditions and type of solvent only, Is this answer better? – Math_Whiz Dec 31 '20 at 3:26
• Well, my comment wasn't to be pedantic. I would rather call that "vapour pressure of a pure substance...." and it is all. No one expect that melting of a ton of stuff happens at higher T than that of melting a kg. That is the point OP should have clear. Afterwards s/he could move to other cases. You can open stating the simple fact. Then go saying" it can also be seen using Raoult law with a component having x = 1......... – Alchimista Dec 31 '20 at 9:26
• Ok, I got it... Thanks :-) – Math_Whiz Dec 31 '20 at 11:09

The vapour pressure under the circumstances of fixed T and n (amount of substance) depends on an additional constraint: either the volume of the container (if rigid) or the value of an applied external pressure.

In a rigid container (controlled fixed volume), as the volume is increased more of the liquid vaporizes but the vapor pressure remains constant until all of the substance is in the gas state. Past that point the pressure will decrease with increasing volume according to Boyle's law.

Under controlled constant pressure there will be vapor present if the applied pressure is smaller that the equilibrium vapor pressure at that temperature. At the boiling point (when applied pressure is equal to the applied pressure) one can transiently alter the applied pressure to allow increased evaporation, thereby increasing the volume for instance, but retaining the same intial and final pressures, at least until all of the substance has evaporated.

All that changing the amount of substance does is change the extent to which the volume will change when the liquid is converted entirely to gas. Between that point and the point at which there is no gas (because applied pressure is higher than vapour pressure, or because available volume is too small, which amounts to the same thing) the vapour pressure will be constant.

• Does it mean that in a two phases system P and V are no longer conjugate variables of the entire system since one can control/fix both P and V? If so, how can this be consistent still with the mathematical structure of thermodynamics? – The Quark Feb 7 at 18:55
• @TheQuark In a two-phase system the pressure would presumably be the same in both phases, if they are to be in mechanical balance. You could set the relative volumes of the two phases arbitrarily by adding or removing energy, provided the (T,P) values correspond to a point on the liquid-vapor coexistence line. – Buck Thorn Feb 8 at 0:42