# Liquid solutions and vapour pressure

"The vapour pressure of water at 80° C is 355 torr. A 100 mL vessel contained water-saturated oxygen at 80° C, the total gas pressure being 760 torr. The contents of the vessel were pumped into a 50 mL vessel at the same temperature. What were the partial pressures of oxygen and of water vapour, what was the total pressure in the final equilibrated state? Neglect the volume of any water which might condense."

This question is given in my module and the solution given to the same is quite vague. Could somebody please explain it to me? Thank you very much. What I could particularly not understand is how the water vapour's pressure in the 50 mL vessel remains unchanged.

• That's precisely because it is saturated vapor. We see it pretty much every day, and still it feels weird. If you put it under the piston and press, its pressure wouldn't grow at all. Aug 21 '19 at 8:47

## 1 Answer

The key to understanding this type of question is that the equilibrium (saturation) vapor pressure of a pure liquid is to a first approximation (that is, assuming ideal behavior) independent of the presence of other gases, such as (in this case) oxygen. Therefore, you can think of $$\pu{355 Torr}$$ as being always the pressure of the vapor when in equilibrium with the pure liquid at $$T=\pu{80 ^\circ C}$$. For equilibrium to be established requires that there is (more than) enough substance to fill the volume with vapor at that pressure. The minimum amount of substance which will fill the volume with vapor at that pressure is, under the assumed ideal conditions, given by the ideal gas law: $$n=\frac{pV}{RT_b}$$where in the present problem $$T_b=\pu{80 ^\circ C}$$, $$p=\pu{355 Torr}$$, and V is the volume occupied by the vapor, which changes from $$V<\pu{100 mL}$$ to $$V<\pu{50 mL}$$. In the present problem you don't know how much of the water is initially a liquid, only that the inequality $$V<\pu{100 mL}$$ is satisfied by the vapor. During the volume reduction some of the vapor condenses into additional liquid in order to retain the equilibrium pressure. The oxygen is treated here as a spectator, although it contributes to the total pressure in the vessel. In practice it also alters the vapor pressure of the liquid, although here by a negligible amount.

• Sir, this is a different question, but related to vapour pressure (I couldn't find any other related question): If we are given a solution of some solutes whose concentrations and temperatures are given, can we compare their vapour pressures? I know that higher temperature means higher VP, but is there an empirical relationship b/w concentration and temp? For example, out of 0.02 M $\ce{NaCl}$ at $50^{\circ}$C , 0.005 M $\ce{CaCl_2}$ at $50^{\circ}$C , or say, 0.03 M sucrose at $15^{\circ}$C. They also have different Van't hoff factors.. so can you explain that in brief? Apr 27 '21 at 12:45
• I am not sure what you mean by "relationship between concentration and temp". I recommend you read about colligative properties and Raoult's law if you haven't already. Raoult's law says that vapour pressure depends on concentration on solute, or rather on mole fraction of solvent: $p=\chi p^\circ$. This will result in an effect on the boiling point T at a particular pressure, known as boiling point elevation. Chem SE has many Q&As on the subjects, and the web is full of resources. Maybe you should take a look at this, for instance: users.stlcc.edu/gkrishnan/solution2.html Apr 27 '21 at 18:41
• Can you have a look at this question I asked? I am not looking for a relation b/w concentration and temperature, but VP and conc. or temp, so that it easy to compare the VP's. I have read colligative properties and Raoult's law, but in this question, since temperatures and concentrations are not same, can we compare VP's? For example, one thing that I am looking for is this.But concentration is still not taken Apr 28 '21 at 2:04
• And congrats for becoming a moderator! I hope, you also answer my question if I have made myself clear enough :). Apr 28 '21 at 2:12