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"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.

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  • $\begingroup$ 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. $\endgroup$ – Ivan Neretin Aug 21 at 8:47
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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.

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