Liquid solutions and vapour pressure
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 $V&lt;\pu{100 mL}$ to $V&lt;\pu{50 mL}$. In the present problem you don't know how much of the water is initially a liquid, only that the inequality $V&lt;\pu{100 mL}$ 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.