A cylinder fitted with a movable piston contains liquid water in equilibrium with water vapour at 25 — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

System: $\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_2\text{O}(g)$ at 25°C. \textbf{Key principle:} The vapour pressure of a \emph{pure} liquid depends \emph{only on temperature}. Mechanical or volumetric changes cause condensation or evaporation, but the system re-establishes the same equilibrium vapour pressure once equilibrium is restored. \textbf{Analysis of each operation:} \begin{enumerate}[(a)] \item \textbf{Moving piston downward:} Compresses the vapour; vapour condenses until equilibrium is re-established at the \emph{same} $P_{\text{vap}}$ (as long as liquid is still present). No change in vapour pressure. ✗ \item \textbf{Removing a small amount of vapour:} Disturbs equilibrium; more liquid evaporates immediately to restore $P_{\text{vap}}$. At equilibrium, vapour pressure is \emph{unchanged}. ✗ \item \textbf{Removing a small amount of liquid:} As long as some liquid remains, equilibrium is re-established at the same $P_{\text{vap}}$. No change. ✗ \item \textbf{Dissolving a non-volatile salt in the water (Raoult's law):} The mole fraction of water decreases: $P_{\text{vap}} = x_{\text{H}_2\text{O}} \cdot P^\circ_{\text{H}_2\text{O}}$. Since $x_{\text{H}_2\text{O}} < 1$, the equilibrium vapour pressure \emph{decreases}. ✓ \end{enumerate} \textbf{Answer: D} — Only dissolving a salt decreases the equilibrium vapour pressure.