A 1-litre flask contains air, water vapour, and a negligible amount of liquid water at a total press — States of Matter and Gaseous State Chemistry Question
Question
A 1-litre flask contains air, water vapour, and a negligible amount of liquid water at a total pressure of $200\text{ mm Hg}$. If this flask is connected to another 1-litre evacuated flask and the stopcock is opened, what will be the final total pressure of the gas mixture at equilibrium? (Assume the temperature is constantly maintained at $50^\circ\text{C}$ and the aqueous tension at $50^\circ\text{C}$ is $93\text{ mm Hg}$).
💡 Solution & Explanation
The initial pressure of the dry air is $P_{dry} = 200 - 93 = 107\text{ mm Hg}$. When connected to the evacuated 1-litre flask, the total volume doubles (from 1L to 2L). By Boyle's law, the partial pressure of the dry air halves to $107 / 2 = 53.5\text{ mm Hg}$. Since liquid water is present and temperature is constant, aqueous tension remains constant at $93\text{ mm Hg}$. Total final pressure = $53.5 + 93 = 146.5\text{ mm Hg}$.