A closed container contains gas and some liquid water, exerting a total pressure of at . The aqueous — States of Matter and Gaseous State Chemistry Question
Question
A closed container contains $O_2$ gas and some liquid water, exerting a total pressure of $740\text{ mm Hg}$ at $27^\circ\text{C}$. The aqueous tension at $27^\circ\text{C}$ is known to be $20\text{ mm Hg}$. If the volume of the container is suddenly reduced to exactly half of its original volume at the same constant temperature, what will be the new total pressure? (Assume the volume of liquid water is completely negligible).
💡 Solution & Explanation
Initial pressure of dry $O_2$ is $P_{dry} = 740 - 20 = 720\text{ mm Hg}$. When the volume is reduced to half, Boyle's law dictates that the partial pressure of the dry $O_2$ will double: $720 \times 2 = 1440\text{ mm Hg}$. Because liquid water is still present and temperature is constant, the aqueous tension remains $20\text{ mm Hg}$. The new total pressure is $1440 + 20 = 1460\text{ mm Hg}$.