Question

The vapour pressure of water at ${80}^{\circ }C$ is 355 torr. A 100 ml vessel contained water - saturated oxygen at ${80}^{\circ }C$, the total gas pressure being 730 torr. The contents of the vessel were pumped into 50.0 ml, vessel at the same temperature. What were partial pressures of oxygen and of vapour and the total pressure in the final equilibrium state ? Neglect the volume of any water might condense.

Answer 1

Vapour pressure of water remains constant in $100ml$ & $50ml$ as temperature remains constant.

$P_T=P_{O_2}+P_{V.P.of{H}_{2}O}$

$\Rightarrow 730=P_{O_2}+355$

$\Rightarrow P_{O_2}=730-355=375mm$

Now, $P_1{V}_1=P_2{V}_2$

$\Rightarrow 375\times 100=P_2\times 50$

$\Rightarrow P_2=\frac{375\times 100}{50}=750mm$= Partial pressure of oxygen

Now, $P_T$ in $50ml$ container

= $P_2+$ Vapour pressure of water

=$750+355$

=$1105mm$.