Question

The vapor pressure of water at ${80}^{\circ }C$ is $355$ torr. A $100ml$ vessel contained water-saturated oxygen at ${80}^{\circ }C$ the total gas pressure being $760$ torr. The contents of the vessel were pumped into a $50.0$ ml, vessel at the same temperature, What were the partial pressure of oxygen and of water vapor, what was the total pressure in the final equilibrated state? Neglect the volume of any water which might condense.

Answer 1

Total pressure $={p_{O_2}}^o+{p_{H_{2}O}}^o$
$\begin{array}{c}760={p_{O_2}}^o+355\\ {p_{O_2}}^o=405torr\end{array}$
Since, $p_1{v}_1=p_2{v}_2$
$\begin{array}{c}{p_{H_{2}O}}^o=\frac{{p_{O_2}}^o\times 100}{50}=\frac{405\times 100}{50}=810\\ {p_{H_{2}O}}^o=810torr\end{array}$
total pressure in $50$ml contain
$=810+355$
$=166torr$