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Q. The vapour pressure of water at $80^{\circ}C$ is $355.5$ mm $Hg. A 100 mL$ vessel contains water saturated with $O_{2}$ at $80^{\circ}C$, the total pressure being $760$ mm of $Hg$. The contents of the vessel were pumped into a $50 mL$ vessel at the same temperature. What is the partial pressure of $O_{2}$?

States of Matter

Solution:

$p_{\text{Total}} = p_{H_2O} + p_{O_2}$ $p_{H_2O} = 760 -355.5 = 404.5\, mm \,Hg$
When the contents were pumped into $50\, mL$ vessel
$p_{1} - 404.5\, mm\, Hg$, $V_{1} = 100 mL, p_{2} = ? V_{2} = 50\, mL$
$P_{1}V_{1} =P_{2}V_{2}$
$p_{2} = \frac{404.5 \times 100}{50} = 809$
Thus, $p_{O_2}$ in $50 mL$ vessel = $809\, mm\, Hg$