The $O_2$ gas is collected over water at $400 \ K$ temperature in a $2 \ L$ vessel. If the total pressure of the mixture is $33.26 \ bar$ and the pressure of dry $O_2$ gas is $32.20 \ bar$,find the vapour pressure of water under the same conditions.

(1.06 BAR) According to Dalton's Law of partial pressures,the total pressure of a gas collected over water is the sum of the pressure of the dry gas and the vapour pressure of water.
$P_{\text{total}} = P_{\text{dry gas}} + P_{\text{water vapour}}$
Given:
$P_{\text{total}} = 33.26 \ bar$
$P_{\text{dry } O_2} = 32.20 \ bar$
Therefore,
$P_{\text{water vapour}} = P_{\text{total}} - P_{\text{dry } O_2}$
$P_{\text{water vapour}} = 33.26 \ bar - 32.20 \ bar = 1.06 \ bar$