Question

A given mass of a gas occupies 919 mL volume in dry state at STP. Same mass of the gas collected over water at ${15}^{\circ }$C & 750 mm of pressure occupies a volume of 1 L. Calculate vapor pressure of water at ${15}^{\circ }$C.

Answer 1

Let us consider the

$P_{drygas}=P_{total}=P_{watervapour}$

Given that:

$\left(AtSTP\right)$

$V_1=919mL$

$V_2=100mL$

$P_1=760mmHg$

$P_2=?\left(drygas\right)$

$T_1=273K$

$T_2=273+15=288K$

$P_{total}=750mmHg$

Using gas equation

$\frac{p_1{v}_1}{T_1}=\frac{p_2{v}_2}{T_2}$

implies that

$\frac{760\times 919}{273}=\frac{p_2\times 1000}{288}$

implies that

$p_2=736.82mmHg$

Hence

$P_{drygas}=P_{total}-P_{watervapour}$

implies that

$P_{watervapour}=750-736.82$

$13.18mm$