A gas occupies $100 \mathrm{mL}$ when collected over water at $15°\mathrm{C}$ and $750 \mathrm{mm}$ pressure. It occupies $92 \mathrm{mL}$ in dry state at $\mathrm{N}.\mathrm{T}.\mathrm{P}$. Find the vapour pressure of water (aqueous tension) at $15°\mathrm{C}$

Given conditions                     At N.T.P 

$\begin{array}{l}{P}_1=750\mathrm{mm}-p\mathrm{mm} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{P}_2=760\mathrm{mm}\\ V_1=100\mathrm{mL} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{V}_2=92\mathrm{mL}\\ T_1=15+273=288\mathrm{K} \hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{T}_2=273\mathrm{K}\end{array}$

  $\begin{array}{c}\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2};\\ \frac{(750-\mathrm{p})\mathrm{mm}\times 100\mathrm{ml}}{288\mathrm{K}}=\frac{760\mathrm{mm}\times 92\mathrm{mL}}{273\mathrm{K}}\end{array}$

$\begin{array}{l} \\ (750\times 100\times 273)-(100\times 273\mathrm{p})=760\times 92\times 288\\ 20475000-27300\mathrm{p}=20136960;27300\mathrm{p}=338040\\ \therefore \mathrm{p}=\frac{338040}{27300}=12.38\mathrm{mm}\end{array}$