At $500 \, K$ temperature in a $2 \, L$ vessel,$0.32 \, g$ of $O_2$ gas is collected over water. If the vapour pressure of water is $32 \, mbar$ at $500 \, K$,find the partial pressure of dry $O_2$ gas. $(R = 8.314 \times 10^{-2} \, L \, bar \, mol^{-1} \, K^{-1})$

$1$. Calculate the moles of $O_2$: $n = \frac{0.32 \, g}{32 \, g/mol} = 0.01 \, mol$.
$2$. Calculate the total pressure $(P_{total})$ using the ideal gas equation $PV = nRT$: $P_{total} = \frac{nRT}{V} = \frac{0.01 \, mol \times 8.314 \times 10^{-2} \, L \, bar \, mol^{-1} \, K^{-1} \times 500 \, K}{2 \, L} = 0.20785 \, bar$.
$3$. Convert water vapour pressure to $bar$: $P_{H_2O} = 32 \, mbar = 0.032 \, bar$.
$4$. Calculate the partial pressure of dry $O_2$ using Dalton's Law: $P_{O_2} = P_{total} - P_{H_2O} = 0.20785 \, bar - 0.032 \, bar = 0.17585 \, bar$.