# Question bfabb

Apr 25, 2014

The dry gas will occupy 108.7 mL.

This is an example of a Combined Gas Laws problem.

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

The "trick" is to calculate the partial pressure of the dry gas. Both water vapour and the gas itself contribute to the pressure of the "wet" gas. According to Dalton's Law of Partial Pressures,

${P}_{\text{gas") + P_("H₂O") = P_("atm}}$

At 25.0 °C, the vapour pressure of water is 23.8 torr.

${P}_{\text{gas") = P_("atm") - P_("H₂O}}$ = (746 – 24.8) torr = 721 torr

${T}_{1}$ = (25.0 + 273.15) K = 298.2 K

Summary:
${P}_{1}$ = 721 torr; ${V}_{1}$ = 125.0 mL; ${T}_{1}$ = 298.2 K
${P}_{2}$ = 760 torr; ${V}_{2}$ = ?; ${T}_{2}$ = 273.15 K

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

V_2 = V_1 × P_1/P_2 × T_2/T_1# =

125.0 mL × $\left(721 \text{ torr")/(760" torr") × (273.15" K")/(298.2" K}\right)$ = 108.7 mL

The volume of the dry gas is 108.7 mL