# Question #4a240

${V}_{2} \cong 40 \cdot L$
$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$
All temperature is reported on the $\text{Absolute scale}$; for $\text{pressure}$ we use the relationship $760$ $m m$ $H g$ $\equiv$ $1 \cdot a t m .$
${V}_{2} = {T}_{2} / \left({P}_{2}\right) \times \frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{256.8 \cdot K}{\frac{385 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1}} \times \frac{1}{303.5 \cdot K} \times \left(\frac{731 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1}\right) \times 26.5 \cdot L \cong 40 \cdot L$