solution:
\[\begin{array}{*{35}{l}}
~ \\
pVM\text{ }=\text{ }mRT \\
~ \\
V=\text{ }mRT/Mp \\
\end{array}\]
Here,
\[\begin{array}{*{35}{l}}
m\text{ }=\text{ }8.8\text{ }g \\
~ \\
R\text{ }=\text{ }0.083\text{ }bar\text{ }LK1\text{ }mol1 \\
~ \\
T\text{ }=\text{ }31.1{}^\circ C\text{ }=\text{ }304.1\text{ }K \\
~ \\
M\text{ }=\text{ }44\text{ }g\text{ }p\text{ }=\text{ }1\text{ }bar \\
\end{array}\]
In this manner, the volume involved is 5.05 L.