An open flask contains air at  ${27}^{0}C.$ To what temperature it must be heated to expel one-fourth of the air?

Suppose the volume  $\left(V_2\right)$ of the air after expansion be  $Vc{m}^3.$
Volume of air expelled  
Therefore, volume of flask, i.e., volume $\left(V_1\right)$  of air at ${27}^{0}C$ is
 $V-\frac{V}{4}=\frac{3}{4}V$
Thus  $V_1=\frac{3}{4}V,T_1=27+273=300K$
$V_2=V,T_2=?$
According to Charles’law,
$\begin{array}{l}\frac{V_1}{T_1}=\frac{V_2}{T_2},i.e.,\frac{\frac{3}{4}{V}_1}{300}=\frac{V}{T_2}\\ T_2=300\times \frac{4}{3}=400K-273={127}^{0}C\end{array}$