Dry air  $(\gamma =\frac{3}{2})$ at atmospheric pressure is suddenly compressed to ${\frac{\text{1}}{\text{4}}}^{\text{th}}$ of its original volume, the pressure will be

$\begin{array}{rcl} & & {\text{Pressure P}}_{\text{1}}\text{=P=1\hspace{0.17em}atm}\\ & & to find {\text{ P}}_{\text{2}}\\ & & given volume {\text{V}}_{\text{1}}{\text{=V and :V}}_{\text{2}}\text{=}\frac{\text{V}}{\text{4}}\\ & & adiabatic process(it is a sudden process) relation between P and V is{\text{PV}}^{\gamma }\text{=Constant}\\ & & \frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{=}{\left(\frac{{\displaystyle {\text{V}}_{\text{2}}}}{{\displaystyle {\text{V}}_{\text{1}}}}\right)}^{\gamma }\\ & & substitute the given values in above equation\\ & & \frac{\text{P}}{{\text{P}}_{\text{2}}}\text{=}{\left(\frac{{\displaystyle \frac{V}{4}}}{{\displaystyle \text{V}}}\right)}^{\frac{3}{2}}\\ & \Rightarrow & \frac{\text{P}}{{\text{P}}_{\text{2}}}\text{=}\frac{1}{8}\\ {\text{P}}_{\text{2}}& =& 8\text{P given P= 1 atm}\\ & \Rightarrow & {\text{P}}_{\text{2}}\text{=8 atm}\end{array}$