At $27^{o}C$ a gas compressed suddenly such that its pressure becomes $\dfrac{1}{32}$ of original pressure. Final temperature in Kelvin will be $\left ( \gamma =5/3 \right )$

Compressed suddenly implies adiabatic process; temperature and pressure are related as

$P^{1-\gamma }T^{\gamma }=$constant

$\displaystyle \left (\dfrac {P_1}{P_2}\right )^{1-\gamma }=\left (\dfrac {T_2}{T_1}\right )^{\gamma }$

$\displaystyle \left (\dfrac {P}{\dfrac {P}{32}}\right )^{1-\dfrac {5}{3}}=\left (\dfrac {T_2}{300}\right )^{5/3}$

$\displaystyle \left (32\right )^{\dfrac {-2}{3}}=\left (\dfrac {T}{300}\right )^{\dfrac {5}{3}}$

$\displaystyle \dfrac {1}{32^2}=\left (\dfrac {T}{300}\right )^5$

$\displaystyle \dfrac {T}{300}=\left (\dfrac {1}{32^2}\right )^{\dfrac {1}{5}}=\dfrac {1}{(2^5)^{2/5}}=\dfrac {1}{4}$

$\displaystyle T=\dfrac {300}{4}=75K$

Option B.