Question

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

• A. 100
• B. 75
• C. 75.83
• D. 82.63

Answer 1

The correct option is B 75

Compressed suddenly implies adiabatic process; temperature and pressure are related as
$P^{1-\gamma }{T}^{\gamma }=$constant
${\left(\frac{P_1}{P_2}\right)}^{1-\gamma }={\left(\frac{T_2}{T_1}\right)}^{\gamma }$
${\left(\frac{P}{\frac{P}{32}}\right)}^{1-\frac{5}{3}}={\left(\frac{T_2}{300}\right)}^{5/3}$
${\left(32\right)}^{\frac{-2}{3}}={\left(\frac{T}{300}\right)}^{\frac{5}{3}}$
$\frac{1}{{32}^2}={\left(\frac{T}{300}\right)}^5$
$\frac{T}{300}={\left(\frac{1}{{32}^2}\right)}^{\frac{1}{5}}=\frac{1}{(2^5{)}^{2/5}}=\frac{1}{4}$
$T=\frac{300}{4}=75K$
Option B.