Q. A certain mass of gas at $273\ K$ is expanded to 81 times its volume under adiabatic condition. If $\gamma = 1.25$ for the gas, then its final temperature is
A) $- 235℃$
B) $- 182℃$
C) $- 91℃$
D) $0℃$
Explanation
(b) For adiabatic process $TV^{\gamma - 1} =$ constant
$\Rightarrow \frac{T_{2}}{T_{1}} = \left( \frac{V_{1}}{V_{2}} \right)^{\gamma - 1} \Rightarrow T_{2} = \left( \frac{V_{1}}{V_{2}} \right)^{\gamma - 1} \times T_{1}$
$\Rightarrow T_{2} = \left( \frac{1}{81} \right)^{1.25 - 1} \times 273 = \left( \frac{1}{81} \right)^{0.25} \times 273$
$= \frac{273}{3} = 91K \rightarrow - 182℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.