Q. At $27℃$ a gas suddenly compressed such that its pressure becomes $\frac{1}{8}th\ $of original pressure. The temperature of the gas will be $(\gamma = 5/3)$

A) $- 142℃$
B) 300K
C) $327{^\circ}$
D) 420 K
Explanation

(a) In adiabatic process, the relation between temperature (T) and pressure(p) is
$\frac{T^{\gamma}}{P^{\gamma - 1}} = constant$
Where$\ \gamma$ is ratio of specific heats.
Given, $T_{1} = 27℃ = 27 + 273 = 300\ K,$
$p_{1} = p,\ p_{2} = \frac{p}{8},\ \gamma = \frac{5}{3}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T_{1}}{T_{2}} = \left( \frac{p_{1}}{p_{2}} \right)^{\frac{\gamma - 1}{\gamma}}$
$\Rightarrow \ \ \ \frac{T_{1}}{T_{2}} = \left( \frac{8}{1} \right)^{\frac{\frac{5}{3} - 1}{5/3}}$
$\ \ \ \ \ = (8)^{0.4} = 2.297$
$\Rightarrow T_{2} = \frac{T_{1}}{2.297} = \frac{300}{2.297}$
$\ \ \ \ \ \ \ \ \ \ \ = 130.6\ K \approx 131K$
$\Rightarrow \ \ T_{2} = 131 - 273$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = - 142℃$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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