Q. In an adiabatic change, the pressure and temperature of a monoatomic gas are related as $p{\propto T}^{- c}$ where $c$ equals
A) $\frac{2}{5}$
B) $\frac{5}{2}$
C) $\frac{3}{5}$
D) $\frac{5}{3}$
Explanation
(a) In an adiabatic change,
$p^{1 - \gamma}T^{\gamma} = constant$
Or $pT^{\gamma/1 - \gamma} = constant$
Or $p \propto T^{(1 - \gamma)/\gamma}$
Thus, $c = \frac{1 - \gamma}{\gamma}$
For a monoatomic gas, $\gamma = \frac{5}{3}$
$\therefore\ - c = \frac{1 - 5/3}{5/3} = - \frac{2}{3}\ \Rightarrow c = \frac{2}{5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.