Q. An ideal gas is expanding such that $pT^{2}$= constant. The coefficient of volume expansion of the gas is

A) $\frac{1}{T}$
B) $\frac{2}{T}$
C) $\frac{3}{T}$
D) $\frac{4}{T}$
Explanation

(c) Given, $pT^{2}$=constant
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \frac{nRT}{V} \right)T^{2} =$constant
or $T^{3}V^{- 1} =$ constant
Differentiating the equation, we get
$\frac{3T^{2}}{V}.dT - \frac{T^{3}}{V^{2}}.dV = 0$
or $3.dT = \frac{T}{V}.dV$
From the equation, $dV = V_{\gamma}.dT$
$\gamma =$ coefficient of volume expansion of gas
$=$ $\frac{dV}{V.dT}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \gamma = \frac{dV}{V.dT} = \frac{3}{T}$

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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