Q. A gas expands with temperature according to the relation $V = kT^{2/3}$. Calculate work done when the temperature changes by 60K?
A) $10\ R$
B) $30\ R$
C) $40\ R$
D) $20\ R$
Explanation
(c) As $V = KT^{2/3}\ \ \therefore\ \ dV = K\frac{2}{3}\ T^{- 1/3}\ dT$
$\therefore\ \frac{dV}{V} = \frac{\frac{2}{3}KT^{- 1/3}dT}{KT^{2/3}} = \frac{2}{3}\frac{dT}{T}$
Work done, $W = \int_{T_{1}}^{T_{2}}{RT\frac{dV}{V} = \int_{T_{1}}^{T_{2}}{RT\frac{2}{3}\frac{dT}{T}}}$
$W = \frac{2}{3}R\left( T_{2} - T_{1} \right) = \frac{2}{3}R \times 60 = 40R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.