Question

A gas expands with temperature according to the relation $V=K{T}^{2/3}$ . Work done when the temperature changes by $60K$ is

• A. $10R$
• B. $30R$
• C. $40R$
• D. $20R$

Answer 1

Answer: (C)

Here,
$dW=PdV=\frac{RT}{V}dV(\because PV=RT)...(i)$
Given $V=K{T}^{2/3}$
$\therefore dV=K\frac{2}{3}{T}^{-1/3}dT$
$\therefore \frac{dV}{V}=\frac{K\frac{2}{3}{T}^{-1/3}dT}{K{T}^{2/3}}$
$=\frac{2}{3}\frac{dT}{T}...(ii)$
From $(i),W=\underset{T_1}{\overset{T_2}{\int }}RT\frac{dV}{V}$
$=\underset{T_1}{\overset{T_2}{\int }}RT\frac{2}{3}\frac{dT}{T}$ (Using $(ii))$
$\therefore W=\frac{2}{3}R(T_2-T_1)$
$\frac{2}{3}R\times 60=40R$