Question

When the gas expands with temperature using the relation $V=KT^{2/3}$ for the temperature change of $40\,K,$ the work done is

• A. $20.1\,\,R$
• B. $30.2\,\,R$
• C. $26.6\,\,R$
• D. $35.6\,\,R$

Answer 1

Answer: (C)

$26.6\,\,R$

Answer 2

$W=\int P d V=\int \frac{\mu R T}{V} d V$
$=\int \frac{\mu R T}{K T^{\frac{2}{3}}} d V$ ...(i)
$\because V=K T^{\frac{2}{3}}$
$d V=K \frac{2}{3} T^{-1 / 3} d T$ ...(ii)
From Eqs. (i) and (ii)
$W=\int \frac{\mu R T}{K T^{\frac{2}{3}}} \times K \frac{2}{3} T^{-1 / 3} d T$
$=\frac{2}{3} R \int T^{1-\frac{1}{3}-\frac{2}{3}} d T$
$=\frac{2}{3} R \int_{0}^{40} T^{0} d T=\frac{2}{3} R(T)_{0}^{40}$
$=\frac{2}{3} R(40-0)=\frac{80 R}{3}=26.6 \,R$