Question

If a gas expands with temperature according to the relation $V=K{T}^{2/3}$, then what will be work done, when the temperature changes by ${30}^{\circ }C$?

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

Answer 1

Answer: (C)

$20R$

Work done,
$W=\int pdv=\int \frac{RT}{V}dV$
Since, $V=K{T}^{2/3}$
$dV=\frac{2}{3}K{T}^{-1/3}dT$
Eliminating $K$, we find $\frac{dV}{V}=\frac{2}{3}\frac{dT}{T}$
Hence, $W={\int }_{T_1}^{T_2}\frac{2}{3}\frac{RT}{T}dT$
$=\frac{2}{3}R(T_2-T_1)=\frac{2}{3}R\left(30\right)=20R$.