Question

A gas expands with temperature according to the relation, $V=k T^{2 / 3}$, where $k$ is a constant. Work done when the temperature changes by $60\, K$ is $( R =$ universal gas constant. $)$

• A. 10 R
• B. 20 R
• C. 50 R
• D. 40 R

Answer 1

Answer: (D)

40 R

Answer 2

Given, $V=k T^{2 / 3}$
From definition of work done,
$dW =P\, dV=\frac{R T}{V} d V$
$=\frac{R T}{k T^{2 / 3}} d V\,\,\,\,\,\,\,\,\,\dots(i)$
Now, $V=k T^{2 / 3}$
Taking derivative on the both sides, we get
$d V=\kappa \frac{2}{3} T^{-1 / 3} d T\,\,\,\,\,\,\,\,\dots(ii)$
Substituting the value of Eqs. (ii) in Eqs. (i) and taking integration on the both sides, we get
$W =\frac{2}{3} R \int\limits_{T_{1}}^{T_{2}} d T=\frac{2}{3} R\left(T_{2}-T_{1}\right)$
$=\frac{2}{3} R(60-0)=\frac{2}{3} \times R \times 60$
$=40 \,R$