Question

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

• A. 10R
• B. 30R
• C. 40R
• D. 20R

Answer 1

The correct option is C 40R

Here

$dW=PdV=\frac{RT}{V}dV$

Given $V=K{T}^{2/3}$

therefore $dV=K\frac{2}{3}\frac{dT}{T}$

From (i) $W={\int }_{T_i}^{T_2}RT\frac{dV}{V}={\int }_{T_i}^{T_2}RT\frac{2}{3}\frac{dT}{T}$

$W=\frac{2}{3}R(T_2-T_1)=\frac{2}{3}R\times 60=40R$