Question

A gas expands with temperature according to the relation $V=k{T}^{2/3}$. What is the work done, when the temperature changes by ${30}^{\circ }\text{C}$?

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

Answer 1

Answer: (B)

$20R$

Given,
$V=k{T}^{2/3}$
Change in Temperature, $\left(\Delta T\right)=T_2-T_1={30}^{\circ }\text{C}$
Now, we know that
$W=\int PdV$
Using,ideal gas equation in the above formula we get,
$W=\int \frac{RT}{V}dV$
Since, $V=k{T}^{2/3}\Rightarrow dV=\frac{2}{3}k{T}^{-1/3}dT$
Eliminating $k$, we find $\frac{dV}{V}=\frac{2}{3}\frac{dT}{T}$
Hence $W={\int }_{T_1}^{T_2}\frac{2}{3}\frac{RT}{T}dT=\frac{2}{3}R(T_2-T_1)=\frac{2}{3}R\left(30\right)=20R$
Thus, option (b) is the correct answer.