Solve this following

Question:

The volume $V$ of a given mass of monoatomic gas changes with temperature $\mathrm{T}$ according to the relation $\mathrm{V}=\mathrm{KT}^{2 / 3}$. The workdone when temperature changes by $90 \mathrm{~K}$ will be $\mathrm{xR}$. The value of $x$ is $[R=$ universal gas constant $]$

Solution:

We know that work done is

$\mathrm{W}=\int \mathrm{PdV}$  ....(1)

$\Rightarrow P=\frac{n R T}{V}$ ..............(2)

$\Rightarrow \mathrm{W}=\int \frac{\mathrm{nRT}}{\mathrm{V}} \mathrm{dv}$ ......................(3)

and $\mathrm{V}=\mathrm{KT}^{2 . / 3}$ ....................(4)

$\Rightarrow \mathrm{W}=\int \frac{\mathrm{nRT}}{\mathrm{KT}^{2 / 3}} \cdot \mathrm{dv}$ ....................(5)

$\Rightarrow$ from (4) : $\mathrm{dv}=\frac{2}{3} \mathrm{KT}^{-1 / 3} \mathrm{dT}$

$\Rightarrow \mathrm{W}=\int_{\mathrm{T}_{1}}^{\mathrm{T}_{2}} \frac{\mathrm{nRT}}{\mathrm{KT}^{2 / 3}} \frac{2}{3} \mathrm{~K} \frac{1}{\mathrm{~T}^{1 / 3}} \mathrm{dT}$

$\Rightarrow \mathrm{W}=\frac{2}{3} \mathrm{nR} \times\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right) \ldots$ (6)

$\Rightarrow \mathrm{T}_{2}-\mathrm{T}_{1}=90 \mathrm{~K}$...................(7)

$\Rightarrow \mathrm{W}=\frac{2}{3} \mathrm{nR} \times 90$

$\Rightarrow \mathrm{W}=60 \mathrm{nR}$

Assuming 1 mole of gas

$\mathrm{n}=1$

So $W=60 R$