The volume V of a given mass of monoatomic gas changes with temperature T according to the relation V = K T…

We know that work done is $W=\int PdV ...\left(1\right)$

$\Rightarrow P=\frac{nRT}{V} ...\left(2\right)$

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

and $V=K{T}^{2/3} ...\left(4\right)$

$\Rightarrow W=\int \frac{nRT}{K{T}^{2/3}}·dv ...\left(5\right)$

$\Rightarrow$ from $\left(4\right) : dv=\frac{2}{3}K{T}^{-1/3}dT$

$\Rightarrow W={\int }_{T_1}^{T_2}\frac{nRT}{K{T}^{2/3}}\frac{2}{3} K\frac{1}{ T^{1/3}}dT$

$\Rightarrow W=\frac{2}{3}nR\times \left(T_2-T_1\right) ...\left(6\right)$

$\Rightarrow T_2-T_1=90 \mathrm{K} ...\left(7\right)$

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

$\Rightarrow W=60nR$

Assuming $1$ mole of gas

$n=1$

So $W=60R$