The volume V of a given mass of monoatomic gas changes with...

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

$⇒ P = \dfrac{nRT}{V} . . . \left(2\right)$

$⇒ W = ∫ \dfrac{nRT}{V}dv . . . \left(3\right)$

and $V = K{T}^{2 / 3} . . . \left(4\right)$

$⇒ W = ∫ \dfrac{nRT}{K{T}^{2 / 3}} \cdot dv . . . \left(5\right)$

$⇒$ from \left(4\right): dv = \dfrac{2}{3}K{T}^{- 1 / 3}dT$</p><p>$⇒ W = {∫}_{{T}_{1}}^{{T}_{2}}\dfrac{nRT}{K{T}^{2 / 3}}\dfrac{2}{3}K\dfrac{1}{{T}^{1 / 3}}dT$</p><p>$⇒ W = \dfrac{2}{3}nR \times \left({T}_{2} - {T}_{1}\right) . . . \left(6\right)$</p><p>$⇒ {T}_{2} - {T}_{1} = 90K . . . \left(7\right)$</p><p>$⇒ W = \dfrac{2}{3}nR \times 90$</p><p>$⇒ W = 60nR$</p><p>Assuming$1$mole of gas</p><p>$n = 1$</p><p>So$W = 60R