1. Given relation

For a given mass of a monoatomic ideal gas, $V = K T^{2/3}$ where $K$ is a constant.

We need the work done when temperature changes by $90\,\text{K}$.

2. Use ideal gas law

For 1 mole of gas, $PV = RT$ So, $P = \frac{RT}{V}$

Using $V = KT^{2/3}$, $P = \frac{RT}{K T^{2/3}} = \frac{R}{K}T^{1/3}$

3. Find infinitesimal work

Work done, $dW = P\,dV$

Now differentiate $V = KT^{2/3}$: $dV = K \cdot \frac{2}{3}T^{-1/3} dT$

Therefore, $dW = \left(\frac{R}{K}T^{1/3}\right)\left(K\frac{2}{3}T^{-1/3}dT\right)$

Simplifying, $dW = \frac{2R}{3}dT$

4. Integrate for temperature change of $90\,\text{K}$

$W = \int dW = \int \frac{2R}{3} dT = \frac{2R}{3}\Delta T$

Given, $\Delta T = 90\,\text{K}$

So, $W = \frac{2R}{3}\times 90 = 60R$

Thus, if work done is written as $xR$, $x = 60$

5. Comparison with stored answer

Stored correct answer: $60$

Our derived answer is also $60$.