A monoatomic gas of mass 4.0 u is kept in an insulated container.

Given that mass of gas is $4 u$ hence its molar mass $M$ is $4 \mathrm{~g} / \mathrm{mol}$

$\therefore \frac{1}{2} \mathrm{mv}^{2}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$

$\frac{1}{2} \mathrm{~m} \times(30)^{2}=\frac{\mathrm{m}}{\mathrm{M}} \times \frac{3 \mathrm{R}}{2} \times \Delta \mathrm{T}$

$\therefore \Delta \mathrm{T}=\frac{3600}{3 \mathrm{R}}$