Answer
$\Delta E_{rot}=1KJ$
Work Step by Step
As $\Delta E_{int}=\Delta E_{trans}+\Delta E_{rot}$
$\implies \Delta E_{rot}=\Delta E_{int}-\Delta E_{trans}$..........eq(1)
We know that the formula of $\Delta E_{trans}$ is:
$\Delta E_{trans}=\frac{3}{2}nR\Delta T$
Substituting the values of $n,R$ and $\Delta T$ in the formula:
$\Delta E_{trans}=\frac{3}{2}(3)(8.314)(40)=1.49\times10^3J$
We already know from part (b) that the change in internal energy is $2.49\times10^3$. Substituting the values of $\Delta E_{int}$ and $\Delta E_{trans}$ in eq (1) and solving:
$\Delta E_{rot}=2.49\times10^3-1.49\times10^3=1\times10^3J=1KJ$
