For $\Delta T=150 K, \quad \Delta U=Q_{v}=6300 \mathrm{J}$
$\therefore \quad 6300=n C_{v}(150)$ $\ldots(1)$
Let the change in internal energy be $\Delta U$ for $\Delta T=300 K$
$\therefore \Delta U=n C_{v}(300)$ $...(2)$
Dividing $( 2 )$ and $( 1 )$ we get $\frac{\Delta U}{6300}=\frac{300}{150}$
$\Longrightarrow \Delta U=12600 \mathrm{J}$



Step $1$ It is first compressed adiabatically from volume $8.0 \,m ^{3}$ to $1.0 \,m ^{3}$.
Step $2$ Then expanded isothermally at temperature $T_{1}$ to volume $10.0 \,m ^{3}$.
Step $3$ Then expanded adiabatically to volume $80.0 \,m ^{3}$.
Step $4$ Then compressed isothermally at temperature $T_{2}$ to volume $8.0 \,m ^{3}$.
Then, $T_{1} / T_{2}$ is
