$\mathrm{R}=8.31 \mathrm{~J} \quad \mathrm{~mol}^{-1} \mathrm{deg}^{-1}=\frac{8.31}{4.18} \mathrm{cal} \mathrm{mol}^{-1} \mathrm{deg}^{-1}$ $=1.99 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{deg}^{-1}$
We know that $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$
$$
\text { or } \mathrm{Cv}=\mathrm{Cp}-\mathrm{R}=7.03-1.99=5.04 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{deg}^{-1}
$$
Heat absorbed by 5 mole of oxygen in heating from $10^{\circ} \mathrm{C}$ to $20^{\circ} \mathrm{C}$
$$
=5 \times \mathrm{C}_{\mathrm{V}} \times \Delta \mathrm{T}=5 \times 5.04 \times 10=252 \mathrm{cal}
$$
Since, the gas is heated at constant volume, no extemal work is done, i.e. $_{n} \mathrm{w}=0$
So, change in internal energy will be equal to heat absorbed,
$$
\Delta \mathrm{U}=\mathrm{q}+\mathrm{w}=252+0=252 \mathrm{cal}
$$