Explanation

(c) For carbon dioxide, number of moles $\left( n_{1} \right) = \frac{22}{44} = \frac{1}{2}\ ;$
molar specific heat of $CO_{2}$ at constant volume ${C_{V}}_{1} = 3\ R$
For oxygen, number of moles $\left( n_{2} \right) = \frac{16}{32} = \frac{1}{2};$
molar specific heat of $O_{2}$ at constant volume ${C_{V}}_{2} = \frac{5R}{2}.$
Let $TK$ be the temperature of mixture.
Heat lost by $O_{2}$ = Heat gained by $CO_{2}.$
$n_{2}{C_{V}}_{2}\mathrm{\Delta}T_{2} = n_{1}{C_{V}}_{1}\mathrm{\Delta}T_{1}$
$\frac{1}{2}\ \left( \frac{5}{2}\ R \right)(310 - T) = \frac{1}{2} \times (3R)(T - 300)$
Or 1550$- 5T = 6T - 1800$
Or $T = 304.54K = 31.5℃$