$22 \mathrm{~g}$ of carbon dioxide at $27^{\circ} \mathrm{C}$ is mixed in closed container with $16…
$22 \mathrm{~g}$ of carbon dioxide at $27^{\circ} \mathrm{C}$ is mixed in closed container with $16 \mathrm{~g}$ of oxygen at $37^{\circ} \mathrm{C}$. If both gases are considered as ideal gases, then the temperature of the mixture is nearly
$22.2^{\circ} \mathrm{C}$
$33.5^{\circ} \mathrm{C}$
$31.5^{\circ} \mathrm{C}$
$28.5^{\circ} \mathrm{C}$
Solution
Let $T^{\circ} \mathrm{C}$ be the temperature of the mixture.
heat given by $\mathrm{O}_2=$ heat absorbed by $\mathrm{CO}_2$
$\begin{aligned} & \Rightarrow \mu_1 C_{V_1} \Delta T=\mu_2 C_{V_2} \Delta T \\ & \Rightarrow \frac{22}{44}(3 R)(T-27)=\frac{16}{32}\left(\frac{5}{2} R\right)(37-T) \\ & \Rightarrow 3(T-27)=\frac{5}{2}(37-T) \\ & \Rightarrow 6 T-162=185-5 T \\ & \Rightarrow 11 T=347 \\ & \Rightarrow T=31.5 \circ \mathrm{C}\end{aligned}$