Question

$22g$ of carbon dioxide at $27^\circ C$ is mixed in a closed container with $16g$ of oxygen at $37^\circ C$ . If both gases are considered as ideal gases, then the temperature of the mixture is (neglect vibrational degree of freedom)

• A. $24.2^\circ C$
• B. $28.5^\circ C$
• C. $31.5 °C$
• D. $33.5^\circ C$

Answer 1

Answer: (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 $C_{V}_{1}=\frac{5 R}{2}$
For oxygen, number of moles $\left(n_{2}\right)=\frac{16}{32}=\frac{1}{2}$
Molar specific heat of $O_{2}$ at constant volume $C_{V}_{2}=\frac{5 R}{2}$
Let $TK$ be the temperature of mixture.
Heat lost by $O_{2}$ = Heat gained by $CO_{2}$
$n_{2}C_{V}_{2}\Delta T_{2}=n_{1}C_{V}_{1}\Delta T_{1}$
$\frac{1}{2} \, \left(\frac{5}{2} \, R\right)\left(310 - T\right)=\frac{1}{2}\times \left(\frac{5 R}{2}\right)\left(T - 300\right)$
$2T=610$
$T=\frac{610}{2}=305K=31.85^\circ C\sim eq31.5^\circ C$