22 g of CO₂ at 27°C is mixed with 16g of O₂ at 37°C. The temperature of the mixture is

To find the equilibrium temperature of the mixture, we can use the principle of conservation of energy. The heat lost by the hotter gas (oxygen) must equal the heat gained by the cooler gas (carbon dioxide) under isolated conditions.

First, let us calculate the number of moles of each gas:

For Carbon Dioxide (CO₂):
Given mass, $m_1=22\text{g}$
Molar mass of CO₂, $M_1=44\text{g/mol}$
Number of moles, $n_1=\frac{22}{44}$
Initial temperature, $T_1=27\text{°C}$

For Oxygen (O₂):
Given mass, $m_2=16\text{g}$
Molar mass of O₂, $M_2=32\text{g/mol}$
Number of moles, $n_2=\frac{16}{32}$
Initial temperature, $T_2=37\text{°C}$

Next, let us consider the molar specific heat capacity at constant volume ($C_v$) for each gas:
O₂ is a diatomic gas, so its degrees of freedom at room temperature are $f=5$.
CO₂ is a linear triatomic gas, which also has $f=5$ degrees of freedom (excluding vibrational modes at moderate temperatures).
Thus, the molar heat capacity at constant volume for both gases is equal:
$C_{v1}=C_{v2}=\frac{5}{2}$

Let $T_{\text{mix}}$ be the equilibrium temperature of the mixture. According to the conservation of internal energy:
$n_1{C}_{v1}(T_{\text{mix}}-T_1)=n_2{C}_{v2}(T_2-T_{\text{mix}})$

Since $n_1=n_2=0.5\text{mol}$ and $C_{v1}=C_{v2}$, the equation simplifies directly to:
$T_{\text{mix}}-T_1=T_2-T_{\text{mix}}$

Solving for $T_{\text{mix}}$:
$2T_{\text{mix}}=T_1+T_2$
$T_{\text{mix}}=\frac{T_1+T_2}{2}$

Substitute the initial temperatures:
$T_{\text{mix}}=\frac{27+37}{2}$

Therefore, the temperature of the mixture is 32°C.