To find the temperature of the mixture of oxygen and nitrogen gases, we will follow these steps:
### Step 1: Calculate the number of moles of each gas
- For oxygen (O2):
\[
\text{Moles of } O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{16 \, \text{g}}{32 \, \text{g/mol}} = 0.5 \, \text{mol}
\]
- For nitrogen (N2):
\[
\text{Moles of } N_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{14 \, \text{g}}{28 \, \text{g/mol}} = 0.5 \, \text{mol}
\]
### Step 2: Convert temperatures from Celsius to Kelvin
- For oxygen at \(37^\circ C\):
\[
T_1 = 37 + 273 = 310 \, \text{K}
\]
- For nitrogen at \(27^\circ C\):
\[
T_2 = 27 + 273 = 300 \, \text{K}
\]
### Step 3: Use the formula for the final temperature of the mixture
Since both gases are diatomic and have the same heat capacity at constant volume (\(C_v\)), we can use the formula:
\[
T_f = \frac{N_1 T_1 + N_2 T_2}{N_1 + N_2}
\]
Where:
- \(N_1\) is the number of moles of oxygen
- \(N_2\) is the number of moles of nitrogen
- \(T_1\) is the initial temperature of oxygen
- \(T_2\) is the initial temperature of nitrogen
Substituting the values:
\[
T_f = \frac{(0.5 \, \text{mol} \times 310 \, \text{K}) + (0.5 \, \text{mol} \times 300 \, \text{K})}{0.5 \, \text{mol} + 0.5 \, \text{mol}}
\]
### Step 4: Calculate the final temperature
Calculating the numerator:
\[
= 155 + 150 = 305 \, \text{K}
\]
Calculating the denominator:
\[
= 1 \, \text{mol}
\]
Thus:
\[
T_f = \frac{305 \, \text{K}}{1} = 305 \, \text{K}
\]
### Step 5: Convert the final temperature back to Celsius
To convert from Kelvin to Celsius:
\[
T_f = 305 \, \text{K} - 273 = 32 \, ^\circ C
\]
### Final Answer
The final temperature of the mixture is \(305 \, \text{K}\) or \(32 \, ^\circ C\).
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