Twenty-two grams of `CO_(2)` at `27^(@)C` is mixed with `16 g` of `O_(2)` at `37^(@)C`. The temperature of the mixture is about

Options

A. `31.5^(@)C`
B. `27^(@)C`
C. `37^(@)C`
D. `30.5^(@)C`

Explanation

To find the temperature of the mixture when 22 grams of CO₂ at 27°C is mixed with 16 grams of O₂ at 37°C, we will use the concept of heat exchange and the specific heat capacities of the gases involved.

### Step-by-Step Solution:

1. **Calculate the number of moles of CO₂ and O₂:**
- The molecular mass of CO₂ = 44 g/mol.
- The number of moles of CO₂ (n₁) = mass / molecular mass = 22 g / 44 g/mol = 0.5 mol.
- The molecular mass of O₂ = 32 g/mol.
- The number of moles of O₂ (n₂) = mass / molecular mass = 16 g / 32 g/mol = 0.5 mol.

2. **Determine the specific heat capacities (Cv) of the gases:**
- For CO₂, degrees of freedom (F) = 6, so Cv = (F/2)R = (6/2)R = 3R.
- For O₂, degrees of freedom (F) = 5, so Cv = (F/2)R = (5/2)R = 2.5R.

3. **Set up the heat exchange equation:**
- Heat gained by CO₂ = Heat lost by O₂.
- The heat gained by CO₂ can be expressed as:
\[ Q_{CO₂} = n₁ \cdot Cv_{CO₂} \cdot (T - T_{CO₂}) \]
where \(T_{CO₂} = 27°C\).
- The heat lost by O₂ can be expressed as:
\[ Q_{O₂} = n₂ \cdot Cv_{O₂} \cdot (T_{O₂} - T) \]
where \(T_{O₂} = 37°C\).

4. **Substitute the values into the equations:**
- For CO₂:
\[ Q_{CO₂} = 0.5 \cdot 3R \cdot (T - 27) \]
- For O₂:
\[ Q_{O₂} = 0.5 \cdot 2.5R \cdot (37 - T) \]

5. **Set the heat gained equal to the heat lost:**
\[ 0.5 \cdot 3R \cdot (T - 27) = 0.5 \cdot 2.5R \cdot (37 - T) \]

6. **Simplify the equation:**
- Cancel out 0.5R from both sides:
\[ 3(T - 27) = 2.5(37 - T) \]
- Expand both sides:
\[ 3T - 81 = 92.5 - 2.5T \]
- Combine like terms:
\[ 3T + 2.5T = 92.5 + 81 \]
\[ 5.5T = 173.5 \]
- Solve for T:
\[ T = \frac{173.5}{5.5} \approx 31.5°C \]

### Final Answer:
The temperature of the mixture is approximately **31.5°C**.