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 final temperature of the mixture of `CO2` and `O2`, we can use the principle of conservation of energy, where the heat gained by `CO2` is equal to the heat lost by `O2`.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Mass of `CO2` (m1) = 22 g
- Temperature of `CO2` (T1) = 27°C
- Mass of `O2` (m2) = 16 g
- Temperature of `O2` (T2) = 37°C

2. **Calculate Moles of Each Gas:**
- Molar mass of `CO2` = 44 g/mol
- Molar mass of `O2` = 32 g/mol
- Number of moles of `CO2` (n1) = m1 / Molar mass of `CO2` = 22 g / 44 g/mol = 0.5 mol
- Number of moles of `O2` (n2) = m2 / Molar mass of `O2` = 16 g / 32 g/mol = 0.5 mol

3. **Use the Heat Transfer Equation:**
- Heat gained by `CO2` = Heat lost by `O2`
- For an ideal gas, the heat transfer can be expressed as:
\[ n_1 C_{v1} (T_f - T_1) = n_2 C_{v2} (T_2 - T_f) \]
- Where \( C_{v1} \) and \( C_{v2} \) are the molar heat capacities at constant volume for `CO2` and `O2`, respectively.

4. **Assign Degrees of Freedom:**
- For `CO2` (a linear molecule), \( C_{v1} = \frac{5}{2} R \) (F = 6)
- For `O2` (a diatomic molecule), \( C_{v2} = \frac{5}{2} R \) (F = 5)

5. **Set Up the Equation:**
\[ 0.5 \cdot \frac{5}{2} R (T_f - 27) = 0.5 \cdot \frac{5}{2} R (37 - T_f) \]
- The \( R \) and \( 0.5 \) cancel out:
\[ \frac{5}{2} (T_f - 27) = \frac{5}{2} (37 - T_f) \]

6. **Simplify the Equation:**
- Cancel \( \frac{5}{2} \):
\[ T_f - 27 = 37 - T_f \]
- Rearranging gives:
\[ 2T_f = 64 \quad \Rightarrow \quad T_f = 32°C \]

7. **Final Temperature:**
- The final temperature of the mixture is approximately **32°C**.