Calculate the average molar heat capacity at constant volume of a mixture containing 2 moles of monoatomic and 3 moles of diatomic ideal gas.

Options

A. R
B. 2.1 R
C. 3.2 R
D. 4 R

Answer

To calculate the average molar heat capacity at constant volume (\(C_{V,\text{avg}}\)) of a mixture containing 2 moles of a monoatomic ideal gas and 3 moles of a diatomic ideal gas, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the number of moles and specific heat capacities:**
- For the monoatomic gas:
- Number of moles (\(n_1\)) = 2
- Molar heat capacity at constant volume (\(C_{V1}\)) = \(\frac{3}{2}R\)
- For the diatomic gas:
- Number of moles (\(n_2\)) = 3
- Molar heat capacity at constant volume (\(C_{V2}\)) = \(\frac{5}{2}R\)

2. **Use the formula for average molar heat capacity of the mixture:**
\[ C_{V,\text{avg}} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2} \]

3. **Substitute the values into the formula:**
\[ C_{V,\text{avg}} = \frac{(2 \cdot \frac{3}{2}R) + (3 \cdot \frac{5}{2}R)}{2 + 3} \]

4. **Calculate the contributions from each gas:**
- Contribution from the monoatomic gas:
\[ 2 \cdot \frac{3}{2}R = 3R \]
- Contribution from the diatomic gas:
\[ 3 \cdot \frac{5}{2}R = \frac{15}{2}R \]

5. **Combine the contributions:**
\[ C_{V,\text{avg}} = \frac{3R + \frac{15}{2}R}{5} \]

6. **Convert \(3R\) into a fraction with a common denominator:**
\[ 3R = \frac{6}{2}R \]
Thus,
\[ C_{V,\text{avg}} = \frac{\frac{6}{2}R + \frac{15}{2}R}{5} = \frac{\frac{21}{2}R}{5} \]

7. **Simplify the expression:**
\[ C_{V,\text{avg}} = \frac{21R}{10} = 2.1R \]

### Final Answer:
The average molar heat capacity at constant volume of the mixture is \(2.1R\).

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