The molar specific heat at constant volume of gas mixture is `(13R)/(6)`. The gas mixture consists of

Options

A. (a) 2 moles of `O_2` and 4 moles of `H_2`
B. (b) 2 moles of `O_2` and 4 moles of argon
C. (c) 2 moles of argon and 4 moles of `O_2`
D. (d) 2 moles of `CO_2` and 4 moles of argon

Explanation

To solve the problem of determining which gas mixture has a molar specific heat at constant volume of \( \frac{13R}{6} \), we will analyze each option using the formula for the molar specific heat of a mixture:

\[ C_{v, \text{mixture}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} \]

Where:
- \( n_1 \) and \( n_2 \) are the number of moles of the two gases.
- \( C_{v1} \) and \( C_{v2} \) are the molar specific heats at constant volume for the respective gases.

### Step 1: Analyze Option 1 - 2 moles of O2 and 4 moles of H2

1. **Identify the specific heats:**
- For \( O_2 \) (diatomic gas), \( C_{v1} = \frac{5R}{2} \).
- For \( H_2 \) (diatomic gas), \( C_{v2} = \frac{5R}{2} \).

2. **Calculate \( C_{v, \text{mixture}} \):**
\[ C_{v, \text{mixture}} = \frac{2 \cdot \frac{5R}{2} + 4 \cdot \frac{5R}{2}}{2 + 4} = \frac{5R + 10R}{6} = \frac{15R}{6} \]

3. **Conclusion:**
- \( C_{v, \text{mixture}} = \frac{15R}{6} \) (not equal to \( \frac{13R}{6} \)).

### Step 2: Analyze Option 2 - 2 moles of O2 and 4 moles of Argon

1. **Identify the specific heats:**
- For \( O_2 \), \( C_{v1} = \frac{5R}{2} \).
- For Argon (monatomic gas), \( C_{v2} = \frac{3R}{2} \).

2. **Calculate \( C_{v, \text{mixture}} \):**
\[ C_{v, \text{mixture}} = \frac{2 \cdot \frac{5R}{2} + 4 \cdot \frac{3R}{2}}{2 + 4} = \frac{5R + 6R}{6} = \frac{11R}{6} \]

3. **Conclusion:**
- \( C_{v, \text{mixture}} = \frac{11R}{6} \) (not equal to \( \frac{13R}{6} \)).

### Step 3: Analyze Option 3 - 2 moles of Argon and 4 moles of O2

1. **Identify the specific heats:**
- For Argon, \( C_{v1} = \frac{3R}{2} \).
- For \( O_2 \), \( C_{v2} = \frac{5R}{2} \).

2. **Calculate \( C_{v, \text{mixture}} \):**
\[ C_{v, \text{mixture}} = \frac{2 \cdot \frac{3R}{2} + 4 \cdot \frac{5R}{2}}{2 + 4} = \frac{3R + 10R}{6} = \frac{13R}{6} \]

3. **Conclusion:**
- \( C_{v, \text{mixture}} = \frac{13R}{6} \) (this is the correct option).

### Step 4: Analyze Option 4 - 2 moles of CO2 and 4 moles of Argon

1. **Identify the specific heats:**
- For \( CO_2 \) (triatomic gas), \( C_{v1} = \frac{6R}{2} = 3R \).
- For Argon, \( C_{v2} = \frac{3R}{2} \).

2. **Calculate \( C_{v, \text{mixture}} \):**
\[ C_{v, \text{mixture}} = \frac{2 \cdot 3R + 4 \cdot \frac{3R}{2}}{2 + 4} = \frac{6R + 6R}{6} = \frac{12R}{6} = 2R \]

3. **Conclusion:**
- \( C_{v, \text{mixture}} = 2R \) (not equal to \( \frac{13R}{6} \)).

### Final Conclusion
The only correct option is **Option 3: 2 moles of Argon and 4 moles of O2**, which gives a molar specific heat at constant volume of \( \frac{13R}{6} \).