If 3 moles of a monoatomic gas is mixed with 6 moles of a diatomic gas at room temperature, then `gamma` of the mixture is

Options

A. 1.49
B. 1.86
C. 1.46
D. 1.89

Answer

To find the value of `gamma` (γ) for the mixture of gases, we will follow these steps:

### Step 1: Identify the properties of the gases
- **Monoatomic gas**:
- Number of moles (n1) = 3
- Degrees of freedom (f1) = 3
- \( C_{v1} = \frac{3}{2} R \)

- **Diatomic gas**:
- Number of moles (n2) = 6
- Degrees of freedom (f2) = 5
- \( C_{v2} = \frac{5}{2} R \)

### Step 2: Calculate the molar heat capacities
- For the monoatomic gas:
\[ C_{v1} = \frac{3}{2} R \]

- For the diatomic gas:
\[ C_{v2} = \frac{5}{2} R \]

### Step 3: Calculate the total heat capacity \( C_v \) for the mixture
Using the formula for the heat capacity of a mixture:
\[ C_v = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} \]
Substituting the values:
\[ C_v = \frac{3 \cdot \frac{3}{2} R + 6 \cdot \frac{5}{2} R}{3 + 6} \]
Calculating the numerator:
\[ = \frac{\frac{9}{2} R + \frac{30}{2} R}{9} = \frac{\frac{39}{2} R}{9} = \frac{39 R}{18} \]

### Step 4: Calculate \( C_p \) for the mixture
Using the relation \( C_p = C_v + R \):
\[ C_p = \frac{39 R}{18} + R = \frac{39 R}{18} + \frac{18 R}{18} = \frac{57 R}{18} \]

### Step 5: Calculate \( \gamma \) for the mixture
Using the formula:
\[ \gamma = \frac{C_p}{C_v} \]
Substituting the values:
\[ \gamma = \frac{\frac{57 R}{18}}{\frac{39 R}{18}} = \frac{57}{39} \]
Calculating this gives:
\[ \gamma = 1.46 \]

### Final Answer
Thus, the value of \( \gamma \) for the mixture is approximately **1.46**.

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