Five moles of helium are mixed with two moles of hydrogen to form a mixture. Take molar mass of helium `M_(1) = 4g` and that of hydrogen `M_(2) = 2g`
The equivalent of `gamma` is

Options

A. `1.59`
B. `1.53`
C. `1.56`
D. None

Explanation

To solve the problem, we need to find the value of gamma (γ), which is defined as the ratio of specific heat at constant pressure (C_P) to specific heat at constant volume (C_V).

### Step-by-Step Solution:

1. **Identify the Moles and Molar Masses**:
- Moles of Helium (He) = 5 moles
- Moles of Hydrogen (H₂) = 2 moles
- Molar mass of Helium (M₁) = 4 g/mol
- Molar mass of Hydrogen (M₂) = 2 g/mol

2. **Calculate the Total Moles**:
- Total moles (n) = Moles of Helium + Moles of Hydrogen = 5 + 2 = 7 moles

3. **Calculate C_P (Specific Heat at Constant Pressure)**:
- For a monoatomic gas like Helium, C_P = (5/2)R
- For a diatomic gas like Hydrogen, C_P = (7/2)R
- Total C_P for the mixture:
\[ C_P = \left( \text{Moles of He} \times C_{P,\text{He}} + \text{Moles of H}_2 \times C_{P,\text{H}_2} \right) / \text{Total moles} \]
\[ C_P = \left( 5 \times \frac{5}{2}R + 2 \times \frac{7}{2}R \right) / 7 \]
\[ C_P = \left( \frac{25R}{2} + \frac{14R}{2} \right) / 7 = \left( \frac{39R}{2} \right) / 7 = \frac{39R}{14} \]

4. **Calculate C_V (Specific Heat at Constant Volume)**:
- For Helium, C_V = (3/2)R
- For Hydrogen, C_V = (5/2)R
- Total C_V for the mixture:
\[ C_V = \left( \text{Moles of He} \times C_{V,\text{He}} + \text{Moles of H}_2 \times C_{V,\text{H}_2} \right) / \text{Total moles} \]
\[ C_V = \left( 5 \times \frac{3}{2}R + 2 \times \frac{5}{2}R \right) / 7 \]
\[ C_V = \left( \frac{15R}{2} + \frac{10R}{2} \right) / 7 = \left( \frac{25R}{2} \right) / 7 = \frac{25R}{14} \]

5. **Calculate Gamma (γ)**:
\[ \gamma = \frac{C_P}{C_V} = \frac{\frac{39R}{14}}{\frac{25R}{14}} = \frac{39}{25} = 1.56 \]

### Final Answer:
The value of gamma (γ) is **1.56**.