Two mole of Hydrogen and three mole of Helium are mixed at room temperature and at atmospheric pressure `P_a` and the mixture occupies a volume V.

Options

A. `C_V` of mixture is 2R
B. `C_P` of mixture is 2.9R
C. `gamma` of mixture is nearly 1.53
D. If the mixture is expanded adiabatically to 2V final pressure is `P_a/(""_(2)^(1.4))`

Explanation

To solve the problem, we need to find the specific heat capacities \( C_v \) and \( C_p \) of the mixture of hydrogen and helium, as well as the ratio \( \gamma \) (gamma). We will then analyze the adiabatic expansion of the mixture.

### Step-by-Step Solution:

1. **Identify the Moles and Degrees of Freedom:**
- For Hydrogen (H₂):
- Number of moles, \( n_1 = 2 \)
- Degrees of freedom, \( f_1 = 5 \) (since it is a diatomic gas)
- For Helium (He):
- Number of moles, \( n_2 = 3 \)
- Degrees of freedom, \( f_2 = 3 \) (since it is a monatomic gas)

2. **Calculate the Total Degrees of Freedom of the Mixture:**
\[ f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} \]
Substituting the values:
\[ f_{\text{mix}} = \frac{2 \cdot 5 + 3 \cdot 3}{2 + 3} = \frac{10 + 9}{5} = \frac{19}{5} \]

3. **Calculate \( C_v \) for the Mixture:**
The formula for \( C_v \) in terms of degrees of freedom is:
\[ C_v = \frac{f_{\text{mix}}}{2} R \]
Substituting the value of \( f_{\text{mix}} \):
\[ C_v = \frac{19/5}{2} R = \frac{19}{10} R = 1.9 R \]

4. **Calculate \( C_p \) for the Mixture:**
The relationship between \( C_p \) and \( C_v \) is given by:
\[ C_p = C_v + R \]
Substituting the value of \( C_v \):
\[ C_p = 1.9 R + R = 2.9 R \]

5. **Calculate the Ratio \( \gamma \):**
The ratio \( \gamma \) is defined as:
\[ \gamma = \frac{C_p}{C_v} \]
Substituting the values:
\[ \gamma = \frac{2.9 R}{1.9 R} = \frac{2.9}{1.9} \approx 1.526 \]

6. **Adiabatic Expansion:**
When the mixture expands adiabatically to a volume of \( 2V \), we use the relation:
\[ P V^\gamma = \text{constant} \]
Initially, we have \( P_a V^\gamma \) and after expansion:
\[ P_f (2V)^\gamma = P_a V^\gamma \]
Rearranging gives:
\[ P_f = P_a \left(\frac{V}{2V}\right)^\gamma = P_a \left(\frac{1}{2}\right)^\gamma \]
Substituting \( \gamma \approx 1.526 \):
\[ P_f = P_a \cdot \frac{1}{2^{1.526}} \approx \frac{P_a}{2^{1.526}} \approx \frac{P_a}{2.828} \approx \frac{P_a}{1.4} \]

### Final Answers:
- \( C_v = 1.9 R \)
- \( C_p = 2.9 R \)
- \( \gamma \approx 1.526 \)
- Final pressure after adiabatic expansion: \( P_f \approx \frac{P_a}{1.4} \)