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 step-by-step, we will analyze the mixture of hydrogen and helium and calculate the required parameters: molar specific heat at constant volume (C_V), molar specific heat at constant pressure (C_P), the adiabatic constant (γ), and the pressure after adiabatic expansion.

### Step 1: Determine the degrees of freedom for each gas
- For hydrogen (H₂), which is diatomic, the degrees of freedom (F₁) is 5.
- For helium (He), which is monoatomic, the degrees of freedom (F₂) is 3.

### Step 2: Calculate the total degrees of freedom for the mixture
Using the formula for the degrees of freedom of the mixture:
\[ F_{\text{mixture}} = \frac{n_1 F_1 + n_2 F_2}{n_1 + n_2} \]
Where:
- \(n_1 = 2\) (moles of H₂)
- \(n_2 = 3\) (moles of He)

Substituting the values:
\[ F_{\text{mixture}} = \frac{2 \times 5 + 3 \times 3}{2 + 3} = \frac{10 + 9}{5} = \frac{19}{5} \]

### Step 3: Calculate the adiabatic constant (γ)
The adiabatic constant is given by:
\[ \gamma = \frac{C_P}{C_V} = 1 + \frac{2}{F_{\text{mixture}}} \]
Substituting \(F_{\text{mixture}} = \frac{19}{5}\):
\[ \gamma = 1 + \frac{2}{\frac{19}{5}} = 1 + \frac{10}{19} = \frac{29}{19} \approx 1.53 \]

### Step 4: Calculate the molar specific heat at constant volume (C_V)
The molar specific heat at constant volume is given by:
\[ C_V = \frac{F_{\text{mixture}}}{2} R \]
Substituting \(F_{\text{mixture}} = \frac{19}{5}\):
\[ C_V = \frac{19/5}{2} R = \frac{19}{10} R \approx 1.9 R \]

### Step 5: Calculate the molar specific heat at constant pressure (C_P)
The molar specific heat at constant pressure is given by:
\[ C_P = C_V + R \]
Substituting \(C_V = \frac{19}{10} R\):
\[ C_P = \frac{19}{10} R + R = \frac{19}{10} R + \frac{10}{10} R = \frac{29}{10} R \approx 2.9 R \]

### Step 6: Calculate the pressure after adiabatic expansion
Using the adiabatic process relation:
\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]
Where:
- \(P_1 = P_a\)
- \(V_1 = V\)
- \(V_2 = 2V\)

Rearranging gives:
\[ P_2 = P_a \left(\frac{V}{2V}\right)^\gamma = P_a \left(\frac{1}{2}\right)^\gamma \]
Substituting \(\gamma \approx 1.53\):
\[ P_2 = P_a \left(\frac{1}{2}\right)^{1.53} \]

### Final Results
- \(C_V \approx 1.9 R\)
- \(C_P \approx 2.9 R\)
- \(\gamma \approx 1.53\)
- \(P_2 = P_a \left(\frac{1}{2}\right)^{1.53}\)