A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure `P_0` volume `V_0` and temperature `T_0`. If the gas mixture is adiabatically compressed to a volume `V_0//4`, then the correct statement(s) is(are) : (Given R is gas constant)

Options

A. Adiabatic constant of the gas mixture is 1.6
B. The work |W| down during the process is `12RT_0`
C. The average kinetic energy of the gas mixture after compression is in between `18RT_(0)` and `19RT_0`
D. The final pressure of the gas mixture after compression is in between `9P_0` and `10P_0`

Explanation

To solve the problem step by step, we will analyze the given information about the gas mixture and the adiabatic process it undergoes.

### Step 1: Determine the Adiabatic Constant (γ) of the Gas Mixture

1. **Identify the components of the gas mixture**:
- Monatomic gas: 5 moles
- Rigid diatomic gas: 1 mole

2. **Calculate the specific heat capacities**:
- For monatomic gas (5 moles):
- \( C_{v1} = \frac{3}{2}R \)
- \( C_{p1} = C_{v1} + R = \frac{3}{2}R + R = \frac{5}{2}R \)

- For diatomic gas (1 mole):
- \( C_{v2} = \frac{5}{2}R \)
- \( C_{p2} = C_{v2} + R = \frac{5}{2}R + R = \frac{7}{2}R \)

3. **Calculate the average \( C_p \) and \( C_v \)**:
- Total \( C_p \):
\[ C_p = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} = \frac{5 \cdot \frac{5}{2}R + 1 \cdot \frac{7}{2}R}{5 + 1} = \frac{\frac{25}{2}R + \frac{7}{2}R}{6} = \frac{32R}{12} = \frac{8R}{3} \]

- Total \( C_v \):
\[ C_v = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{5 \cdot \frac{3}{2}R + 1 \cdot \frac{5}{2}R}{6} = \frac{\frac{15}{2}R + \frac{5}{2}R}{6} = \frac{20R}{12} = \frac{5R}{3} \]

4. **Calculate the adiabatic constant (γ)**:
\[ \gamma = \frac{C_p}{C_v} = \frac{\frac{8R}{3}}{\frac{5R}{3}} = \frac{8}{5} = 1.6 \]

### Step 2: Calculate the Work Done During the Compression

1. **Use the formula for work done in an adiabatic process**:
\[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} \]

2. **Find the initial and final pressures**:
- Initial state: \( P_0, V_0, T_0 \)
- Final state after compression to \( V = \frac{V_0}{4} \):
\[ P_2 = P_0 \left( \frac{V_0}{V_0/4} \right)^{\gamma} = P_0 \left( 4 \right)^{1.6} \]

3. **Calculate the work done**:
\[ W = \frac{P_0 V_0 - P_0 \cdot 4^{1.6} \cdot \frac{V_0}{4}}{1.6 - 1} \]
\[ W = \frac{P_0 V_0 (1 - 4^{0.6})}{0.6} \]

### Step 3: Calculate the Average Kinetic Energy of the Gas Mixture

1. **Average kinetic energy**:
\[ KE = \frac{3}{2} nRT \]
where \( n \) is the total number of moles and \( T \) is the temperature after compression.

2. **Calculate the final temperature \( T_2 \)**:
\[ T_2 = T_0 \left( \frac{V_0}{\frac{V_0}{4}} \right)^{\gamma - 1} = T_0 \cdot 4^{0.6} \]

3. **Substituting values**:
\[ KE = \frac{3}{2} \cdot 6 \cdot R \cdot T_0 \cdot 4^{0.6} \]

### Step 4: Determine the Final Pressure of the Gas Mixture

1. **Final pressure calculation**:
\[ P_2 = P_0 \cdot 4^{1.6} \]

### Conclusion

- The adiabatic constant \( \gamma \) is 1.6.
- The work done during the process is not equal to \( 12RT_0 \).
- The average kinetic energy of the gas mixture can be calculated using the final temperature.
- The final pressure of the gas mixture is approximately between \( 9P_0 \) and \( 10P_0 \).