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, we will follow these steps:

### Step 1: Determine the degrees of freedom for the gas mixture
For a monatomic gas, the degrees of freedom (F1) is 3. For a rigid diatomic gas, the degrees of freedom (F2) is 5.

- Number of moles of monatomic gas (N1) = 5
- Number of moles of diatomic gas (N2) = 1

Using the formula for the degrees of freedom of the mixture:
\[ F_{mixture} = \frac{N_1 F_1 + N_2 F_2}{N_1 + N_2} \]
Substituting the values:
\[ F_{mixture} = \frac{5 \cdot 3 + 1 \cdot 5}{5 + 1} = \frac{15 + 5}{6} = \frac{20}{6} = \frac{10}{3} \]

### Step 2: Calculate the value of gamma (γ)
The value of gamma (γ) is given by:
\[ \gamma = 1 + \frac{F_{mixture}}{2} \]
Substituting the value of \(F_{mixture}\):
\[ \gamma = 1 + \frac{10/3}{2} = 1 + \frac{10}{6} = 1 + \frac{5}{3} = \frac{8}{3} \approx 1.67 \]

### Step 3: Determine the final temperature (Tf) after adiabatic compression
For an adiabatic process, we use the relation:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
Where:
- \(T_1 = T_0\)
- \(V_1 = V_0\)
- \(V_2 = \frac{V_0}{4}\)

Rearranging gives us:
\[ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} \]
Substituting the values:
\[ T_2 = T_0 \left(\frac{V_0}{\frac{V_0}{4}}\right)^{\gamma - 1} = T_0 \left(4\right)^{\gamma - 1} \]
Calculating \(\gamma - 1\):
\[ \gamma - 1 = \frac{8}{3} - 1 = \frac{5}{3} \]
Thus:
\[ T_2 = T_0 \cdot 4^{\frac{5}{3}} = T_0 \cdot (2^2)^{\frac{5}{3}} = T_0 \cdot 2^{\frac{10}{3}} = T_0 \cdot 2^{3.33} \approx 10.08 T_0 \]

### Step 4: Calculate the work done during the adiabatic process
The work done (W) in an adiabatic process can be calculated using:
\[ W = \frac{n R (T_f - T_i)}{\gamma - 1} \]
Where:
- Total moles \(n = N_1 + N_2 = 5 + 1 = 6\)
- \(T_f = T_2\) and \(T_i = T_0\)

Substituting the values:
\[ W = \frac{6R (10.08 T_0 - T_0)}{\frac{5}{3}} = \frac{6R (9.08 T_0)}{\frac{5}{3}} = \frac{6 \cdot 3R \cdot 9.08 T_0}{5} = \frac{54.48 R T_0}{5} \approx 10.896 R T_0 \]

### Step 5: Determine the final pressure (Pf)
Using the relation:
\[ P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \]
Rearranging gives:
\[ P_2 = P_1 \left(\frac{V_1}{V_2}\right)^{\gamma} \]
Substituting the values:
\[ P_2 = P_0 \left(\frac{V_0}{\frac{V_0}{4}}\right)^{\gamma} = P_0 (4)^{\gamma} = P_0 \cdot 4^{\frac{8}{3}} \approx P_0 \cdot 10.08 \]

### Conclusion
The correct statements regarding the gas mixture after adiabatic compression are:
1. The value of gamma (γ) is approximately 1.67.
2. The work done during the process is approximately \(10.896 R T_0\).
3. The final pressure of the gas mixture is approximately \(10.08 P_0\).