Adiabatic Compression of Monatomic and Diatomic Gas Mixture

To evaluate the correct options for the adiabatic compression of the gas mixture, we calculate the effective molar specific heat capacity at constant volume and the adiabatic index, then apply the adiabatic state equations to determine the final pressure, temperature, and work done.

First, the total molar heat capacity at constant volume $C_V$ for the mixture of $n_1 = 5\text{ moles}$ of monatomic gas ($C_{V1} = \frac{3}{2}R$) and $n_2 = 1\text{ mole}$ of rigid diatomic gas ($C_{V2} = \frac{5}{2}R$) is: $C_V = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2} = \frac{5\left(\frac{3}{2}R\right) + 1\left(\frac{5}{2}R\right)}{5 + 1} = \frac{5}{3}R$

The adiabatic index $\gamma$ of the mixture is: $\gamma = 1 + \frac{R}{C_V} = 1 + \frac{3}{5} = 1.6$ This matches statement (D).

For an adiabatic process, $P V^\gamma = \text{constant}$. The final pressure $P_f$ after compressing to $V_f = V_0/4$ is: $P_f = P_0 \left(\frac{V_0}{V_f}\right)^\gamma = P_0 (4)^{1.6} = P_0 \cdot 2^{3.2} = 9.2 P_0$ Since $9 P_0 < 9.2 P_0 < 10 P_0$, statement (C) is correct.

The final temperature $T_f$ is given by $T V^{\gamma-1} = \text{constant}$: $T_f = T_0 \left(\frac{V_0}{V_f}\right)^{\gamma-1} = T_0 (4)^{0.6} = T_0 \cdot 2^{1.2} = 2.3 T_0$

The magnitude of work done during the adiabatic process is: $|W| = |\Delta U| = (n_1 + n_2) C_V (T_f - T_0) = 6 \cdot \left(\frac{5}{3}R\right) (2.3 T_0 - T_0) = 10 R (1.3 T_0) = 13 R T_0$ This matches statement (A).

The total translational and rotational kinetic energy of the mixture post-compression is: $E_k = (n_1 + n_2) C_V T_f = 6 \cdot \left(\frac{5}{3}R\right) (2.3 T_0) = 23 R T_0$ Thus, statement (B) is incorrect.

Hence, the correct options are (A), (C), and (D).