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(D) For the mixture: $n_1 = 5$ (monatomic),$C_{v1} = 3R/2$; $n_2 = 1$ (diatomic),$C_{v2} = 5R/2$.Total moles $n = 5 + 1 = 6$.$(C_v)_m = \frac{n_1 C_{v

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$1, 4$

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(D) For the mixture: $n_1 = 5$ (monatomic),$C_{v1} = 3R/2$; $n_2 = 1$ (diatomic),$C_{v2} = 5R/2$.Total moles $n = 5 + 1 = 6$.$(C_v)_m = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{5(3R/2) + 1(5R/2)}{6} = \frac{10R}{6} = \frac{5R}{3}$.$(C_p)_m = (C_v)_m + R = \frac{5R}{3} + R = \frac{8R}{3}$.Adiabatic constant $\gamma_m = \frac{(C_p)_m}{(C_v)_m} = \frac{8/3}{5/3} = 1.6$. Thus,$(4)$ is correct.For adiabatic compression: $P_0 V_0^{\gamma} = P_f (V_0/4)^{\gamma}$.$P_f = P_0 (4)^{1.6} = P_0 (2^2)^{1.6} = P_0 (2^{3.2}) = 9.2 P_0$. Thus,$(1)$ is correct.Using $T_f V_f^{\gamma-1} = T_0 V_0^{\gamma-1}$,$T_f = T_0 (V_0 / (V_0/4))^{0.6} = T_0 (4)^{0.6} = T_0 (2^{1.2}) = 2.3 T_0$.Total internal energy $U = n_1 (3/2 RT) + n_2 (5/2 RT) = 5(1.5 RT) + 1(2.5 RT) = 10 RT$.$U_f = 10 R (2.3 T_0) = 23 RT_0$. Thus,$(2)$ is incorrect.Work done $|W| = |\Delta U| = |n_f C_{vm} T_f - n_i C_{vm} T_i| = |6 	imes (5R/6) 	imes (2.3 T_0 - T_0)| = |5R 	imes 1.3 T_0| = 6.5 RT_0$. Thus,$(3)$ is incorrect.Correct statements are $(1)$ and $(4)$.

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