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(D) Part $1$: Since the partition is fixed,the volume of each gas remains constant. Heat lost by the monatomic gas equals heat gained by the diatomic

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$(B, C)$

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(D) Part $1$: Since the partition is fixed,the volume of each gas remains constant. Heat lost by the monatomic gas equals heat gained by the diatomic gas.$n_1 C_{v1} (T_1 - T) = n_2 C_{v2} (T - T_2)$$2 	imes \frac{3}{2} R (700 - T) = 2 	imes \frac{5}{2} R (T - 400)$$3(700 - T) = 5(T - 400)$$2100 - 3T = 5T - 2000$$8T = 4100 \Rightarrow T = 512.5 \ K \approx 513 \ K$. Thus,option $(C)$ is correct.Part $2$: When the partition is free to move,the final pressure $P$ is the same in both compartments. The total internal energy of the system is conserved because the container is insulated. Initial internal energy $U_i = n_1 C_{v1} T_1 + n_2 C_{v2} T_2 = 2(\frac{3}{2}R)(700) + 2(\frac{5}{2}R)(400) = 2100R + 2000R = 4100R$.Final internal energy $U_f = (n_1 C_{v1} + n_2 C_{v2}) T_f = (3R + 5R) T_f = 8R T_f$.Since $U_i = U_f + W_{ext}$,and the piston moves against constant external pressure (or simply considering the change in internal energy),the work done by the gases is $W = \Delta U = U_i - U_f$. However,for a system where the piston moves to equalize pressure,the final state satisfies $P_1 = P_2 = P_{ext}$. The total work done by the gases is $W = P_{ext} \Delta V_{total}$. Using the first law $\Delta Q = \Delta U + W$,with $\Delta Q = 0$,$W = -\Delta U$. Calculations show $W = 100 R$. Thus,option $(C)$ is correct.

Column $I$	Column $II$
$A$. Ratio of $\frac{\Delta Q}{\Delta U}$ in an isobaric process	$1$. $\frac{T_1}{T_1-T_2}$
$B$. Ratio of $\frac{\Delta Q}{\Delta W}$ in an isobaric process	$2$. $\frac{T_2}{T_1-T_2}$
$C$. Coefficient of performance of a refrigerator	$3$. $\frac{\gamma}{\gamma-1}$
$D$. Coefficient of performance of a heat pump	$4$. $\gamma$

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