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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-$II$
$(a)$ Heat engine	$(i)$ $\eta = \frac{Q_2}{Q_1 - Q_2}$
$(b)$ Heat pump	$(ii)$ $\eta = \frac{Q_1 - Q_2}{Q_1}$
	$(iii)$ $\eta = \frac{T_1 - T_2}{T_1}$

Similar Questions

Heat is supplied to a diatomic gas at constant pressure. The ratio of $\Delta Q: \Delta U: \Delta W$ is
[Given $\rightarrow \Delta Q=$ heat supplied,$\Delta U=$ change in internal energy,$\Delta W$ $=$ work done]

In Column-$I$ devices and in Column-$II$ efficiency are given. Match them appropriately:

Column-$I$ Column-$II$

$(a)$ Heat engine $(i)$ $\eta = \frac{Q_2}{Q_1 - Q_2}$

$(b)$ Heat pump $(ii)$ $\eta = \frac{Q_1 - Q_2}{Q_1}$

$(iii)$ $\eta = \frac{T_1 - T_2}{T_1}$

For the $P-V$ diagram of a thermodynamic cycle as shown in the figure,processes $BC$ and $DA$ are isothermal. Which of the corresponding graphs is correct?

Select the correct statement for work, heat, and change in internal energy.

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