1. Understand the system

• The whole container is perfectly insulated from the outside, so $Q_{\text{total}}=0.$
• The partition between the two gases is thermally conducting, so heat can flow between the gases.
• The partition is rigid and fixed, so the lower gas has constant volume.
• The upper gas has a movable frictionless piston, so its pressure can adjust; as heat is transferred, it undergoes a process at constant pressure (external pressure on the piston remains fixed).
• At equilibrium, both gases reach the same final temperature $T$.

2. Initial data

• Lower compartment: 2 moles of ideal monatomic gas $n_1=2,\quad T_1=700\,\text{K},\quad C_{v1}=\frac{3}{2}R$
• Upper compartment: 2 moles of ideal diatomic gas $n_2=2,\quad T_2=400\,\text{K},\quad C_{p2}=\frac{7}{2}R$

3. Apply energy balance

Since no heat enters or leaves the combined system, $\Delta U_1 + \Delta H_2 = 0$ if we directly use:

• constant volume for lower gas: heat exchanged by it equals $Q_1=n_1 C_{v1}(T-700)$
• constant pressure for upper gas: heat exchanged by it equals $Q_2=n_2 C_{p2}(T-400)$

And because heat lost by one gas equals heat gained by the other, $Q_1+Q_2=0.$

So, $2\left(\frac{3}{2}R\right)(T-700)+2\left(\frac{7}{2}R\right)(T-400)=0$

4. Simplify

$3R(T-700)+7R(T-400)=0$ Divide by $R$: $3(T-700)+7(T-400)=0$ $3T-2100+7T-2800=0$ $10T-4900=0$ $T=490\,\text{K}$

5. Check with options

$T=490\,\text{K}$ So the correct option is D.

6. Compare with stored answer

Stored correct answer: D

This matches our derived answer.