Final Equilibrium Temperature of Gas System with Rigid Partition

To find the final equilibrium temperature of the gas system, we analyze the thermodynamics of both compartments upon reaching thermal equilibrium at temperature $T$.

1. Analysis of the Lower Compartment

• Gas type: Monatomic ideal gas
• Number of moles: $n_1 = 2\text{ moles}$
• Initial temperature: $T_1 = 700\text{ K}$
• Molar heat capacity at constant volume: $C_{v1} = \frac{3}{2}R$

Since the dividing partition is rigidly fixed, the volume of the lower compartment remains constant. The heat lost by the lower gas as its temperature decreases from $T_1$ to the final temperature $T$ is given by: $Q_{\text{lost}} = n_1 C_{v1} (T_1 - T)$

2. Analysis of the Upper Compartment

• Gas type: Diatomic ideal gas
• Number of moles: $n_2 = 2\text{ moles}$
• Initial temperature: $T_2 = 400\text{ K}$
• Molar heat capacity at constant pressure: $C_{p2} = \frac{7}{2}R$

The top piston is frictionless and free to move, which keeps the pressure in the upper compartment constant throughout the process. The heat gained by the upper gas as its temperature increases from $T_2$ to $T$ at constant pressure is given by: $Q_{\text{gained}} = n_2 C_{p2} (T - T_2)$

3. Energy Conservation

Since the outer container and the piston are perfectly insulated, no heat is exchanged with the surroundings: $Q_{\text{lost}} = Q_{\text{gained}}$

Substituting the expressions for $Q_{\text{lost}}$ and $Q_{\text{gained}}$: $n_1 C_{v1} (T_1 - T) = n_2 C_{p2} (T - T_2)$

Substitute the given values: $2 \times \left(\frac{3}{2}R\right) \times (700 - T) = 2 \times \left(\frac{7}{2}R\right) \times (T - 400)$

Simplify the equation: $3R(700 - T) = 7R(T - 400)$

Divide both sides by $R$: $2100 - 3T = 7T - 2800$

Rearrange terms to solve for $T$: $10T = 4900$ $T = 490\text{ K}$

Thus, the final equilibrium temperature of the gas system is $490\text{ K}$.

Correct Answer: D