Equilibrium Pressure in Thermally Isolated Compartmentalized Container

To find the value of $\alpha$, we analyze the state of thermodynamic and mechanical equilibrium of the system.

1. Thermal Equilibrium: Since the movable piston is thermally conducting, the temperature of the gas in both the left and right compartments must be equal at thermodynamic equilibrium: $T_L = T_R = T$

2. Ideal Gas Equation: At equilibrium, both compartments have equal length $\frac{L}{2}$ and equal cross-sectional area $A$. Thus, their volumes are equal: $V_L = V_R = A \left(\frac{L}{2}\right)$

Using the ideal gas equation $PV = nRT$ for each compartment: For the left compartment with $n_L = \frac{3}{2}$ moles: $P_L V_L = n_L R T \implies P_L \left(\frac{AL}{2}\right) = \frac{3}{2} R T \implies P_L = \frac{3RT}{AL}$

For the right compartment with $n_R = 1$ mole: $P_R V_R = n_R R T \implies P_R \left(\frac{AL}{2}\right) = 1 \cdot R T \implies P_R = \frac{2RT}{AL}$

From these two expressions, the ratio of pressures is: $\frac{P_L}{P_R} = \frac{n_L}{n_R} = \frac{3/2}{1} = \frac{3}{2}$

Thus, $P_L = \frac{3}{2} P_R$

3. Mechanical Equilibrium of the Piston: The spring has a natural length $L_0 = \frac{2L}{5}$ and its current length at equilibrium is $x = \frac{L}{2}$.

The extension of the spring is: $\Delta x = \frac{L}{2} - \frac{2L}{5} = \frac{5L - 4L}{10} = \frac{L}{10}$

Since the spring is stretched, it exerts a restoring force $F_s = k \Delta x = k\left(\frac{L}{10}\right)$ directed towards the left.

Balancing the horizontal forces acting on the piston: $P_L A = P_R A + F_s$ $P_L A = P_R A + k \left(\frac{L}{10}\right)$ $P_L - P_R = \frac{kL}{10A}$

4. Calculating $P_R$ and $\alpha$: Substitute $P_L = \frac{3}{2} P_R$ into the force balance equation: $\frac{3}{2} P_R - P_R = \frac{kL}{10A}$ $\frac{1}{2} P_R = \frac{kL}{10A}$ $P_R = \frac{2kL}{10A} = \frac{kL}{A} (0.2)$

Given that the pressure in the right compartment is $P = \frac{kL}{A} \alpha$: $\frac{kL}{A} \alpha = 0.2 \frac{kL}{A} \implies \alpha = 0.2$