$\text{The left and right compartments of a thermally isolated container of length}L\text{are separated by a thermally}$

$\text{conducting, movable piston of area}\mathrm{ A}.\text{The left and right compartments are filled with }\frac{3}{2}\text{ and}1\text{moles of an ideal}$

$\text{gas, respectively.}\text{In the left compartment the piston is attached by a spring with spring constant}k\text{and natural}$

$\text{length }\frac{2L}{5}.\text{In thermodynamic equilibrium, the piston is at a distancen }\frac{L}{2}\text{ from the left and right edges of the}$

$\text{container as shown in the figure.}\text{ Under the above conditions, if the pressure in the right compartment is}\mathrm{ P}=\frac{kL}{A}\alpha ,$

$\text{then the value of}\mathrm{ \alpha }\text{ is}\text{\_\_\_\_\_}$

The correct answer is 0.20.

1. Understanding the Given Parameters:
From the text and the diagram, we can identify the following parameters:

• The total length of the thermally isolated container is $L$ (indicated by the dashed arrow spanning the container's length in the diagram).
• The movable piston has area $A$ and divides the container into two compartments. At equilibrium, the piston lies at a distance of $\frac{L}{2}$ from the left wall (labeled as $L/2$ in the diagram) and thus also $\frac{L}{2}$ from the right wall.
• The piston is thermally conducting, meaning both compartments reach the same absolute temperature $T$ in thermodynamic equilibrium.
• The left compartment contains $n_1=\frac{3}{2}$ moles of an ideal gas.
• The right compartment contains $n_2=1$ mole of the ideal gas.
• A spring with spring constant $k$ and natural length $l_0=\frac{2L}{5}$ is connected between the left wall and the piston inside the left compartment.

2. Pressures in the Compartments:
The volume of the left compartment $V_1$ and the right compartment $V_2$ are equal since the piston is at the midpoint:
$V_1=V_2=A\cdot \frac{L}{2}$
Using the ideal gas law ($PV=nRT$), the pressure in the left compartment $P_1$ is:
$P_1=\frac{n_{1}RT}{V_1}=\frac{\left(\frac{3}{2}\right)RT}{A\cdot \frac{L}{2}}=\frac{3RT}{AL}$
Similarly, the pressure in the right compartment $P$ is:
$P=\frac{n_{2}RT}{V_2}=\frac{1\cdot RT}{A\cdot \frac{L}{2}}=\frac{2RT}{AL}$
Taking the ratio of the two equations, we can express $P_1$ in terms of $P$:
$P_1=\frac{3}{2}P$

3. Forces on the Piston:
The current length of the spring in equilibrium is $x=\frac{L}{2}$. The extension of the spring from its natural length $l_0$ is:
$\Delta x=x-l_0=\frac{L}{2}-\frac{2L}{5}=\frac{L}{10}$
Since $\Delta x>0$, the spring is stretched and exerts a restoring force pulling the piston to the left:
$F_s=k\Delta x=k\frac{L}{10}$

4. Equilibrium Condition:
The net force acting on the piston along the horizontal direction must be zero at equilibrium:
$P_{1}A-PA-F_s=0$
Substitute the expressions for $P_1$ and $F_s$:
$\left(\frac{3}{2}P-P\right)A=k\frac{L}{10}$
$\frac{1}{2}PA=\frac{kL}{10}$
$P=\frac{kL}{5A}=0.20\frac{kL}{A}$

5. Finding α:
Comparing the derived pressure with the given expression:
$P=\frac{kL}{A}\alpha$
We get:
$\alpha =0.20$