1. Given data

• Left compartment contains $n_L=\dfrac{3}{2}$ moles.
• Right compartment contains $n_R=1$ mole.
• The piston is thermally conducting and movable.
• Entire container is thermally isolated, but at equilibrium the conducting piston ensures $T_L=T_R=T.$
• In equilibrium, piston is at the middle, so each side has length $\frac{L}{2}.$
• Cross-sectional area of piston/container is $A$, hence volumes are $V_L=V_R=A\frac{L}{2}.$
• Spring constant is $k$.
• Spring natural length is $\dfrac{2L}{5}$.

2. Use ideal gas law on both sides

Since $T_L=T_R=T$ and volumes are equal, $P_LV_L=n_LRT, \qquad P_RV_R=n_RRT.$

Because $V_L=V_R$, dividing, $\frac{P_L}{P_R}=\frac{n_L}{n_R}=\frac{3/2}{1}=\frac{3}{2}.$

So, $P_L=\frac{3}{2}P_R.$

3. Spring extension/compression

The piston is attached to the left wall by a spring.

• Actual spring length at equilibrium = left compartment length = $\dfrac{L}{2}$.
• Natural length = $\dfrac{2L}{5}$.

So extension is $x=\frac{L}{2}-\frac{2L}{5}=\frac{L}{10}.$

Hence spring force magnitude is $F_s=kx=k\frac{L}{10}.$

This spring is stretched, so it pulls the piston toward the left.

4. Mechanical equilibrium of piston

Forces on piston:

• Left gas pushes rightward with force $P_LA$.
• Right gas pushes leftward with force $P_RA$.
• Spring pulls leftward with force $kL/10$.

Equilibrium gives $P_LA=P_RA+\frac{kL}{10}.$

Thus, $A(P_L-P_R)=\frac{kL}{10}.$

Using $P_L=\dfrac{3}{2}P_R$, $A\left(\frac{3}{2}P_R-P_R\right)=\frac{kL}{10}$ $A\left(\frac{1}{2}P_R\right)=\frac{kL}{10}$ $P_R=\frac{kL}{5A}.$

5. Find $\alpha$

Given $P_R=\frac{kL}{A}\,\alpha,$ so comparing with $P_R=\frac{kL}{5A},$ we get $\alpha=\frac{1}{5}=0.2.$

6. Comparison with stored answer

Derived answer: $0.2$

Stored correct answer: $0.2$

They match.