Question

The left and right compartments of a thermally isolated container of length $L$ are separated by a thermally conducting, movable piston of area $A$. The left and right compartments are filled with $\frac{3}{2}$ and 1 moles of an ideal gas, respectively. In the left compartment the piston is attached by a spring with spring constant $k$ and natural length $\frac{2L}{5}$. In thermodynamic equilibrium, the piston is at a distance $\frac{L}{2}$ from the left and right edges of the container as shown in the figure. Under the above conditions, if the pressure in the right compartment is $P = \frac{kL}{A} \alpha$, then the value of $\alpha$ is ____

Hint:

Balance piston forces from gas pressures and spring force; use ideal gas law to relate pressure, moles, and volume in compartments.

Answer 1

Step 1: Spring Extension
At equilibrium:
\[ x = \frac{L}{2} - \frac{2L}{5} = \frac{L}{10} \]
Step 2: Force Balance
Forces on piston:
\[ P_{\text{right}}A = P_{\text{left}}A - kx \] \[ P_{\text{right}} = P_{\text{left}} - \frac{kx}{A} = P_{\text{left}} - \frac{kL}{10A} \]
Step 3: Ideal Gas Law
Both sides at same temperature $T$:
\[ P_{\text{left}}\left(\frac{AL}{2}\right) = \frac{3}{2}RT \Rightarrow P_{\text{left}} = \frac{3RT}{AL} \] \[ P_{\text{right}}\left(\frac{AL}{2}\right) = RT \Rightarrow P_{\text{right}} = \frac{2RT}{AL} \]
Step 4: Solve
Substitute into force balance:
\[ \frac{2RT}{AL} = \frac{3RT}{AL} - \frac{kL}{10A} \] \[ -\frac{RT}{AL} = -\frac{kL}{10A} \] \[ RT = \frac{kL^2}{10} \] Now find $P_{\text{right}}$: \[ P_{\text{right}} = \frac{2RT}{AL} = \frac{2}{AL}\left(\frac{kL^2}{10}\right) = \frac{kL}{5A} \] Given $P_{\text{right}} = \frac{kL}{A}\alpha$: \[ \frac{kL}{A}\alpha = \frac{kL}{5A} \] \[ \alpha = \boxed{\dfrac{1}{5}} \]