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JEE ADVANCED 2025
Paper-2 2025
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 $\frac{2 L}{5}$. 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 $P=\frac{k L}{A} \alpha$, then the value of $\alpha$ is $\_\_\_\_$ .
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Solution

$$ \begin{aligned} & \Delta x=\frac{L}{2}-\frac{2 L}{5}=\frac{L}{10}, \\ & p v=n R T \\ & P_0=\frac{3}{2} \frac{k l \alpha}{A} \end{aligned} $$ On piston

$$ P_0 A=\frac{3}{2} K L \alpha $$ At equilibrium $$ \begin{aligned} & \frac{K L}{10}-\frac{3}{2} K L \alpha=-K L \alpha \\ & \alpha=\frac{2}{10}=0.2 \\ & \Rightarrow 00.20 \end{aligned} $$
Question Tags
JEE Advance
Physics
Easy
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