Physics · Kinetic Theory of Gases

JEE Advanced 2025 — Paper 2 — Question 13

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 $P=\frac{\mathrm{k} L}{\mathrm{~A}} \alpha$ , then the value of $\alpha$ is $\qquad$

Answer: 0.2

Numerical answer — enter this value.

Step-by-step solution

Extension in spring $\mathrm{x}=0.5 \mathrm{~L}-0.4 \mathrm{~L}$

$=0.1 \mathrm{~L}$

FBD of piston

$\mathrm{kx}+\mathrm{P}_{2} \mathrm{~A}=\mathrm{P}_{1} \mathrm{~A}$

$\mathrm{P}_{2} \mathrm{~A}=\mathrm{P}_{1} \mathrm{~A}-\mathrm{kx}$

$P_{2}=P_{1}-\frac{k L}{A(10)}$

$\mathrm{P}_{1} \mathrm{~V}=\mathrm{n}_{1} \mathrm{RT}$

$\mathrm{P}_{2} \mathrm{~V}=\mathrm{n}_{2} \mathrm{RT}$

$\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{3}{2}$

$\mathrm{P}_{1}=\frac{3}{2} \mathrm{P}_{2}$

$\mathrm{P}_{2}=\frac{3}{2} \mathrm{P}_{2}-\frac{\mathrm{kL}}{10 \mathrm{~A}}$

$\frac{\mathrm{P}_{2}}{2}=\frac{\mathrm{kL}}{10 \mathrm{~A}}$

$P_{2}=\frac{k L}{5 A}=\frac{k L}{A} \alpha$

$\alpha=\frac{1}{5}=0.2$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 2
Subject: Physics
Chapter: Kinetic Theory of Gases
Topic: Energy of Gas and Gas Laws and Miscellaneous Problems