JEE Advanced 2025 — Paper 2 — PHY-III solution

JEE Advanced2025Paper 2PHY-III

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 $P = \frac{k L}{A} α$ , then the value of $α$ is _____

Your answer

Solution

The piston sits $L /2$ from the left wall, so the spring (natural length $2 L /5$ ) is stretched by

$x = \frac{L}{2} - \frac{2 L}{5} = \frac{L}{10}$

Force balance on the piston: the spring pulls it leftward with $k x$ , the left gas pushes right with $P_{1} A$ , the right gas pushes left with $P_{2} A$ :

$P_{1} A = P_{2} A + k x \Rightarrow P_{1} - P_{2} = \frac{k L}{10 A}$ ... (i)

Both compartments have equal volume $A L /2$ , and the piston is thermally conducting, so the gases share the same temperature. Hence $P V = n R T$ gives

$\frac{P _{1}}{P _{2}} = \frac{n _{1}}{n _{2}} = \frac{3/2}{1} = \frac{3}{2} \Rightarrow P_{1} = \frac{3}{2} P_{2}$ ... (ii)

Substituting (ii) into (i):

$\frac{3}{2} P_{2} - P_{2} = \frac{P _{2}}{2} = \frac{k L}{10 A} \Rightarrow P_{2} = \frac{k L}{5 A}$

Comparing with $P_{2} = \frac{k L}{A} α$ gives $α = \frac{1}{5} = 0.20$ .

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