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JEE Advance - Physics (2025 - Paper 2 Online - No. 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 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 __________. JEE Advanced 2025 Paper 2 Online Physics - Heat and Thermodynamics Question 1 English
Answer
0.20

Explanation

JEE Advanced 2025 Paper 2 Online Physics - Heat and Thermodynamics Question 1 English Explanation 1

Extension in spring

$$ \begin{aligned} & \mathrm{x}=0.5 \mathrm{~L}-0.4 \mathrm{~L} \\ & =0.1 \mathrm{~L} \end{aligned} $$

FBD of piston

JEE Advanced 2025 Paper 2 Online Physics - Heat and Thermodynamics Question 1 English Explanation 2

$$ \begin{aligned} & \mathrm{kx}+\mathrm{P}_2 \mathrm{~A}=\mathrm{P}_1 \mathrm{~A} \\ & \mathrm{P}_2 \mathrm{~A}=\mathrm{P}_1 \mathrm{~A}-\mathrm{kx} \\ & \mathrm{P}_2=\mathrm{P}_1-\frac{\mathrm{kL}}{\mathrm{~A}(10)} ........(i) \\ & \mathrm{P}_1 \mathrm{~V}=\mathrm{n}_1 \mathrm{RT} \\ & \mathrm{P}_2 \mathrm{~V}=\mathrm{n}_2 \mathrm{RT} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{3}{2} \end{aligned} $$

$$ \begin{aligned} & \mathrm{P}_1=\frac{3}{2} \mathrm{P}_2 ........(ii) \\ & \mathrm{P}_2=\frac{3}{2} \mathrm{P}_2-\frac{\mathrm{kL}}{10 \mathrm{~A}} \\ & \frac{\mathrm{P}_2}{2}=\frac{\mathrm{kL}}{10 \mathrm{~A}} \\ & \mathrm{P}_2=\frac{\mathrm{kL}}{5 \mathrm{~A}}=\frac{\mathrm{kL}}{\mathrm{~A}} \alpha \\ & \alpha=\frac{1}{5}=0.2 \end{aligned} $$

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