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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2025 · Paper 2 · Physics · Thermodynamics

The left and right compartments of a thermally isolated container...

The left and right compartments of a thermally isolated container of length LL are separated by a thermally conducting, movable piston of area AA. The left and right compartments are filled with 32\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 2L5\frac{2 L}{5}. In thermodynamic equilibrium, the piston is at a distance L2\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=kL AαP=\frac{\mathrm{k} L}{\mathrm{~A}} \alpha, then the value of α\alpha is \qquad The left and right compartments of a thermally isolated container of length L are separated by a thermally conducting,

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The left and right compartments of a thermally isolated container of length L are separated by a thermally conducting,
Extension in spring x=0.5 L−0.4 L=0.1 L\begin{aligned} & x=0.5 \mathrm{~L}-0.4 \mathrm{~L} \\ & =0.1 \mathrm{~L} \end{aligned} FBD of pistonThe left and right compartments of a thermally isolated container of length L are separated by a thermally conducting,
kx+P2 A=P1 AP2 A=P1 A−kx\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}\end{aligned}P_2=P_1-\frac{k L}{A(10)} \qquad ....(i) P1 V=n1RTP2 V=n2RTP1P2=n1n2=32\begin{aligned} & \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} \mathrm{P}_1=\frac{3}{2} \mathrm{P}_2 \qquad ....(ii) P2=32P2−kL10AP22=kL10AP2=kL5A=kLAαα=15=0.2\begin{aligned} & P_2=\frac{3}{2} P_2-\frac{k L}{10 A} \\ & \frac{P_2}{2}=\frac{k L}{10 A} \\ & P_2=\frac{k L}{5 A}=\frac{k L}{A} \alpha \\ & \alpha=\frac{1}{5}=0.2\end{aligned}$
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