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JEE Advanced 2020Paper 2
Exam
JEE Advanced
Year
2020
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2020 · Paper 2 · Physics · Thermodynamics

A thermally isolated cylindrical closed vessel of height 8m is kept...

A thermally isolated cylindrical closed vessel of height 8m8m is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass 8.3kg8 . 3kg. Thus the partition is held initially at a distance of 4m4m from the top, as shown in the schematic figure below. Each of the two parts of the vessel contains 0.10 . 1 mole of an ideal gas at temperature 300K300K. The partition is now released and moves without any gas leaking from one part of the vessel to the other. When equilibrium is reached, the distance of the partition from the top (in mm) will be ____________ (take the acceleration due to gravity =10ms−2= 10m{s}^{- 2} and the universal gas constant=8.3Jmol−1K−1= 8 . 3J{mol}^{- 1}{K}^{- 1}).

A thermally isolated cylindrical closed vessel of height 8m is kept vertically. It is divided into two equal parts by a

Correct answer

0

Step-by-step solution

InitiallyPA(4)=0.1×R×300...(i)PA ( 4 ) = 0 . 1 \times R \times 300 . . . ( i ) for both part
Let final distance of piston from top is ′y′' y '

0.1×R×300=P1×A⋅y=P2A(8−y)…(ii)0 . 1 \times R \times 300 = {P}_{1} \times A \cdot y = {P}_{2}A ( 8 - y ) \ldots \left(ii\right)

30Ry+8.3g=30R8−y\dfrac{30R}{y} + 8 . 3g = \dfrac{30R}{8 - y}

30y+10=308−y\dfrac{30}{y} + 10 = \dfrac{30}{8 - y}

3+yy=38−y\dfrac{3 + y}{y} = \dfrac{3}{8 - y}

24+8y−3y−y2=3y24 + 8y - 3y - {y}^{2} = 3y

y2−2y−24=0{y}^{2} - 2y - 24 = 0

y=2±4+962×1=2±102=6my = \dfrac{2 \pm \sqrt{4 + 96}}{2 \times 1} = \dfrac{2 \pm 10}{2} = 6m.

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