JEE Advanced 2020 · Paper 2 · Physics Q6

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PhysicsInteger+3 / -1Question 6 of 54

A thermally isolated cylindrical closed vessel of height 8 m is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass 8.3 kg. Thus the partition is held initially at a distance of 4 m from the top, as shown in the schematic figure below. Each of the two parts of the vessel contains 0.1 mole of an ideal gas at temperature 300 K. 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 m) will be ______ (take the acceleration due to gravity $= 10\ \mathrm{m\,s^{-2}}$ and the universal gas constant $= 8.3\ \mathrm{J\,mol^{-1}K^{-1}}$ ).

A closed vertical cylinder of height 8 m with a horizontal partition (black disc) held at mid-height.

Your answer

Answer and worked solution

Official answer: Bonus: full marks to everyone

This question was given marks to all.

Expected analysis. Let the partition settle at distance $x$ from the top, and let $A$ be its area. The lower gas pushes up more than the upper gas pushes down, by the partition's weight (83 N):
$\frac{nRT}{A(8-x)}A-\frac{nRT}{Ax}A=83\implies0.83\,T\left(\frac1{8-x}-\frac1x\right)=83,$
since $nR=0.1\times8.3=0.83$ (the area cancels). If the temperature stays at 300 K:
$3\left(\frac1{8-x}-\frac1x\right)=1\implies x^2-2x-24=0\implies x=6\text{ m}.$

Why it was dropped. The vessel is thermally isolated, so the energy released as the partition falls ( $83\times2=166$ J) must heat the gas, and $T$ does not stay at 300 K. The final position then depends on the gas's heat capacity, and the problem does not say whether the gas is monatomic or diatomic. With no unique integer answer, marks were given to all.

Topics:Ideal gas equation First law and internal energy

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JEE Advanced 2020 · Paper 2, Physics question 6