Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

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 (inm) will be _______

(take the acceleration due to gravity =10 ms−2 and the universal gas constant = 8.3 J mol−1K-1)

solution : volume of of each portion is half of total volume of container. let volume of container is V₀.

pressures are P₁ and P₂ respectively.

see diagram,

assuming temperature remains constant at 300K,

from Boyles law, PV = constant

i.e., P₁V₁ = P₂V₂

here V₁ = V₀/2 and V₂ = (V₀/2 - Ax) , P₁ = P₁ , P₂ = P'₁

now, P₁V₀/2 = P'₁(V₀/2 - Ax)

so, P₁' = (P₁V₀/2)/(V₀/2 - Ax) ........(1)

similarly for 2nd part,

V₁ = V₀/2, V₂ = V₀/2 + Ax , P₁ = P₂ and P₂ = P₂'

so, P₂' = P₂V₀/2/(V₀/2 + Ax) .......(2)

here (P₁' - P₂')A = mg

so, ⇒(P₁V₀/2)/(V₀/2 - Ax) - P₂V₀/2/(V₀/2 + Ax) = mg

but P₁V₀/2 = P₂V₀/2 = nRT

so, nRT[1/(V₀/2A - x) - 1/(V₀/2A + x) ] = mg

here V₀/2A = 8/2 = 4 , n = 0.1 , R = 8.3 and T = 300

⇒0.1 × 8.3 × 300 [1/(4 - x) - 1/(4 + x)] = 8.3 × 10

⇒3[4 + x - 4 + x)/(16 - x²)] = 1

⇒3 × 2x = 16 - x²

⇒x² - 6x - 16 = 0

⇒x = 2

so the distance from the top will be 4 + 2 = 6 m