Q48 / 54 · JEE Advanced 2020 (Paper 1) · Physics

As shown schematically in the figure, two vessels contain water solutions (at temperature T) of potassium permanganate KMnO4 of different concentrations n1 and n2n1>n2 molecules per unit volume with Δn=n1-n2≪n1. When they are connected by a tube of small length l and cross-sectional area S, KMnO4 starts to diffuse from the left to the right vessel through the tube. Consider the collection of molecules to behave as dilute ideal gases and the difference in their partial pressure in the two vessels causing the diffusion. The speed v of the molecules is limited by the viscous force -βv on each molecule, where β is a constant. Neglecting all terms of the order Δn2. Which of the following is/are correct? (kB is the Boltzmann constant)

Figure

  • Correct answer
    A
    the force causing the molecules to move across the tube is ΔnkBTS
  • Correct answer
    B
    force balance implies n1βvl=ΔnkBT
  • Correct answer
    C
    total number of molecules going across the tube per sec is ΔnlkBTβS
  • D
    rate of molecules getting transferred through the tube does not change with time

Solution

(a) Figure

F=ΔP.A=P1-P2A

=n1×RTA-n2RT.A

F=ΔnRTA

(b) βv×n1×A×ℓ=ΔnkBTA

βvn1ℓ=ΔnkBT

(c) dNdt=Svn1

=sn1×ΔnkBTβn1ℓ=S×ΔnkBTSβℓ

(d) Δn→decreasing then dNdt→decreasing

Hence, the correct answer is options: 'A', 'B', 'C'.

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