Verified Solution • 2025–26 Session moderate

QJEEAdv20Pha0d2: JEE Advanced 2020 Physics - Class 11 JEEAdv20Pha0d2 Solution: As shown schematically in the figure, two vessels contain water …

As shown schematically in the figure, two vessels contain water solutions (at temperature T) of potassium permanganate (KMnO4) of different concentrations n1 and n2 (n1 > n2) molecules per unit volume with Δ\Delta n = (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 − β\beta v on each molecule, where β\beta is a constant. Neglecting all terms of the order ( Δ\Delta n)2, which of the following is/are correct? (kB is the Boltzmann constant) JEE Advanced 2020 Paper 1 Offline Physics - Properties of Matter Question 33 English

Multiple Choice Options

  • A the force causing the molecules to move across the tube is ΔnkbTS\Delta n{k_b}TS
  • B force balance implies n1βvl=ΔnkBT{n_1}\beta vl = \Delta n{k_B}T
  • C total number of molecules going across the tube per sec is (Δnl)(kBTβ)S\left( {{{\Delta n} \over l}} \right)\left( {{{{k_B}T} \over \beta }} \right)S
  • D rate of molecules getting transferred through the tube does not change with time

Step-by-Step Verified Solution

{n_1} &gt; &gt; ({n_1} - {n_2}) = \Delta n

∵\because p1=n1RTNA{p_1} = {{{n_1}RT} \over {{N_A}}} and p2=n2RTNA{p_2} = {{{n_2}RT} \over {{N_A}}}

F=(n1−n2)kBTS=ΔnkBTSF = ({n_1} - {n_2}){k_B}TS = \Delta n{k_B}TS

V=ΔnkBTSβV = {{\Delta n{k_B}TS} \over \beta }

ΔnkBTS=ln⁡1Sβv\Delta n{k_B}TS = {\ln _1}S\beta v

⇒n1βvl=ΔnkBT\Rightarrow {n_1}\beta vl = \Delta n{k_B}T

Total number of molecules/second

=(n1vdt)Sdt= {{({n_1}vdt)S} \over {dt}}

=n1vS=ΔnkBTvSβvl=(Δnl)(kBTβ)S= {n_1}vS = {{\Delta n{k_B}TvS} \over {\beta vl}} = \left( {{{\Delta n} \over l}} \right)\left( {{{{k_B}T} \over \beta }} \right)S

As Δ\Delta n will decrease with time, therefore rate of molecules getting transfer decreases with time.

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