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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 $Δ n = (n_{1} - n_{2}) ≪ n_{1} .$ . 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 $- β v$ on each molecule, where $β$ is a constant. Neglecting all terms of the order $(Δ n)^{2}$ , which of the following is/are correct? (𝑘_𝐵 is the Boltzmann constant)-

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rate of molecules getting transferred through the tube does not change with time

total number of molecules going across the tube per sec is $(\frac{Δn}{ℓ}) (\frac{k_{B} T}{β}) S$

$force balance implies n_{1} β v ℓ = Δ n k_{B} T$

the force causing the molecules to move across the tube is $Δ n k_{B} T S$

answer is A, B, C.

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Detailed Solution

$[\begin{matrix} PV = N k_{B} T \\ \Rightarrow P = \frac{N}{V} k_{B} T = {nk}_{B} T \end{matrix}]$

$\begin{array}{l} n_{1} >> (n_{1} - n_{2}) = Δn \\ F = (P_{1} - P_{2}) S = (n_{1} - n_{2}) k_{B} TS = Δ {nk}_{B} TS \\ Option (A) is correct . \end{array}$

$Viscous Force = Total number of molecules \times β v = (n_{1} l S) βv$

$If viscous force is balanced by pressure force, then$

$\begin{array}{l} {Δnk}_{B} TS = ℓ n_{1} S β v \\ \Rightarrow n_{1} β v ℓ = {Δnk}_{B} T \\ Option (B) is correct . \end{array}$

$Total number of molecules per \sec = \frac{(n_{1} vdt) S}{dt} = n_{1} vS = \frac{{Δnk}_{B} TvS}{β v ℓ} = (\frac{Δn}{ℓ}) (\frac{k_{B} T}{β}) S$

Option (C) is correct.

As $∆$ n will decrease with time therefore rate of molecules getting transfer decreases with time.

Option (D) is incorrect.

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