As shown schematically in the figure, two vessels contain water solutions (at temperature ) of potassium permanganate of different concentrations and molecules per unit volume with When they are connected by a tube of small length and cross-sectional

Question

As shown schematically in the figure, two vessels contain water solutions (at temperature $T$ ) of potassium permanganate $({KMnO}_{4})$ of different concentrations $n_{1}$ and $n_{2}$ $(n_{1} > n_{2})$ 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 $S$ , ${KMnO}_{4}$ 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) [2013]

Smart Achievers · Accepted Answer

(1, 2, 3)

$Force = Pressure \times Area$

$p_{1} = \frac{n_{1} R T}{N_{A}} and p_{2} = \frac{n_{2} R T}{N_{A}}$

$F = Δ p \cdot A = (\frac{n_{1} R T}{N_{A}} - \frac{n_{2} R T}{N_{A}}) S$

$F = (n_{1} - n_{2}) k_{B} T S = Δ n k_{B} T S$ $(∵ \frac{R}{N_{A}} = K_{B} (Boltzmann constant))$

Hence, option (1) is correct.

$V = \frac{Δ n k_{B} T S}{β}$

Force balance = pressure $\times$ area = total number of molecules $\times β v$

$Δ n k_{B} T S = ℓ n_{1} S β v$ $[β v = viscous force (given)]$

$\Rightarrow n_{1} β v ℓ = Δ n k_{B} T$

So option (2) is correct.

Total number of molecules/sec, $\frac{Δ N}{Δ t} = \frac{(n_{1} v d t) S}{Δ t}$

$= n_{1} v S = \frac{Δ n k_{B} T v S}{β v ℓ} = (\frac{Δ n}{ℓ}) (\frac{k_{B} T}{β}) S$

Option (3) is correct.

As $Δ n$ will decrease with time, so rate of molecules getting transferred through the tube decreases with time.

Hence option (4) is incorrect.

Solution

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

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

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

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

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Solution

1 the force causing the molecules to move across the tube is ΔnkBTS

2 force balance implies n1βvl=ΔnkBT

3 total number of molecules going across the tube per sec is (Δnl)(kBTβ)S

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

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

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

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