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Q. As shown schematically in the figure, two vessels contain water solutions (at temperature $T$ ) of potassium permanganate $\left( KMnO _{4}\right)$ of different concentrations $n _{1}$ and $n _{2}\left( n _{1} > n _{2}\right)$ molecules per unit volume with $\Delta n =\left( n _{1}- n _{2}\right) < < n _{1}$. When they are connected by a tube of small length $\ell$ 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 $-\beta v$ on each molecule, where $\beta$ is a constant. Neglecting all terms of the order $(\Delta n )^{2}$, which of the following is/are correct? ( $k_{B}$ is the Boltzmann constant)

JEE AdvancedJEE Advanced 2020

Solution:

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$PV = NKT$
$P _{1}= n _{1} K _{ B } T$
$P _{2}= n _{2} K _{ B } T$
$n _{1}, n _{2} \rightarrow$ no. of molecules per unit volume
Force acting on the molecules
$F =\Delta PS$
$=\left( P _{1}- P _{2}\right) S$
$F =\left( n _{1}- n _{2}\right) K _{ B } TS =\Delta nk _{ B } TS \ldots$..(i)
So Option (A) is correct
Total no. of molecules in the tube $=\left( n _{1} S \ell\right)$
Force acting on each molecules
$F =-\beta v$
So total force acting on ( $n _{1} S \ell$ ) molecules $=-$
$(\beta v)( n _{1} s \ell$ )...(ii)
From (i) and (ii)
$\Delta nK _{ B } TS =(\beta v )\left( n _{1} S \ell\right)$
So, $\Delta nK _{ B } T =\left(\beta vn _{1} \ell\right) \ldots$ (iii)
So, Option (B) is correct.
No. of molecules going across the tube per second
$=\left({svn}_{1}\right)$
$=\left( sn _{1}\right)\left(\frac{\Delta nK _{\beta} T }{\beta n _{1} \ell}\right)=\left(\frac{ S \Delta n K _{\beta} T }{\beta \ell}\right)$
So, Option (C) is correct.
With respect to time $\Delta$ changes hence rate of molecules getting transferred through the tube changes with time.
So, Option (D) is incorrect.