Q: As shown schematically in the figure , two vessels contain water solution (temperature T) of potassium Permanganate (KMnO_{4}) of different concentration n_{1} and n_{2} (n_{1} >n_{2}) molecules per unit volume with Δn =(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 left to 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 molecules is limited by the viscous force -β v on each molecule , where β is constant . Neglecting all terms of the order (Δn)^2 , Which of the following is/are correct ? (k_{B} is the Boltzmann constant)

(a) the force causing the molecules to move across the tube is Δn k_{B} T S

(b) force balance implies n_{1}βvl = Δn k_{B} T

(c) total number of molecules going across the tube per sec is $(\frac{\Delta n}{l}) (\frac{k_B T}{\beta})S$

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

**Click to See Solution : **

Sol: n_{1} >> (n_{1}-n_{2}) = Δn

$p_1 = \frac{n_1 R T}{N_A}$ and $p_2 = \frac{n_2 R T}{N_A}$

$F = (n_1 -n_2)k_B T S = \Delta n k_B T S$

$V = \frac{\Delta n k_B T S}{\beta} $

$\Delta n k_B T S = l n_1 S \beta v $

$n_1 \beta v l = \Delta n k_B T $

Total number of molecules/second $= \frac{(n_1 v dt)S}{dt}$

$= n_1 v S = \frac{\Delta n K_B T v S}{\beta v l}$

$= (\frac{\Delta n}{l}) (\frac{k_B T}{\beta})S $

As Δn will decrease with time , therefore rate of molecules getting transfer decreases with time