JEE Advance - Physics (2020 - Paper 1 Offline - No. 5)
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 $$\Delta $$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 −$$\beta $$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? (kB is the Boltzmann constant)
the force causing the molecules to move across the tube is $$\Delta n{k_b}TS$$
force balance implies $${n_1}\beta vl = \Delta n{k_B}T$$
total number of molecules going across the
tube per sec is $$\left( {{{\Delta n} \over l}} \right)\left( {{{{k_B}T} \over \beta }} \right)S$$
tube per sec is $$\left( {{{\Delta n} \over l}} \right)\left( {{{{k_B}T} \over \beta }} \right)S$$
rate of molecules getting transferred through the tube does not change with time
Explanation
$${n_1} > > ({n_1} - {n_2}) = \Delta n$$
$$ \because $$ $${p_1} = {{{n_1}RT} \over {{N_A}}}$$ and $${p_2} = {{{n_2}RT} \over {{N_A}}}$$
$$F = ({n_1} - {n_2}){k_B}TS = \Delta n{k_B}TS$$
$$V = {{\Delta n{k_B}TS} \over \beta }$$
$$\Delta n{k_B}TS = {\ln _1}S\beta v$$
$$ \Rightarrow {n_1}\beta vl = \Delta n{k_B}T$$
Total number of molecules/second
$$ = {{({n_1}vdt)S} \over {dt}}$$
$$ = {n_1}vS = {{\Delta n{k_B}TvS} \over {\beta vl}} = \left( {{{\Delta n} \over l}} \right)\left( {{{{k_B}T} \over \beta }} \right)S$$
As $$\Delta $$n will decrease with time, therefore rate of molecules getting transfer decreases with time.
$$ \because $$ $${p_1} = {{{n_1}RT} \over {{N_A}}}$$ and $${p_2} = {{{n_2}RT} \over {{N_A}}}$$
$$F = ({n_1} - {n_2}){k_B}TS = \Delta n{k_B}TS$$
$$V = {{\Delta n{k_B}TS} \over \beta }$$
$$\Delta n{k_B}TS = {\ln _1}S\beta v$$
$$ \Rightarrow {n_1}\beta vl = \Delta n{k_B}T$$
Total number of molecules/second
$$ = {{({n_1}vdt)S} \over {dt}}$$
$$ = {n_1}vS = {{\Delta n{k_B}TvS} \over {\beta vl}} = \left( {{{\Delta n} \over l}} \right)\left( {{{{k_B}T} \over \beta }} \right)S$$
As $$\Delta $$n will decrease with time, therefore rate of molecules getting transfer decreases with time.
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