Physics · Motion in one Dimension

JEE Advanced 2020 — Paper 1 — Question 11

As shown schematically in the figure, two vessels contain water solutions (at temperature T) of potassium permanganate $\left(\mathrm{KMnO}_{4}\right)$ of different concentrations $\mathrm{n}_{1}$ and $\mathrm{n}_{2}\left(\mathrm{n}_{1}>\mathrm{n}_{2}\right)$ molecules per unit volume with $\Delta \mathrm{n}=\left(\mathrm{n}_{1}-\mathrm{n}_{2}\right) \ll \mathrm{n}_{1}$ . When they are connected by a tube of small length $\ell$ and cross-sectional area $\mathrm{S}, \mathrm{KMnO}_{4}$ 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 $(\Delta \mathrm{n})^{2}$ , which of the following is/are correct? ( $k_{B}$ is the Boltzmann constant)

Option A:
the force causing the molecules to move across the tube is $\Delta n k_{B} T S$
Correct
Option B:
force balance implies $n_{1} \beta v l=\Delta n k_{B} T$
Correct
Option C:
total number of molecules going across the tube per sec is $\left(\frac{\Delta n}{l}\right)\left(\frac{k_{B} t}{\beta}\right) \mathrm{S}$
Correct
Option D:
rate of molecules getting transferred through the tube does not change with time

Answer: A, B, C

Step-by-step solution

$\mathrm{PV}=\mathrm{NKT}$

$\mathrm{P}_{1}=\mathrm{n}_{1} \mathrm{~K}_{\mathrm{B}} \mathrm{T}$

$\mathrm{P}_{2}=\mathrm{n}_{2} \mathrm{~K}_{\mathrm{B}} \mathrm{T}$

$\mathrm{n}_{1}, \mathrm{n}_{2} \rightarrow$ no. of molecules per unit volume

Force acting on the molecules

$\mathrm{F}=\Delta \mathrm{PS}$

$=\left(\mathrm{P}_{1}-\mathrm{P}_{2}\right) \mathrm{S}$

$\mathrm{F}=\left(\mathrm{n}_{1}-\mathrm{n}_{2}\right) \mathrm{K}_{\mathrm{B}} \mathrm{TS}=\Delta \mathrm{nk}_{\mathrm{B}} \mathrm{TS} \ldots$ (i)

So Option (A) is correct

Total no. of molecules in the tube $=\left(\mathrm{n}_{1} \mathrm{~S} \ell\right)$

Force acting on each molecules

$\mathrm{F}=-\beta \mathrm{v}$ $\ell$

So total force acting on ( $\mathrm{n}_{1} \mathrm{~S} \ell$ ) molecules $=-$

$(\beta \mathrm{v})\left(\mathrm{n}_{1} \mathrm{~s} \ell\right) \ldots(\mathrm{ii})$

From (i) and (ii)

$\Delta \mathrm{nK}_{\mathrm{B}} \mathrm{TS}=(\beta \mathrm{v})\left(\mathrm{n}_{1} \mathrm{~S} \ell\right)$

So, $\Delta \mathrm{nK}_{\mathrm{B}} \mathrm{T}=\left(\beta \mathrm{vn}_{1} \ell\right) \ldots$ (iii)

So, Option (B) is correct.

No. of molecules going across the tube per second

$=\left(\mathrm{svn}_{1}\right)$

$=\left(\mathrm{sn}_{1}\right)\left(\frac{\Delta \mathrm{nK}_{\beta} \mathrm{T}}{\beta \mathrm{n}_{1} \ell}\right)=\left(\frac{\mathrm{S} \Delta \mathrm{nK}_{\beta} \mathrm{T}}{\beta \ell}\right)$

So, Option (C) is correct.

With respect to time $\Delta \mathrm{n}$ changes hence rate of molecules getting transferred through the tube

changes with time.

So, Option (D) is incorrect.

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 1
Subject: Physics
Chapter: Motion in one Dimension
Topic: Non-Uniformly Accelerated Motion