QJEEM19Chabe1: JEE Main 2019 Chemistry - Class 12 JEEM19Chabe1 Solution: Adsorption of a gas follows Freundlich adsorption isotherm. In…

According to Freundlich adsorption isotherm, in the median range of pressure

        ${x \over m} \propto {P^{{1 \over n}}}$

$\Rightarrow$    ${x \over m}$ = kP $^{{1 \over n}}$

taking log both sides, we get,

log ${x \over m}$ = logk + ${1 \over n}$ logP

Here in graph between log ${x \over m}$ and logP, slope is ${1 \over n}$ and intercepts = log k.

From the given graph,

slope = ${2 \over 4}$ = ${1 \over 2}$

$\therefore\,\,\,$ ${1 \over n}$ = ${1 \over 2}$

$\Rightarrow$    n = 2

$\therefore\,\,\,$ ${x \over m}$ = kP $^{{1 \over 2}}$

$\therefore$ ${x \over m} \propto {P^{{1 \over 2}}}$