JEE PYQ: Surface Chemistry Question 32
Question 32 - 2019 (09 Jan Shift 1)
Type: MCQ
Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, $x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent at pressure $p$. $\frac{x}{m}$ is proportional to:
[Graph shows $\log \frac{x}{m}$ vs $\log P$ with slope $= \frac{2}{4} = \frac{1}{2}$]
(a) $p^2$
(b) $p^{1/4}$
(c) $p^{1/2}$
(d) $p$
Show Answer
Answer: (b)
Solution
In Freundlich adsorption isotherm the extent of adsorption ($x/m$) of a gas on the surface of a solid is related to the pressure of the gas ($P$) which can be formulated as:
$\frac{x}{m} = K(P)^{1/n}$
In the given plot, the slope between $\log \frac{x}{m}$ versus $\log P = \frac{2}{4} = \frac{1}{2}$
$\therefore \frac{x}{m} \propto P^{1/2}$
BETA
