JEE PYQ: Surface Chemistry Question 23
Question 23 - 2019 (08 Apr Shift 1)
Type: MCQ
Adsorption of a gas follows Freundlich adsorption isotherm. $x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent. The plot of $\frac{x}{m}$ versus $\log p$ is shown in the given graph. $\frac{x}{m}$ is proportional to:
[Graph shows $\log \frac{x}{m}$ vs $\log p$ with slope $= \frac{2}{3}$]
(a) $p^{2/3}$
(b) $p^{3/2}$
(c) $p^3$
(d) $p^2$
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Answer: 0
Solution
According to Freundlich adsorption isotherm: $\frac{x}{m} \propto kp^{1/n}$
$\log \frac{x}{m} = \log k + \frac{1}{n} \log p$
Slope $= \frac{1}{n} = \frac{2}{3}$
$\therefore \frac{x}{m} \propto p^{2/3}$
BETA
