Surface Chemistry - Result Question 8
####8. 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 $\log \frac{x}{m}$ versus $\log p$ is shown in the given graph. $\frac{x}{m}$ is proportional to
(2019 Main, 8 April I)
(a) $p^{2 / 3}$
(b) $p^{3 / 2}$
(c) $p^{3}$
(d) $p^{2}$
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Answer:
Correct Answer: 8. (d)
Solution:
$$ \begin{aligned} & \text { Key Idea According to Freundlich, } \ & \qquad \frac{x}{m}=K p^{1 / n}[n>1] \end{aligned} $$
where, $m=$ mass of adsorbent, $x=$ mass of the gas adsorbed, $\frac{x}{m}=$ amount of gas adsorbed per unit mass of solid adsorbent, $p=$ pressure, $K$ and $n=$ constants.
The logarithm equation of Freundlich adsorption isotherm is
$$ \log \frac{x}{m}=\log K+\frac{1}{n} \log p $$
On comparing the above equation with straight line equation, $(y=m x+c)$
we get
$$ m=\text { slope }=\frac{1}{n} \quad \text { and } \quad c=\log K $$
From the given plot,
$$ \begin{array}{rlrl} & & m=\frac{y _2-y _1}{x _2-x _1} & =\frac{1}{n}=\frac{2}{3} \ \therefore & \frac{x}{m} & =K p^{2 / 3} \end{array} $$
