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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 :
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$\frac{x}{m}$ is proportional to :
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The correct answer is:
$p^{2 / 3}$
According to Freundlich adsorption isotherm $\frac{x}{m} \propto p^{\frac{1}{n}}, \frac{x}{m}=k p^{\frac{1}{n}}$
Slope $=\frac{2}{3}$
$\log \frac{x}{m}=\log k+\frac{1}{n} \log p$
Slope $=\frac{1}{n}=\frac{2}{3}$
$\frac{x}{m} \propto p^{\frac{2}{3}}$
Slope $=\frac{2}{3}$
$\log \frac{x}{m}=\log k+\frac{1}{n} \log p$
Slope $=\frac{1}{n}=\frac{2}{3}$
$\frac{x}{m} \propto p^{\frac{2}{3}}$
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