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(A) According to the Freundlich adsorption isotherm,the relationship is given by: $\frac{x}{m} = k p^{1/n}$ Taking the logarithm on both sides: $\log(

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$p^{2/3}$

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(A) According to the Freundlich adsorption isotherm,the relationship is given by: $\frac{x}{m} = k p^{1/n}$ Taking the logarithm on both sides: $\log(\frac{x}{m}) = \log k + \frac{1}{n} \log p$ This equation represents a straight line of the form $y = mx + c$,where the slope is $\frac{1}{n}$. From the given graph,the slope is calculated as: $	ext{Slope} = \frac{	ext{change in } \log(\frac{x}{m})}{	ext{change in } \log p} = \frac{2}{3}$ Therefore,$\frac{1}{n} = \frac{2}{3}$. Substituting this back into the original expression: $\frac{x}{m} \propto p^{1/n} \implies \frac{x}{m} \propto p^{2/3}$.

Adsorption of a gas follows the 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 :

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