Adsorption of a gas follows the Freundlich ad | vedclass

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(A) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k P^{1/n}$ Taking the logarithm on both sides,we get: $\log \frac{x}{m

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$p^{2/3}$

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(A) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k P^{1/n}$ Taking the logarithm on both sides,we get: $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$ 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{rise}}{	ext{run}} = \frac{2}{3}$. Therefore,$\frac{1}{n} = \frac{2}{3}$. Substituting this back into the original equation,we get: $\frac{x}{m} = k P^{2/3}$. Thus,$\frac{x}{m}$ is proportional to $p^{2/3}$.

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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