Adsorption of a gas follows the Freundlich ad | vedclass

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(C) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = kP^{1/n}$.Taking logarithm on both sides: $\log(\frac{x}{m}) = \log k

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$p^{1/2}$

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(C) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = kP^{1/n}$.Taking logarithm on both sides: $\log(\frac{x}{m}) = \log k + \frac{1}{n} \log P$.This is in the form of a straight line equation $y = mx + c$,where the slope is $\frac{1}{n}$.From the given plot,the slope is calculated as: $	ext{Slope} = \frac{	ext{change in } y}{	ext{change in } x} = \frac{2}{4} = \frac{1}{2}$.Therefore,$\frac{1}{n} = \frac{1}{2}$.Substituting this back into the original equation,we get: $\frac{x}{m} \propto P^{1/2}$.

Adsorption of a gas follows the Freundlich adsorption isotherm. In the given plot,$x$ is the mass of gas adsorbed on mass $m$ of the adsorbent at pressure $p$. $\frac{x}{m}$ is proportional to:

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