Adsorption of a gas on a solid adsorbent foll | vedclass

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Question

(A) According to the Freundlich adsorption isotherm: $\frac{x}{m} = k \cdot p^{1/n}$ Taking logarithm on both sides: $\log(\frac{x}{m}) = \log k + \fr

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$p^{1/2}$

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(A) According to the Freundlich adsorption isotherm: $\frac{x}{m} = k \cdot p^{1/n}$ Taking 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{3}{6} = \frac{1}{2}$ Therefore, $\frac{1}{n} = \frac{1}{2}$. Substituting this back into the Freundlich equation: $\frac{x}{m} = k \cdot p^{1/2}$ Thus, the extent of adsorption is proportional to $p^{1/2}$.

Adsorption of a gas on a solid adsorbent follows the Freundlich adsorption isotherm. If $x$ is the mass of the gas adsorbed on mass '$m$' of the adsorbent at pressure $p$, then from the given graph, the extent of adsorption is proportional to:

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