Adsorption of a gas on a surface follows the | vedclass

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(C) According to the Freundlich adsorption isotherm:$\log \frac{x}{m} = \log k + \frac{1}{n} \log P$Comparing this with the equation of a straight lin

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Adsorption is proportional to the square root of pressure.

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(C) According to the Freundlich adsorption isotherm:$\log \frac{x}{m} = \log k + \frac{1}{n} \log P$Comparing this with the equation of a straight line $y = mx + c$,the slope is $\frac{1}{n}$.Given that the slope is $0.5$,we have $\frac{1}{n} = 0.5 = \frac{1}{2}$.Substituting this into the adsorption equation $\frac{x}{m} = k \cdot P^{1/n}$,we get:$\frac{x}{m} = k \cdot P^{1/2}$This implies that the adsorption is proportional to the square root of pressure $(P^{1/2})$.

Adsorption of a gas on a surface follows the Freundlich adsorption isotherm. The plot of $\log \frac{x}{m}$ versus $\log P$ gives a straight line with a slope equal to $0.5$. Then,(where $\frac{x}{m}$ is the mass of the gas adsorbed per gram of adsorbent):

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