A plot of log (x/m) versus log p for the adso | vedclass

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(D) According to the Freundlich adsorption isotherm,the relation is given by $x/m = K p^{1/n}$.Taking logarithm on both sides,we get $\log(x/m) = \log

Vedclass · Accepted Answer

$1/n$

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(D) According to the Freundlich adsorption isotherm,the relation is given by $x/m = K p^{1/n}$.Taking logarithm on both sides,we get $\log(x/m) = \log K + \frac{1}{n} \log p$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \log(x/m)$,$x = \log p$,slope $m = 1/n$,and intercept $c = \log K$.Therefore,the slope of the plot is $1/n$.

$A$ plot of $\log (x/m)$ versus $\log p$ for the adsorption of a gas on a solid gives a straight line with slope equal to

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