In a plot of log left( frac{x}{m} right) vs l | vedclass

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Question

(A) According to the Freundlich adsorption isotherm,the relationship is given by $\frac{x}{m} = K p^{\frac{1}{n}}$.Taking the logarithm on both sides,

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$\frac{1}{n}$

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

In a plot of $\log \left( \frac{x}{m} \right)$ $vs$ $\log p$,for the adsorption of a gas on a solid,a straight line is obtained. The slope of that straight line is

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