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A plot of $\log\dfrac{x}{m}$ versus $\log p$ for the adsorption of a gas on a solid gives a straight line with slope equal to :
A
$-\log K$
B
$n$
C
$\dfrac{1}{n}$
D
$\log K$
Detailed Solution
Freundlich adsorption isotherm: $\dfrac{x}{m} = K\,p^{1/n}$ (with $n \gt 1$)
Taking logarithms on both sides: $\log\dfrac{x}{m} = \log K + \dfrac{1}{n}\log p$
This is of the form $y = c + mx$, the equation of a straight line.
So a plot of $\log\dfrac{x}{m}$ against $\log p$ is a straight line with slope $= \dfrac{1}{n}$
and intercept on the y-axis $= \log K$.
Hence the slope is equal to $\dfrac{1}{n}$.
Taking logarithms on both sides: $\log\dfrac{x}{m} = \log K + \dfrac{1}{n}\log p$
This is of the form $y = c + mx$, the equation of a straight line.
So a plot of $\log\dfrac{x}{m}$ against $\log p$ is a straight line with slope $= \dfrac{1}{n}$
and intercept on the y-axis $= \log K$.
Hence the slope is equal to $\dfrac{1}{n}$.