Question

From Freundlich adsorption isotherm, if $log \frac{x}{m}$ is plotted in $y$ -axis and $log\,p$ in $x$ -axis, the slope will be (Where $'k'$ and $'n'$ are constants that depend on the nature of the adsorbent and the gas at a particular temperature, $'x'$ is the mass of gas adsorbed on mass $'m'$ )

• A. $n$
• B. $\frac{1}{n}$
• C. $log\,k$
• D. $\frac{1}{log\,k}$

Answer 1

Answer: (B)

$\frac{1}{n}$

Answer 2

According to Freundlich adsorption isotherm :
$\frac{x}{m}=k\cdot p^{\frac{1}{n}}$
$log \frac{x}{m}=log\,k+\frac{1}{n}log\,p$