Question

In the plot of ${\log }_e\left(\frac{x}{m}\right)$ vs ${\log }_{e}P$ when a gas follows Freundlich adsorption isotherm
Here
$x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent. $P$ is the pressure of the gas
$s=\text{slope}k=\text{constant}$

• A. $s=\frac{1}{k}$
• B. None of the above
• C. $s={\log }_e\left(k\right)$
• D. $s=\frac{1}{n}$

Answer 1

The correct option is D $s=\frac{1}{n}$

Freundlich adsorption isotherm:
$\frac{x}{m}=k{P}^{\frac{1}{n}}..\left(i\right)$
Taking log both sides
${\log }_e\left(\frac{x}{m}\right)={\log }_e\left(k{P}^{\frac{1}{n}}\right)$
${\log }_e\left(\frac{x}{m}\right)=\frac{1}{n}{\log }_{e}P+{\log }_e\left(k\right)$
Comparing with
$y=sx+c$
$s=\frac{1}{n}$
$c={\log }_e\left(k\right)$

Hence, option B is correct.