Question

Adsorption of a gas follows Freundlich adsorption isotherm
$x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent. $P$ is the pressure of the gas

In the plot of ${\log }_e\left(\frac{x}{m}\right)$ vs ${\log }_{e}P$:
Here
$m=\text{slope}c=\text{Intercept}$

• A. $m=\frac{1}{k}c={\log }_e\left(n\right)$
• B. $m=\frac{1}{n}c={\log }_e\left(k\right)$
• C. $m={\log }_e\left(k\right)c=\frac{1}{n}$
• D. $m={\log }_e\left(n\right)c=\frac{1}{k}$

Answer 1

The correct option is B $m=\frac{1}{n}c={\log }_e\left(k\right)$

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
$y=mx+c$
$m=\frac{1}{n}$
$c={\log }_e\left(k\right)$

Hence, option B is correct.