Question

A plot of $\log \left(\frac{x}{m}\right)$ vs $\log P$ for the adsorption of a gas on a solid gives a straight line with slope equal to :

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

Answer 1

The correct option is A $\frac{1}{n}$

$x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent. $P$ is the pressure of the gas
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
$y=mx+c$
$\text{slope}=\frac{1}{n}$
$c=\text{Intercept=}{\log }_e\left(k\right)$

Hence, option (c) is correct.