Question

For a gas solid adsorbent system, the adsorption isotherm $\frac{x}{m}=k{P}^{\frac{1}{n}}$ is applicable, where $n=5$ in the specified case. Point out the correct statement:

• A. $\log \left(\frac{x}{m}\right)$ vs $\log P$ plot is a linear graph with slope equal to 5
• B. $\log \left(\frac{x}{m}\right)$ vs $\log P$ plot is a linear graph with slope equal to 0.2
• C. $\left(\frac{x}{m}\right)$ vs $P$ plot is a linear graph with slope equal to 5
• D. None of these

Answer 1

The correct option is B $\log \left(\frac{x}{m}\right)$ vs $\log P$ plot is a linear graph with slope equal to 0.2

$x$ 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, $\log \left(\frac{x}{m}\right)$ vs $\log P$ plot is a linear graph with slope :
$\text{Slope}=\frac{1}{n}=\frac{1}{5}=0.2$
Option (c) is correct