Question

The graph between log⁡ $\left(\frac{x}{m}\right)$ and log $p$ forms a straight line at an angle of $45^{\circ} $ with the intercept of $0.6020\,$ .

The extent of adsorption $\left(\frac{x}{m}\right)$ at a pressure of $1\,atm$ is

Answer 1

Freundlich adsorption isotherm is expressed using the following equation,
$\frac{x}{m}=k \cdot(p)^{\frac{1}{n}}($ where, $n>1)$
Taking log on both sides,
$\log \frac{x}{m}=\log k+\frac{1}{n} \log p$
Comparing with the graph,
Slope $=\frac{1}{n}=\tan 45^{\circ}=1$
Intercept $=\log k=0.6020$
Substituting the values along with the given value of pressure.
$\log \frac{x}{m}=0.6020+1 \log 1=0.602 \log \frac{x}{m}=\log 4$
$\therefore \frac{x}{m}=4$