Question

Adsorption of a gas follows Freundlich adsorption isotherm. $x$ is the mass of the gas adsorbed on mass ${}^{\prime }{m}^{\prime }$ of the adsorbent. The plot of $\log \frac{x}{m}$ versus $\log p$ is shown in the given graph. $\frac{x}{m}$ is proportional to

Answer 1

According to Freundlich,
$\frac{x}{m}=k{p}^{\frac{1}{n}}[n>1]$
where, $m=$ mass of adsorbent, $x=$ mass of the gas adsorbed, $\frac{x}{m}=$ amount of gas adsorbed per unit mass of solid adsorbent, $p=$ pressure, $k$ and $n=$ constants.
The logarithm equation of Freundlich adsorption isotherm is
$\log \frac{x}{m}=\log k+\frac{1}{n}\log p$
On comparing the above equation with straight line equation $(y=mx+c)$
we get
$m=slope=\frac{1}{n}andc=\log k$
From the given plot,
$m=\frac{y_2-y_1}{x_2-x_1}=\frac{1}{n}=\frac{2}{3}$
$\frac{x}{m}=k{p}^{\frac{2}{3}}$