Question

For a gas adsorbed on a particular adsorbent at
$O^{\circ }C,$ the plot of $\log \frac{x}{m}$ versus log $P$ where $P$ is
in atm has a slope and intercept as shown in the fig.
Find the mass of the gas adsorbed by $10$ g of the adsorbent at $0.2$ atm.

Answer 1

$\frac{x}{m}=k.p^{v_n}$

$\log \left(\frac{x}{m}\right)=\frac{1}{n}[\log k.p]$

$\Rightarrow \log \left(\frac{x}{m}\right)=\frac{1}{n}[\log p+\log k]$

$\log \frac{x}{m}=\frac{1}{n}\log p+\frac{1}{n}\log k$

$y=mx+c$

$\Rightarrow m=\tan \theta$

$\to \frac{1}{n}=\tan {45}^{\circ }$

$\to \frac{1}{n}=1$

$\to n=1$

$\log k=0.30$

$\log \frac{x}{10}=\log 0.2+0.30$

$\log \frac{x}{10}=\log \frac{2}{10}+0.30$

$\log x=\log 2+0.30$

$\log \frac{x}{2}=0.30$

$\frac{x}{2}=2$

$\Rightarrow x=4$