Question

Graph between log $\left(\frac{x}{m}\right)$ and $logp$ is a straight line at an angle ${45}^{\circ }$ with intercept on y-axis at 0.3010. Calculate the amount of gas adsorbed in gram per gram of the adsorbent when pressure is $0.2atm$.

• A. 0.4
• B. 0.6
• C. 0.8
• D. 0.2

Answer 1

The correct option is A 0.4

By Freundlich Adsorption Isotherm,
$\frac{x}{m}\propto p^{\frac{1}{n}}$
n > 1
$\frac{x}{m}$ is the extent of adsorption which means the amount of gas adsorbed by the adsorbent with pressure at constant temperature.
$\frac{x}{m}=k{p}^{\frac{1}{n}}....\left(1\right)$
k and n are constants that depends on the nature of gas and adsorbent.
Taking log of equation (1) gives,

log $\left(\frac{x}{m}\right)=logk+\frac{1}{n}logP$
This is a straight line equation.
$\therefore$
$\text{Slope}=\frac{1}{n}$
Inclined angle of straight line is ${45}^o$
$\text{Slope}=tan{45}^o=1$
$\therefore \frac{1}{n}=tan{45}^o=1$

$\text{Intercept}=logk$
$\therefore$
log k= 0.3
$\Rightarrow k\approx 2$

Now, $\frac{x}{m}=k{P}^{1/n}=2\times (0.2{)}^1=0.4$