For Freundlich adsorption isotherm,


For Freundlich adsorption isotherm, a plot of $\log (x / \mathrm{m})$ ( $y$-axis) and $\log \mathrm{p}(x$-axis) gives a straight line. The intercept and slope for the line is $0.4771$ and 2, respectively. The mass of gas, adsorbed per gram of adsorbent if the initial pressure is $0.04 \mathrm{~atm}$, is________$\times 10^{-4} \mathrm{~g}$. $(\log 3=0.4771)$


(48) Freundlich adsorption isotherm :

$\frac{x}{m}=k_{p}^{1 / n}$

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

Slope $\left(\frac{1}{n}\right)=2$

Intercept $=\log k=0.4771$, so $k=$ Antilog $(0.4771)=3$

So, $\left(\frac{x}{m}\right)=k(p)^{1 / n}$

$\frac{x}{m}=3 \cdot p^{2}(p=0.04 \mathrm{~atm})=3 \times(0.04)^{2}=48 \times 10^{-4}$

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