For Freundlich adsorption isotherm,

Question:

For Freundlich adsorption isotherm, a plot of $\log (\mathrm{x} / \mathrm{m})$ ( $\mathrm{y}$-axis) and $\log \mathrm{p}$ ( $\mathrm{x}$-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__________$\times 10^{-4 \mathrm{~g}}$

$(\log 3=0.4771)$

Solution:

$\frac{\mathrm{X}}{\mathrm{m}}=\mathrm{KP}^{1 / \mathrm{n}}$

$\log \left(\frac{x}{m}\right)=\frac{1}{n} \log P+\log K$

slope $=\frac{1}{n}=2$

intercept $=\log \mathrm{K}=0.4771$

$\mathrm{K}=3$

mass of gas adsorbed per gm of adsorbent $=\frac{x}{m}$

$\frac{x}{m}=3 \times(0.04)^{2}=48 \times 10^{-4}$

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