Adsorption of a gas on solids follows Freundl | vedclass

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(C) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k \cdot p^{1/n}$.Taking logarithm on both sides,we get: $\log \frac{x}

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k \cdot p^{1/n}$.Taking logarithm on both sides,we get: $\log \frac{x}{m} = \log k + \frac{1}{n} \log p$.Comparing this with the equation of a straight line $y = mx + c$,we have the slope equal to $\frac{1}{n} = 3$ and the intercept equal to $\log k = 0.30$.Since $\log k = 0.30$ and we are given $\log 2 = 0.3$,we find that $k = 2$.Now,substituting the values into the equation at $p = 2 \ atm$:$\log \frac{x}{m} = \log k + \frac{1}{n} \log p = 0.30 + 3 	imes \log 2$.$\log \frac{x}{m} = 0.30 + 3 	imes 0.3 = 0.30 + 0.90 = 1.20$.Therefore,$\frac{x}{m} = 	ext{antilog}(1.20) = 10^{1.20} = 10^{1} 	imes 10^{0.20}$.Using $10^{0.3} = 2$,we approximate $10^{0.2} \approx 1.58$,so $\frac{x}{m} \approx 10 	imes 1.58 = 15.8 \approx 16$.

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