The graph of log(x/m) versus log p is a strai | vedclass

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Question

(A) The Freundlich adsorption isotherm is given by $\frac{x}{m} = k p^{1/n}$.Taking logarithm on both sides: $\log(x/m) = \log k + \frac{1}{n} \log p$

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(A) The Freundlich adsorption isotherm is given by $\frac{x}{m} = k p^{1/n}$.Taking logarithm on both sides: $\log(x/m) = \log k + \frac{1}{n} \log p$.This is in the form of a straight line equation $y = mx + c$,where $y = \log(x/m)$,$m = 	ext{slope} = \frac{1}{n}$,and $c = 	ext{intercept} = \log k$.Given: $	ext{Angle} = 45^o$,so $	ext{slope} = 	an(45^o) = 1$. Thus,$\frac{1}{n} = 1$.Given: $	ext{Intercept} = \log k = 0.3010$. Since $\log(2) \approx 0.3010$,we have $k = 2$.The equation becomes $\log(x/m) = 0.3010 + 1 	imes \log(0.2)$.$\log(x/m) = \log(2) + \log(0.2) = \log(2 	imes 0.2) = \log(0.4)$.Therefore,$\frac{x}{m} = 0.4$.

The graph of $\log(x/m)$ versus $\log p$ is a straight line with an angle of $45^o$ and an intercept of $0.3010$ on the $y$-axis. Calculate the amount of gas adsorbed per gram at a pressure of $0.2 \, atm$.

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