The graph of log frac{x}{m} vs log p for an a | vedclass

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(D) The Freundlich adsorption isotherm is given by the equation: $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$. Comparing this with the equation of

Vedclass · Accepted Answer

$16$

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(D) The Freundlich adsorption isotherm is given by the equation: $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$. Comparing this with the equation of a straight line $y = mx + c$,we have the slope $\frac{1}{n} = 	an 45^{\circ} = 1$ and the intercept $c = \log k = 0.6020$. Since $\log 2 = 0.3010$,then $2 	imes \log 2 = 0.6020$,which implies $\log 2^2 = \log 4 = 0.6020$. Thus,$k = 4$. Substituting these values into the isotherm equation: $\frac{x}{m} = k \cdot P^{1/n} = 4 	imes (0.4)^1 = 1.6$. Expressing $1.6$ in the form $......... 	imes 10^{-1}$,we get $1.6 = 16 	imes 10^{-1}$. Therefore,the nearest integer is $16$.

Gas	$H_2$	$CH_4$	$CO_2$	$SO_2$
Critical Temp $/ K$	$33$	$190$	$304$	$630$

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