The graph drawn between log left(frac{x}{m}ri | vedclass

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

Vedclass · Accepted Answer

$0.6$

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(A) The Freundlich adsorption isotherm is given by $\frac{x}{m} = K P^{1/n}$.Taking logarithm on both sides: $\log \left(\frac{x}{m}ight) = \log K + \left(\frac{1}{n}ight) \log P$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \left(\frac{x}{m}ight)$,$x = \log P$,slope $m = \frac{1}{n}$,and intercept $c = \log K$.From the graph,the intercept $c = \log K = 0.3010$.Since $\log 2 = 0.3010$,we have $K = 2$.The slope of the graph is $	an 45^{\circ} = 1$,so $\frac{1}{n} = 1$.Now,at pressure $P = 0.3 \ atm$,$\log P = \log(0.3) = \log(3 	imes 10^{-1}) = \log 3 + \log 10^{-1} = 0.477 - 1 = -0.523$.Substituting these values into the equation: $\log \left(\frac{x}{m}ight) = 0.3010 + (1)(-0.523) = -0.222$.Therefore,$\frac{x}{m} = 10^{-0.222} \approx 0.6$.

The graph drawn between $\log \left(\frac{x}{m}\right)$ and $\log (p)$ is shown below with intercept $OA = 0.3010$. What is the value of $\left(\frac{x}{m}\right)$ at a pressure of $0.3 \ atm$ ? (Given: $\log 2 = 0.3010, \log 3 = 0.477$)

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