The plot of log(frac{x}{m}) against log P is | vedclass

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(D) The Freundlich adsorption isotherm is given by $\frac{x}{m} = K(P)^{1/n}$.Taking logarithm on both sides,we get $\log(\frac{x}{m}) = \log K + \fra

Vedclass · Accepted Answer

$5$

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(D) The Freundlich adsorption isotherm is given by $\frac{x}{m} = K(P)^{1/n}$.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 $y = mx + c$,the slope is $\frac{1}{n}$.Given that the line is inclined at an angle of $45^{\circ}$,the slope is $	an 45^{\circ} = 1$.Therefore,$\frac{1}{n} = 1$,which implies $n = 1$.Substituting the values $K = 10$,$P = 0.5 \ atm$,and $n = 1$ into the isotherm equation:$\frac{x}{m} = 10 	imes (0.5)^{1} = 5$.Since $m = 1 \ g$,the amount of solute adsorbed $x$ is $5 \ g$.

The plot of $\log(\frac{x}{m})$ against $\log P$ is a straight line inclined at an angle of $45^{\circ}$. When the pressure is $0.5 \ atm$ and the Freundlich parameter $(K)$ is $10$,the amount of solute adsorbed per gram of adsorbent will be $(\log 5 = 0.699)$ .......... $g$.

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