If x gram of gas is adsorbed by m gram of ads | vedclass

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(B) According to the Freundlich adsorption isotherm,the relationship is given by:$\frac{x}{m} = kP^{\frac{1}{n}}$Taking the logarithm on both sides:$\

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$\frac{1}{n}$

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(B) According to the Freundlich adsorption isotherm,the relationship is given by:$\frac{x}{m} = kP^{\frac{1}{n}}$Taking the logarithm on both sides:$\log_{10} \frac{x}{m} = \log_{10} (kP^{\frac{1}{n}})$$\log_{10} \frac{x}{m} = \frac{1}{n} \log_{10} P + \log_{10} k$This equation follows the linear form $y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log P$,the slope $m = \frac{1}{n}$,and the intercept $c = \log k$.Therefore,the slope of the plot is $\frac{1}{n}$.

If $x$ gram of gas is adsorbed by $m$ gram of adsorbent at pressure $P$,the plot of $\log \frac{x}{m}$ versus $\log P$ is linear. The slope of the plot is: ($n$ and $k$ are constants and $n > 1$ )

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