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Answer:
† Question :-
Explain Freundlich adsorption isotherm briefly with mathematical calculations.
\large \dag† Answer :-
\rm \purple \maltese \: \: \: \red{ \underline{ \underline{Freundlich \: Adsorption \: Isotherm }}}✠
FreundlichAdsorptionIsotherm
According to Freundlich adsorption isotherm the extent of adsorption is directly proportional to fractional power of pressure.
In mathematical form :-
\bf \red\bigstar \: \: \orange{ \pmb{ \underbrace{ \underline{ \blue{ \frac{x}{m} \propto p {}^{ \ 1/n} \: \: \: }}}}}★
m
x
∝p
1/n
m
x
∝p
1/n
\begin{gathered}:\longmapsto \rm \frac{ x }{m} = kp {}^{1/n} \: ; \small0 < n < 1\: \: - - - (1) \\ \end{gathered}
:⟼
m
x
=kp
1/n
;0<n<1−−−(1)
❒ Taking log on both sides of (1) :
\begin{gathered}:\longmapsto \rm log \frac{x}{m} = log( kp {}^{1/n} )\\ \end{gathered}
:⟼log
m
x
=log(kp
1/n
)
\begin{gathered}:\longmapsto \rm log \frac{x}{m} = log(k) + log( p {}^{1/n} ) \\ \end{gathered}
:⟼log
m
x
=log(k)+log(p
1/n
)
\begin{gathered}:\longmapsto \rm log \frac{x}{m} = log(k) + \frac{1}{n} log( p ) - - - (2) \\ \end{gathered}
:⟼log
m
x
=log(k)+
n
1
log(p)−−−(2)
Also graph of both equations are given in attachment.