A) \[\log \left( \frac{x}{m} \right)=\log \left( \frac{a}{b} \right)+\frac{1}{a}\log \,P\]
B) \[\frac{x}{m}=\frac{b}{a}+\frac{1}{ap}\]
C) \[\frac{1}{({\scriptstyle{}^{x}/{}_{m}})}=\frac{b}{a}+\frac{1}{aP}\]
D) \[\frac{x}{m}=\frac{1+bP}{aP}\]
Correct Answer: C
Solution :
If \[{\scriptstyle{}^{x}/{}_{m}}\] is the mass of adsorbate per unit mass of adsorbent .P is the pressure of the adsorbate gas and ?a? and ?b? are constant. Then Langmuir adsorption isotherm is given as: \[\frac{x}{m}=\frac{aP}{1+bP}\] \[\frac{1}{{\scriptstyle{}^{x}/{}_{m}}}=\frac{1+bP}{aP}\] \[\frac{x}{{\scriptstyle{}^{x}/{}_{m}}}=\frac{1}{aP}+\frac{b}{a}\]
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