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) \[\frac{x}{m}=\frac{1+bp}{ap}\]
D) \[\frac{1}{(x/m)}=\frac{a}{b}+\frac{p}{a}\]
E) \[\frac{1}{(x/m)}=\frac{b}{a}+\frac{1}{ap}\]
Correct Answer: E
Solution :
If,\[x/m\]is the mass of adsorbate per unit mass of adsorbent. P is the pressure of the adsorbate gas and a and b are constants. Then Langmuir adsorption isotherm is given as: \[\frac{x}{m}=\frac{ap}{1+bp}\] Or \[\frac{1}{x/m}=\frac{1+bp}{ap}\] \[\frac{1}{x/m}=\frac{1}{ap}+\frac{b}{a}\]
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