Question

Which of these are correct regarding Freundlich adsorption isotherm, if $\frac{x}{m}$ is the amount adsorbed per unit mass of adsorbent?

• A. $\frac{X}{m}\alpha$ $(P{)}^{1/n}$
• B. $log\frac{X}{m}=logK+\frac{1}{n}logP$
• C. At low $P$, $\frac{X}{m}\alpha$ $P_{\left(V\right)}$
• D. At high $P$, $\frac{X}{m}=K_{\left(V\right)}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\frac{X}{m}\alpha$ $(P{)}^{1/n}$
B $log\frac{X}{m}=logK+\frac{1}{n}logP$
C At low $P$, $\frac{X}{m}\alpha$ $P_{\left(V\right)}$
D At high $P$, $\frac{X}{m}=K_{\left(V\right)}$

According to Freundlich adsorption isotherm, $\frac{x}{m}=K{P}^{1/n}$.

Hence, all are correct.