According to the Freundlich adsorption isothe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The Freundlich adsorption isotherm is given by the equation: $x/m = k \cdot p^{1/n}$ (where $n > 1$).Taking the logarithm on both sides,we get: $\

Vedclass · Accepted Answer

$\log(x/m) 	ext{ vs. } \log(p)$

Vedclass · Answer

(B) The Freundlich adsorption isotherm is given by the equation: $x/m = k \cdot p^{1/n}$ (where $n > 1$).Taking the logarithm on both sides,we get: $\log(x/m) = \log(k) + (1/n) \log(p)$.This equation is in the form of a straight line $y = mx + c$,where $y = \log(x/m)$,$x = \log(p)$,slope $m = 1/n$,and intercept $c = \log(k)$.Therefore,the plot of $\log(x/m)$ versus $\log(p)$ gives a straight line.

According to the Freundlich adsorption isotherm,a straight line is obtained for the plot of:

Similar Questions

Which of the following is correct for Freundlich adsorption isotherm?

Which of the following is incorrect for physisorption?

In the Freundlich adsorption isotherm equation $(x/m) = kp^{1/n}$,the value of $n$ is:

The critical temperatures of gases $A, B, C,$ and $D$ are $190 \ K, 630 \ K, 261 \ K,$ and $400 \ K$ respectively. The quantity of gas adsorbed per gram of charcoal at the same pressure is least for the gas:

In the process of adsorption,which of the following values is less than zero?

Vedclass Test Series

Exam Paper Generator

Online Exam Module