Adsorption of a gas on an adsorbent follows F | vedclass

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

Question

(C) The Freundlich adsorption isotherm equation is given by: $\log \frac{x}{m} = \log k + \frac{1}{n} \log p$. Comparing this with the equation of a s

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(C) The Freundlich adsorption isotherm equation is given by: $\log \frac{x}{m} = \log k + \frac{1}{n} \log p$. Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log p$,slope $= \frac{1}{n}$,and intercept $= \log k$. Given: Slope $(\frac{1}{n}) = 0.5$ and Intercept $(\log k) = 1.0$. Pressure $(p) = 100 \ atm$. Substituting the values into the equation: $\log \frac{x}{m} = 1.0 + 0.5 	imes \log(100)$. Since $\log(100) = 2$,we get: $\log \frac{x}{m} = 1.0 + 0.5 	imes 2 = 1.0 + 1.0 = 2.0$. Therefore,$\frac{x}{m} = 10^2 = 100$.

Adsorption of a gas on an adsorbent follows Freundlich adsorption isotherm. The slope and intercept (on $y$-axis) of the isotherm are $0.5$ and $1.0$ respectively. What is the value of $\frac{x}{m}$,when the pressure of the gas $(p)$ is $100 \ atm$?

Similar Questions

The rate of physical adsorption increases with......

The Freundlich adsorption isotherm is given by:

If $x$ is the amount of adsorbate and $m$ is the amount of adsorbent,which of the following relations is not related to the adsorption process?

Choose the correct statement for chemical adsorption.

Assertion : In chemisorption,adsorption first increases with temperature and then decreases.
Reason : Heat keeps on providing more and more activation energy.

Vedclass Test Series

Exam Paper Generator

Online Exam Module