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

Which of the following is not a characteristic of chemisorption?

Identify the factors which favour the physical adsorption from the following.
$(I)$ High surface area
$(II)$ Low temperatures
$(III)$ High temperatures
$(IV)$ Low pressures
$(V)$ High pressures

Adsorption of a gas on a surface follows the Freundlich adsorption isotherm. The plot of $\log \frac{x}{m}$ versus $\log P$ gives a straight line with a slope equal to $0.5$. Then,(where $\frac{x}{m}$ is the mass of the gas adsorbed per gram of adsorbent):

Adsorption due to strong chemical forces is called

Which of the following statements is correct for the spontaneous adsorption of a gas on a solid surface?

Vedclass Test Series

Exam Paper Generator

Online Exam Module