Which of the following is correct for Freundl | vedclass

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Question

(B) According to the Freundlich adsorption isotherm,the relationship is given by $x/m = K P^{1/n}$.Here,$x/m$ is the amount of gas adsorbed per unit m

Vedclass · Accepted Answer

$x/m = K P^{1/n}; 0 \le 1/n \le 1$

Vedclass · Answer

(B) According to the Freundlich adsorption isotherm,the relationship is given by $x/m = K P^{1/n}$.Here,$x/m$ is the amount of gas adsorbed per unit mass of the adsorbent at pressure $P$.$K$ and $n$ are constants that depend on the nature of the gas and the adsorbent.The value of $1/n$ ranges between $0$ and $1$ (i.e.,$0 \le 1/n \le 1$),where $n > 1$.At low pressure,$1/n = 1$,so $x/m \propto P^1$.At high pressure,$1/n = 0$,so $x/m \propto P^0$ (independent of pressure).Thus,the general expression is $x/m = K P^{1/n}$ with $0 \le 1/n \le 1$.

Gas	$H_2$	$CH_4$	$CO_2$	$SO_2$
Critical Temp $/ K$	$33$	$190$	$304$	$630$

Which of the following is correct for Freundlich adsorption isotherm?

Similar Questions

Which one of the following gives a straight line for Freundlich adsorption isotherm?

Adsorption increases when:

When a graph is plotted between $\log (x/m)$ and $\log P$,it is a straight line with an angle of $45^o$ and an intercept of $0.3010$ on the $y-$axis. If the initial pressure is $0.3 \ atm$,what will be the amount of gas adsorbed per $g$ of adsorbent $(x/m)$?

Given the following data:
Gas $H_2$ $CH_4$ $CO_2$ $SO_2$
Critical Temp $/ K$ $33$ $190$ $304$ $630$
Based on the data provided,predict which of the following gases shows the least adsorption on a definite amount of charcoal.

Assertion $(A)$: Finely divided charcoal can act as an adsorbent.
Reason $(R)$: Finely divided material has a small surface area.

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