Question

For the adsorption of a gas on a solid, the plot of $\log (\frac{x}{m})$ versus $\log P$ linear with slope equal to:
A. $K$
B. $\log K$
C. $N$
D. $\frac{1}{n}$

Answer 1

Hint: The plot of $\log (\frac{x}{m})$ versus $\log P$ forms a $y=mx+c$ type of curve where the slope is $m$. The slope can be determined by the Freundlich adsorption isotherm.

Complete Step by Step Solution:
Freundlich adsorption isotherm provided an empirical relation between the amount of gas adsorbed by a solid adsorbent at a pressure and at a particular temperature. It is expressed as follows:
$\frac{x}{m}=k.{{P}^{\frac{1}{n}}}$

Taking $\log $ at both sides we get-
$\log (\frac{x}{m})=\log k+\frac{1}{n}\log P$

Here the term $x$ is the mass of the gas adsorbed. Here $m$ is the mass of the adsorbent at pressure $P$. $k,n$ are the constants that depend on the nature of the adsorbent and the gas at a given temperature. Thus the plot of $\log (\frac{x}{m})$ versus $\log P$ linear with slope equal to $\frac{1}{n}$ .
Thus the correct option is D.

Note: This curve is linearly increasing with slope $\frac{1}{n}$ . If the curve is $y=mx$ type then it will pass through the origin with a slope of $\frac{1}{n}$ .