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# If x gram of gas is adsorbed by m gram of adsorbent at pressure P, the plot of log $\dfrac{x}{m}$ versus P is linear. The slope of the plot is: (n and k are constants and n>1)A. log kB. $\dfrac{1}{n}$C. 2kD. n

Last updated date: 02nd Aug 2024
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Hint: We know that surface chemistry is the study of phenomenon occurring in the surface of substances. The term adsorption is used when solute particles assimilated at the surface of substances.

Let’s discuss two terms, adsorbate and adsorbent in surface chemistry. The solute particles that adsorbed at the surface of substances are termed as adsorbate and the substance on which adsorbate accumulated is termed as adsorbent.
The given question is on Freundlich adsorption isotherm. Now, we understand about the Freundlich adsorption isotherm. It is the mathematical representation of the variation of extent of adsorption (x/m) with respect to pressure at a given temperature. This shows the mass of gas adsorbed in one gram of adsorbent with pressure at constant temperature.
The representation of Freundlich adsorption isotherm is,

$\dfrac{x}{m} = K{P^{1/n}}$ …… (1)

Where, x is the mass of adsorbate, m is the mass of adsorbent, P is pressure and K and n are constants for the adsorbate and adsorbent at particular temperature.
Now, we take logarithm in equation (1).

$log\dfrac{x}{m} = log K{P^{1/n}}$

$log \dfrac{x}{m} = \dfrac{1}{n}\log P + \log K$ …… (2)

Now, we have to compare equation (2) with the equation of the straight line, that is, $y = mx + c$, where m is the slope of the line and c is y intercept.

Therefore, the slope of the plot is $\dfrac{1}{n}$ and y intercept is $\log \,K$.

Hence, the correct option is B.

Note: It is to be noted that Fruendlich adsorption isotherm is not applicable for adsorption of solids at higher pressure. At higher pressure the value of $\dfrac{1}{n}$ is zero. Therefore, the extent of adsorption does not depend on pressure.