QJEEM17Chcf9f: JEE Main 2017 Chemistry - Class 12 JEEM17Chcf9f Solution: Adsorption of a gas on a surface follows Freundlich adsorption…

Question

Verified Solution • 2025–26 Session moderate

Adsorption of a gas on a surface follows Freundlich adsorption isotherm. Plot of log

\frac{x}{m}

versus log p gives a straight line with slope equal to 0.5, then:

(

\frac{x}{m}

is the mass of the gas adsorbed per gram of adsorbent)

Accepted Answer

According to Freundlich adsorption isotherm, in the median range of pressure

\frac{x}{m} \propto P^{\frac{1}{n}}

\Rightarrow

\frac{x}{m}

= kP

^{\frac{1}{n}}

taking log both sides, we get,

log

\frac{x}{m}

= logk +

\frac{1}{n}

logP

Here in graph between log

\frac{x}{m}

and logP, slope is

\frac{1}{n}

and intercepts = log k.

Given that,

slope = 0.5

∴

\frac{1}{n}

= 0.5

\Rightarrow

n = 2

∴

\frac{x}{m}

= kP

^{\frac{1}{2}}

= k

\sqrt{P}

∴

Adsorption is proportional to the square root of pressure.

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