JEE Advanced 2022 Paper 1 · Chemistry · Solutions, Solid State and Surface Chemistry · medium

The adsorption of a gas on a solid obeys the Langmuir isotherm v=vm(KP)/(1+KP), where v is the volume (at STP) adsorbed per gram of the solid. At a…

Numerical answer

The adsorption of a gas on a solid obeys the Langmuir isotherm

v=v_m\dfrac{KP}{1+KP}

, where

v

is the volume (at STP) adsorbed per gram of the solid. At a fixed temperature,

v=12\ \mathrm{mL\,g^{-1}}

P=0.2\ \mathrm{bar}

and

v=18\ \mathrm{mL\,g^{-1}}

P=0.6\ \mathrm{bar}

. The pressure, in

\mathrm{bar}

, at which the volume adsorbed is

21\ \mathrm{mL\,g^{-1}}

is (give the answer correct to two decimal places)

Show answer and solution

Answer: 1.40

Divide the two isotherm equations to eliminate

v_m

\dfrac{18}{12}=\dfrac{0.6(1+0.2K)}{0.2(1+0.6K)}=\dfrac{3(1+0.2K)}{1+0.6K}

. Hence

1.5+0.9K=3+0.6K

, so

K=5\ \mathrm{bar^{-1}}

. Substituting in the first equation,

12=v_m\dfrac{5(0.2)}{1+5(0.2)}=v_m\cdot\dfrac12

, so

v_m=24\ \mathrm{mL\,g^{-1}}

(monolayer capacity). For

v=21

, the fractional coverage is

\theta=21/24=0.875

. From

\theta=\dfrac{KP}{1+KP}

we get

KP=\dfrac{\theta}{1-\theta}=7

, so

P=7/5=1.40\ \mathrm{bar}

. Check: at

P=1.4

KP=7

and

v=24\times7/8=21

. A common error is to treat the isotherm as the Freundlich equation, which has no saturation limit. Answer: 1.40.

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