In Langmuir adsorption isotherm, the extent of adsorption increases linearly with pressure at:Option: 1 Low pressure range<div class='q

In Langmuir adsorption isotherm, the extent of adsorption increases linearly with pressure at:
Option: 1
Low pressure range

Option: 2
High pressure range

Option: 3
At intermediate pressure range

Option: 4
All pressure range.

Answers (1)

In Langmuir adsorption isotherm,
$\theta$ : Fraction of surface covered on adsorbent
$1-\theta:$ Free surface are
$\mathrm{P}:$ Pressure of gas

By langmuir adsorption isotherm,

$\mathrm{\text{Rate of adsorption }=k_{a} P(1-\Theta)}$
$\mathrm{\text{Rate of desorption }=k_{d} \Theta}$
$\mathrm{k_{a}}$ and $\mathrm{k_{d}}$ are rate constants.

At equilibrium condition,
Rate of adsorption $=$ Rate of desorption

$\mathrm{k_{a} P(1-\Theta)=k_{d} \Theta}$
$\mathrm{\Theta=\frac{K P}{K P+1}}$

According to Langmuir, extent of adsorption is proportional to $\mathrm{\Theta}$
$\mathrm{\frac{x}{m} \propto \Theta}$
$\mathrm{\frac{x}{m} \propto \frac{K P}{1+K P}}$
$\mathrm{\frac{x}{m}=K^{\prime} \frac{K P}{1+K P}}$
$\mathrm{\frac{x}{m}=\frac{K^{\prime} K P}{1+K P}}$
$\mathrm{\frac{x}{m}=\frac{a P}{1+b P}}$

Where,
$\mathrm{a=K^{\prime} K=K^{\prime} \frac{k_{a}}{k_{d}}}$
$\mathrm{b=K=\frac{k_{a}}{k_{d}}}$
$\mathrm{\frac{x}{m}=\frac{a P}{1+b P}}$

At low pressure, $\mathrm{b P<<1}$
so, $\mathrm{1+b P \equiv 1}$
$\mathrm{\frac{x}{m}=a P}$

Thus, $\mathrm{\frac{x}{m}}$ increases linearly with pressure.