Question

In Langmuir adsorption isotherm, the extent of adsoption increases linearly with pressure at

• A. Low pressure range
• B. High pressure range
• C. At inremediate pressure range
• D. All pressure range

Answer 1

The correct option is A Low pressure range

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

By langmuir adsorption isotherm,
$\text{Rate of adsorption}=k_{a}P(1-\Theta )$
$\text{Rate of desorption}=k_d\Theta$
$k_a$ and $k_d$ are rate constants.

At equilibrium condition,
$\text{Rate of adsorption}=\text{Rate of desorption}$
$k_{a}P(1-\Theta )=k_d\Theta$
$\Theta =\frac{KP}{KP+1}$

According to Langmuir, extent of adsorption is proportional to $\Theta$
$\frac{x}{m}\propto \Theta$
$\frac{x}{m}\propto \frac{KP}{1+KP}$

$\frac{x}{m}=K^{\prime }\frac{KP}{1+KP}$

$\frac{x}{m}=\frac{K^{\prime }KP}{1+KP}$
$\frac{x}{m}=\frac{aP}{1+bP}$
where,
$a=K^{\prime }K=K^{\prime }\frac{k_a}{k_d}$
$b=K=\frac{k_a}{k_d}$

$\frac{x}{m}=\frac{aP}{1+bP}$

At low pressure, $bP<<1$
so, $1+bP\equiv 1$
$\frac{x}{m}=aP$
Thus, $\frac{x}{m}$ increases linearly with pressure.