In Freundlich adsorption isotherm, slope of AB line is:

Freundlich adsorption isotherm is :

$\dfrac{x}{m} = {kp}^{1 / n}$

$x =$ mass of adsorbate
$m =$ mass of adsorbent
$P =$ eq. pressure

${k}_{1}n = \dfrac{1}{n}logp + logk$
$y = mx + c$

compairing

$m = \dfrac{1}{n} =$slope \left[\dfrac{1}{n} = 0to1\right]$</p><p>$n > 1$</p><p>$\log \left(\dfrac{x}{m}\right) = logK + \dfrac{1}{n}logP$</p><p>$y = c + mx$</p><p>$m = 1 / n$so slope will be equal to$1 / n$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/6867faff-a5c9-4a8d-848c-6450aeff9f69-image.png" style="width: 251px; height: 226px;" /></p><p>Hence$0 \le \dfrac{1}{n} \le 1