$\frac{x}{m}=4 k$

$\frac{x}{m}=\frac{1.414}{k}$

$\frac{x}{m}=\frac{k}{1.414}$

$\frac{x}{m}=1.414 k$

Gel

Emulsion

Foam

Aerosol

$\log \frac{x}{m}=\log p+\frac{1}{n} \log k$

$\log \frac{x}{m}=\log k+\frac{1}{n} \log p$

$\frac{x}{m}=k+\frac{1}{n} \log p$

$\frac{x}{m}=\log p+\frac{1}{n} \log k$

It is not specific in nature.

Enthalpy of adsorption of this is low.

It increases with increase of temperature.

It is a multilayer adsorption under high pressure.