If $$x$$ is the mass of the gas adsorbed on mass $$m$$ of the adsorbent at pressure $$p$$, Freundlich adsorption isotherm gives a straight line on plotting

The Freundlich adsorption isotherm is an empirical relation that connects the extent of adsorption with the equilibrium pressure (or concentration) of the adsorbate. It is written as

$$\frac{x}{m}=k\,p^{1/n} \qquad (k \gt 0,\; n \gt 1) \;$$ $$-(1)$$

Here

• $$x$$ = mass of gas adsorbed (at equilibrium)
• $$m$$ = mass of adsorbent
• $$p$$ = equilibrium pressure of the gas
• $$k$$ and $$n$$ are adsorption constants that depend on the nature of the adsorbent-adsorbate pair and temperature.

To transform $$-(1)$$ into the equation of a straight line, take logarithms (base 10 or natural - either gives linearity). Using common logarithms:

$$\log\!\left(\frac{x}{m}\right)=\log k + \frac{1}{n}\,\log p \qquad -(2)$$

Equation $$-(2)$$ is of the form $$y = c + m\,x$$, where

• dependent variable, $$y = \log(x/m)$$,
• independent variable, $$x = \log p$$,
• slope, $$m = \dfrac{1}{n}$$,
• intercept, $$c = \log k$$.

Hence, a plot of $$\log(x/m)$$ on the ordinate versus $$\log p$$ on the abscissa yields a straight line with slope $$1/n$$ and intercept $$\log k$$.

Therefore, the correct choice is:
Option C which is: $$\log x/m$$ vs $$\log p$$