The physical adsorption of a gas follows the Freundlich adsorption isotherm. In an experiment, of a — Surface Chemistry Chemistry Question
Question
The physical adsorption of a gas follows the Freundlich adsorption isotherm. In an experiment, $x\text{ grams}$ of a gas is adsorbed on $1\text{ g}$ of a solid adsorbent at an equilibrium pressure of $2\text{ atm}$. When the pressure is increased to $32\text{ atm}$ at the same temperature, the amount of gas adsorbed becomes $4x\text{ grams}$. What is the value of $1/n$ for this specific adsorption isotherm?
💡 Solution & Explanation
According to the Freundlich isotherm, $\frac{x}{m} \propto p^{1/n}$. Here, $m=1\text{ g}$ is constant. So, $\frac{x_2}{x_1} = \left(\frac{p_2}{p_1}\right)^{1/n}$. Substituting the given values: $\frac{4x}{x} = \left(\frac{32}{2}\right)^{1/n} \implies 4 = (16)^{1/n}$. Since $16 = 4^2$, we have $4 = (4^2)^{1/n} = 4^{2/n}$. Equating the exponents gives $1 = \frac{2}{n} \implies \frac{1}{n} = 0.5$.