Freundlich adsorption isotherms for the physical adsorption of a gas at temperature $T_1, T_2$ and $T_3$ are…
Freundlich adsorption isotherms for the physical adsorption of a gas at temperature $T_1, T_2$ and $T_3$ are shown in the graph given below. The correct relationship between $T_1, T_2$ and $T_3$ is
$T_1 < T_2 < T_3$
$T_3 < T_1 < T_2$
$T_3 < T_2 < T_1$
$T_2 < T_1 < T_3$
Solution
Lower the temperature more the adsorption. So, as the temperature increases adsorption decreases. According to Freundlich adsorption isotherm.
$
\frac{x}{m}=k \cdot p^{1 / n}(n>1)
$
where, $x$ is the mass of gas adsorbed.
$
\begin{aligned}
& \text { or } \quad \log \frac{x}{m}=\log k+\frac{1}{n} \log p \\
& \therefore \quad T_3 < T_2 < T_1
\end{aligned}
$
