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$10 \mathrm{~g}$ of a gas is adsorbed on $500 \mathrm{~g}$ of solid at $10 \mathrm{bar}$. If the pressure is increased at $20 \mathrm{bar}, 14 \mathrm{~g}$ of the gas is adsorbed by the same solid at the same temperature. What is the slope of Freundlich adsorption isotherm?
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0.5
Freundlich Adsorption Isotherm:

(ii) $\div(\mathrm{i}) \Rightarrow \frac{14 / 500}{10 / 500}=\frac{\mathrm{k}(20)^{\frac{1}{\mathrm{n}}}}{\mathrm{k}(10)^{\frac{1}{\mathrm{n}}}}$
$\begin{aligned} & \Rightarrow \frac{14}{10}=2^{1 / n}, \\ & \therefore \quad \frac{1}{n}=0.485 \approx 0.5\end{aligned}$

(ii) $\div(\mathrm{i}) \Rightarrow \frac{14 / 500}{10 / 500}=\frac{\mathrm{k}(20)^{\frac{1}{\mathrm{n}}}}{\mathrm{k}(10)^{\frac{1}{\mathrm{n}}}}$
$\begin{aligned} & \Rightarrow \frac{14}{10}=2^{1 / n}, \\ & \therefore \quad \frac{1}{n}=0.485 \approx 0.5\end{aligned}$
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