Adsorption of a gas on an adsorbent follows F | vedclass

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(C) The Freundlich adsorption isotherm equation is given by: $\log \frac{x}{m} = \log k + \frac{1}{n} \log p$. Comparing this with the equation of a s

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$100$

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(C) The Freundlich adsorption isotherm equation is given by: $\log \frac{x}{m} = \log k + \frac{1}{n} \log p$. Comparing this with the equation of a straight line $y = mx + c$, where $y = \log \frac{x}{m}$, $x = \log p$, slope $= \frac{1}{n}$, and intercept $= \log k$. Given: Slope $(\frac{1}{n}) = 0.5$ and Intercept $(\log k) = 1.0$. Pressure $(p) = 100 \ atm$. Substituting the values into the equation: $\log \frac{x}{m} = 1.0 + 0.5 	imes \log(100)$. Since $\log(100) = 2$, we get: $\log \frac{x}{m} = 1.0 + 0.5 	imes 2 = 1.0 + 1.0 = 2.0$. Therefore, $\frac{x}{m} = 10^2 = 100$.

Adsorption of a gas on an adsorbent follows Freundlich adsorption isotherm. The slope and intercept (on $y$-axis) of the isotherm are $0.5$ and $1.0$ respectively. What is the value of $\frac{x}{m}$, when the pressure of the gas $(p)$ is $100 \ atm$?

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