SATHEE: Surface Chemistry - Result Question 17

Surface Chemistry - Result Question 17

####17. Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, $x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent at pressure $p \cdot \frac{x}{m}$ is proportional to

(2019 Main, 9 Jan I)

(a) $p^{2}$

(b) $p^{1 / 4}$

(d) $p$

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Answer:

Correct Answer: 17. (d)

Solution:

According to Freundlich adsorption isotherm,

$\frac{x}{m} \propto p^{1 / n} \Rightarrow \frac{x}{m} = K p^{1 / n}$

On taking $\log$ on both sides, we get

$\log (\frac{x}{m}) = \log K + \frac{1}{n} \log p$

On comparing with equation of straight line, $y = m x + c$ , plot of $\log \frac{x}{m} v_{s} \log p$ gives,

$\begin{array}{lc} Slope \frac{(y_{2} - y_{1})}{(x_{2} - x_{1})} = \frac{1}{n} \Rightarrow \frac{2}{4} = \frac{1}{2} ∴ & \frac{x}{m} \propto p^{1 / 2} \end{array}$

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Surface Chemistry - Result Question 17

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