Question

Adsorption of a gas on a solid adsorbent follows the Freundlich adsorption isotherm. If \(x\) is the mass of the gas adsorbed on mass \(m\) of the adsorbent at pressure \(p\), and from the graph of \(\log\left(\frac{x}{m}\right)\) vs. \(\log p\) we obtain a slope of \(\tfrac{1}{2}\), then the extent of adsorption is proportional to: 

• A. \(p^{\tfrac{1}{2}}\)
• B. \(p^2\)
• C. \(p\)
• D. \(p^{\tfrac{1}{4}}\)

Hint:

The Freundlich isotherm is an empirical relation: \(\tfrac{x}{m} = k\,p^{1/n}\). - The slope in a \(\log(\tfrac{x}{m})\) vs. \(\log p\) plot directly gives \(\tfrac{1}{n}\).

Answer 1

The Correct Option is A

Step 1: Freundlich Adsorption Isotherm \[ \frac{x}{m} = k\, p^{\,\frac{1}{n}}. \] 

Step 2: Taking Logarithms \[ \log\left(\frac{x}{m}\right) = \log(k) + \frac{1}{n}\,\log(p). \] A plot of \(\log\left(\frac{x}{m}\right)\) vs. \(\log(p)\) is a straight line with slope \(\frac{1}{n}\). 

Step 3: Slope Equals \(\frac{1}{2}\) If the slope is \(\frac{1}{2}\), then \[ \frac{1}{n} = \frac{1}{2} \quad \Longrightarrow \quad n = 2. \] Hence, \[ \frac{x}{m} \propto p^{\frac{1}{2}}, \] or \(\frac{x}{m} = k\, p^{1/2}\).