Answer the question on the basis of passage given below:
Adsorption is a surface phenomenon which can be expressed by means of an empirical relationship known as Freundlich Adsorption isotherm. The relationship is given by
\(\frac{x}{m} = Kp^{\frac{1}{n}}\left(n > 1\right)\)
\(x = \) Mass of the gas adsorbed
\(m = \) Mass of adsorbent
\(p = \) Pressure at which adsorption takes place
\(K\) and \(n\) are constant
The curve of above relationship indicates that at a fixed pressure there is a decrease in physical adsorption with increase in temperature. At high pressure, curves become parallel to pressure axis, indicating saturation of the surface of adsorbent.
Adsorption of a gas follows Freundlich isotherm shown below. \(\frac{x}{m}\) is proportional to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(p^{1/3}\)
Taking logarithms of the Freundlich equation x/m = k p1/n gives a straight line:
\( \displaystyle \log\frac{x}{m} = \log k + \frac{1}{n}\log p \)
The slope of the plot of log x/m against log p is 1/n. In the graph the line rises 1 unit for every 3 units along log p:
\( \displaystyle \text{slope} = \frac{1}{3} = \frac{1}{n} \;\Rightarrow\; \frac{x}{m} = k\,p^{1/3} \)
So x/m is proportional to p1/3.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Adsorption Isotherms – “… The validity of Freundlich isotherm can be verified by plotting log \( \frac{x}{m} \) on y-axis (ordinate) and log P on x-axis (abscissa). If it comes to be a straight line, the Freundlich isotherm is valid, otherwise not (Fig. 5.2). The slope of the straight line gives the value of \( \frac{1}{n} \). The intercept on the y-axis gives the value of log k.”
