Answer the question on the basis of the passage given below:
Adsorption is a surface phenomenon and it differs from absorption which occurs throughout the body of the substance which absorbs. In physisorption, the attractive forces are mainly van der Waals forces while in chemisorption actual bonding occurs between the particles of adsorbent and adsorbate. Gases which are easily liquefiable are adsorbed more easily on the surface of a solid as compared to the gases which are liquefied with difficulty. Adsorption increases with increase in pressure and decreases as the temperature is increased.
Freundlich adsorption isotherm gives a straight line on plotting
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(log\, \ x/m\) vs\(log\, \ p\)
The Freundlich adsorption isotherm relates the mass of gas adsorbed per gram of adsorbent (x/m) to the pressure at a fixed temperature:
\( \displaystyle \frac{x}{m} = k\,p^{1/n} \quad (n \gt 1) \)
This is a curve, not a straight line. Taking logarithms on both sides gives a linear form:
\( \displaystyle \log \frac{x}{m} = \log k + \frac{1}{n}\log p \)
This has the form y = c + mx, so a plot of log x/m against log p is a straight line with slope 1/n and intercept log k. If the experimental points fall on a straight line, the Freundlich isotherm holds.
Why the other options are incorrect
Option (1) \(x/m\) vs\(p\) — Incorrect. This is the adsorption isotherm itself; it is a curve that rises and then levels off towards saturation at high pressure.
Option (2) \(log\, \ x/m\) vs\(p\) — Incorrect. Only one axis is logarithmic, so the plot stays curved.
Option (4) \(x/m\) vs \(1/p\) — Incorrect. x/m falls as 1/p grows; the plot is a curve, not the straight line of the Freundlich equation.
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 – …
