Which of the following is the correct expression for adsorption isotherm. (x = mass of gas adsorbed, m = mass of adsorbent, p = pressure, k and n = constants)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \log \displaystyle\frac{x}{m} = \log k + \displaystyle\frac{1}{n} \log p \)
The Freundlich adsorption isotherm relates the mass of gas adsorbed per gram of adsorbent to the pressure at a given temperature:
\( \displaystyle \frac{x}{m} = k\,p^{1/n} \qquad (n > 1) \)
Taking logarithms on both sides:
\( \displaystyle \log \frac{x}{m} = \log k + \frac{1}{n}\log p \)
A plot of log(x/m) against log p is a straight line with slope 1/n and intercept log k.
Why the other options are incorrect
Option (2) — Incorrect. It uses m/x instead of x/m; the amount adsorbed is x per unit mass m of adsorbent.
Option (3) — Incorrect. The roles of p and x/m are interchanged.
Option (4) — Incorrect. The exponent is 1/n, not n; with n in the slope, adsorption would rise faster than the pressure, which is not what is observed.
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.