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Q.Observe the given figure and answer the following questions : The figure is a straight-line plot of log⁡xm\log \frac{x}{m} (y-axis) against log⁡p\log p (x-axis) with a positive slope and a positive intercept on the log⁡xm\log \frac{x}{m} axis (the Freundlich adsorption isotherm, where xm\frac{x}{m} is the mass of gas adsorbed per gram of adsorbent and pp is the pressure).

(i) Write the expression for adsorption of gases on solids in the form of an equation.
(ii) What is the slope of the graph ?
(iii) What does the intercept of the line represent ?
CBSECBSE Class XII Board 2022Subjective· 3mImportance★★★★★est
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The Freundlich adsorption isotherm gives xm=kp1/n\frac{x}{m} = k p^{1/n}. The plot of log⁡xm\log \frac{x}{m} vs log⁡p\log p is a straight line with slope 1/n1/n and intercept log⁡k\log k. The slope is positive and less than 1, and the intercept represents the adsorption capacity constant kk.

The Freundlich isotherm is an empirical relationship that describes how the amount of gas adsorbed per unit mass of adsorbent changes with pressure. When you take logarithms on both sides, the equation becomes linear — that’s why plotting log⁡(x/m)\log (x/m) against log⁡p\log p gives a straight line. The slope and intercept of this line each carry a physical meaning.

Let’s go through the three parts step by step.


1. Writing the Freundlich adsorption equation

The Freundlich isotherm states that at a fixed temperature, the mass of gas adsorbed per gram of adsorbent (x/mx/m) is proportional to the pressure raised to some power 1/n1/n, where n>1n > 1:

xm=k p1/n\frac{x}{m} = k \, p^{1/n}

Here:

  • xx = mass of gas adsorbed
  • mm = mass of adsorbent
  • pp = equilibrium pressure
  • kk and nn are constants that depend on the adsorbent, gas, and temperature. kk is related to the adsorption capacity, and nn indicates how favourable the adsorption is.

xm=kp1/n\frac{x}{m} = k p^{1/n}

This is the required expression for part (i).


2. Finding the slope of the graph

Take the natural (or common) logarithm of both sides of the Freundlich equation:

log⁡xm=log⁡k+1nlog⁡p\log \frac{x}{m} = \log k + \frac{1}{n} \log p

Compare this with the equation of a straight line y=mx+cy = mx + c, where y=log⁡(x/m)y = \log (x/m) and x=log⁡px = \log p:

  • Slope m=1nm = \frac{1}{n}
  • Intercept c=log⁡kc = \log k

Since the problem states the graph has a positive slope, and n>1n > 1 for typical adsorption, the slope 1/n1/n is positive and lies between 0 and 1. …

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