The following problem was asked in JEE Mains 2020 (Sept 2, Shift 1),

The mass of gas adsorbed, $x$, per unit mass of adsorbate, $m$, was measured at various pressures, $p$. A graph between $\log\frac xm$ and $\log p$ gives a straight line with slope equal to $2$ and the intercept equal to $0.4771$. The value of $\frac xm$ at a pressure of $\pu{4 atm}$ is: (Given $\log 3 = 0.4771$)

I know the Freundlich adsorption isotherm formula, but I decided to write the equation of given line $( y = c + mx)$ as,

$$ \log \frac xm = \log 3 + 2\log p $$

So, the relation comes out to be,

$$ \frac xm = 3p^2 $$

On plugging $p = 4$, we get,

$$ \frac xm = 3(16) = 48 $$

But, the answer was wrong as per the official key, which mostly has genuine answers. Where am I wrong?

The answer given is,


  • $\begingroup$ $1/n$ lies between 0 and 1.. $\endgroup$ Sep 8 '20 at 14:38
  • $\begingroup$ @Safdar: Yes, I'd doubt over this during the exam. But, if I take $n=2$ and then, draw for $\log p$ v/s $\log x/m$, then it goes very complex. $\endgroup$ Sep 8 '20 at 14:41
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    $\begingroup$ I think the question has a mistake. You should take $n=2$ then equation is: $$\log \frac{x}{m} = \log 3 + \frac{1}{n} \log p$$ and you got the given answer. $\endgroup$ Sep 8 '20 at 15:20
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    $\begingroup$ @MathewMahindaratne, Great, I never thought that. Actually, I have to challenge the given key, that's why I asked. Thanks for pointing this thing :) $\endgroup$ Sep 8 '20 at 15:30

One version of Freundlich adsorption isotherm equation is: $$ \frac xm = Kp^{\frac12}, $$ which can also be written as: $$ \log \frac xm = \log K + \frac12\log p $$ This is a straight line equation of type $( y = c + mx)$ as given in the question. However, I think the question has made a mistake saying the slope is $2$, but instead it should be $\frac12$. Accordingly, the equation with given numeric values should be:

$$ \log \frac xm = \log 3 + \frac12\log p $$

Or without logarithm:

$$ \frac xm = 3p^\frac12 $$

On plugging $p = 4$, You'll get:

$$ \frac xm = 3 \times 4^\frac12 = 3 \times 2 = 6 $$

Hence, you get the given answer.


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