CO$$_2$$ gas adsorbs on charcoal following Freundlich adsorption isotherm. For a given amount of charcoal, the mass of CO$$_2$$ adsorbed becomes 64 times when the pressure of CO$$_2$$ is doubled.
The value of n in the Freundlich isotherm equation is _________ $$\times 10^2$$. (Round off to the Nearest Integer)

In adsorption studies, the Freundlich adsorption isotherm for gases is usually written as

$$\frac{x}{m}=k\,P^{1/n}$$

where

$$\frac{x}{m}$$ is the mass of gas adsorbed per unit mass of the adsorbent,

$$P$$ is the equilibrium pressure of the gas,

$$k$$ and $$n$$ are empirical constants for the particular adsorbate-adsorbent system.

For the same mass of charcoal, let us denote the initial pressure as $$P_1$$ and the corresponding amount adsorbed as $$\left(\dfrac{x}{m}\right)_1$$. Thus, by substituting into the Freundlich equation, we have

$$\left(\dfrac{x}{m}\right)_1 = k\,P_1^{1/n}\;. \quad -(1)$$

The pressure is now doubled, so the new pressure becomes

$$P_2 = 2P_1$$.

The mass of CO$$_2$$ adsorbed at this new pressure is given to be 64 times the original amount. Therefore, writing the Freundlich expression for the new conditions, we get

$$\left(\dfrac{x}{m}\right)_2 = k\,P_2^{1/n} = k\,(2P_1)^{1/n} = k\,2^{1/n}\,P_1^{1/n}\;. \quad -(2)$$

We are told that

$$\left(\dfrac{x}{m}\right)_2 = 64\;\left(\dfrac{x}{m}\right)_1\;.$$

Now we take the ratio of Eq. (2) to Eq. (1):

$$\frac{\left(\dfrac{x}{m}\right)_2}{\left(\dfrac{x}{m}\right)_1} =\frac{k\,2^{1/n}\,P_1^{1/n}}{k\,P_1^{1/n}} = 2^{1/n}\;.$$

But this same ratio is numerically equal to 64, so we set

$$2^{1/n}=64\;.$$

Next we express 64 as a power of 2, recognizing that

$$64 = 2^{6}\;.$$

Hence, equating the exponents of 2 on both sides, we get

$$\frac{1}{n}=6 \quad\Longrightarrow\quad n=\frac{1}{6}\;.$$

Evaluating the numeric value,

$$n = 0.166666\dots\;.$$

Multiplying by $$10^{2}$$ as required in the question statement, we obtain

$$n \times 10^{2} = 0.166666\dots \times 100 = 16.6666\dots\;.$$

Rounding this to the nearest whole number gives

$$17\;.$$

Hence, the correct answer is Option 17.