$CO_2$ 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)
Show Hint
When dealing with ratio problems involving Freundlich or Langmuir isotherms, it is often easiest to write the equation for the two conditions and then divide one by the other. This cancels out the constants (like k) and simplifies the algebra.
The Freundlich adsorption isotherm equation is given by:
$\frac{x}{m} = k P^{1/n}$
where $\frac{x}{m}$ is the mass of gas adsorbed per unit mass of adsorbent, P is the pressure, and k and n are constants.
Let the initial state be denoted by subscript 1 and the final state by subscript 2.
Initial state: $\left(\frac{x}{m}\right)_1 = k P_1^{1/n}$
Final state: $\left(\frac{x}{m}\right)_2 = k P_2^{1/n}$
We are given the following conditions:
The mass of $CO_2$ adsorbed becomes 64 times: $\left(\frac{x}{m}\right)_2 = 64 \left(\frac{x}{m}\right)_1$.
The pressure is doubled: $P_2 = 2P_1$.
Substitute these into the equations:
$64 \left(\frac{x}{m}\right)_1 = k (2P_1)^{1/n}$
Now, divide the final state equation by the initial state equation:
$\frac{64 \left(\frac{x}{m}\right)_1}{\left(\frac{x}{m}\right)_1} = \frac{k (2P_1)^{1/n}}{k P_1^{1/n}}$
$64 = \frac{(2)^{1/n} (P_1)^{1/n}}{(P_1)^{1/n}} = 2^{1/n}$
We need to solve for n. We can write 64 as a power of 2: $64 = 2^6$.
$2^6 = 2^{1/n}$
Equating the exponents:
$6 = \frac{1}{n} \implies n = \frac{1}{6}$
Now, we need to express the answer in the required format: ________ $\times 10^{-2}$.
$n = \frac{1}{6} \approx 0.1666...$
$0.1666... = 16.66... \times 10^{-2}$.
Rounding off to the nearest integer, we get 17.
So, the value is $17 \times 10^{-2}$.
