Question

# In an adsorption experiment, a graph between $\log \left( \dfrac{x}{m} \right)$versus $\log P$ is found to be linear with slope of ${{45}^{{}^\circ }}$. The intercept on $\log \left( \dfrac{x}{m} \right)$ axis was found to be 0.3010. The amount of gas adsorbed per gram of charcoal under the pressure of 0.5atm will be:[A] 1.0[B] 2.0[C] 0.50[D] 0.25

Hint: Use the Freundlich adsorption isotherm equation. This equation will give us a relation between the gas adsorbed per gram and the pressure. We can use the equation of straight line to find out other terms i.e. ‘K’ and ‘n’ in the isotherm equation.

We know that the Freundlich adsorption isotherm gives us a relation between the extent of adsorption i.e. the gas adsorbed per unit mass and the pressure at a given temperature.
We can write the Freundlich adsorption equation as-
$\dfrac{x}{m}=K\times {{P}^{\dfrac{1}{n}}}$
Where, $\dfrac{x}{m}$is the extent of adsorption and ‘x’ is the gas adsorbed and ‘m’ is the mass. Therefore, the extent of adsorption is gas adsorbed per unit mass.
P is the pressure and K is the
In the question, the slope of a $\log \left( \dfrac{x}{m} \right)$ versus $\log P$ graph is given to us. The graph is a straight line with an angle of ${{45}^{{}^\circ }}$.
So, taking log on both sides of the Freundlich adsorption equation, we will get-
$\log \dfrac{x}{m}=\log K+\dfrac{1}{n}\log P$
We know that the equation of a straight line is- y = mx + c
Where, m is the slope and c is the intercept and x and y depicts the two axes.
Comparing the equation of a straight line with the logarithmic Freundlich adsorption equation, we will get that the slope (m) is $\dfrac{1}{n}$ and the intercept (c) is log k.
We also know that the slope of a line can be written as $\tan \theta$. The slope of the linear graph is given to us which is ${{45}^{{}^\circ }}$ and the intercept is also given to us which is 0.3010. Therefore,

we can write that-
$m=\tan {{45}^{{}^\circ }}=\dfrac{1}{n}$ and $\log k=C=0.3010$.
We know that $\tan {{45}^{{}^\circ }}=1$. Therefore, n = 1 and
We also know that log 2 = 0.3010. Therefore, K=2.

Putting these values in the Freundlich adsorption equation, we can write that-
\begin{align} & \dfrac{x}{m}=2\times {{\left( 0.5 \right)}^{\dfrac{1}{1}}} \\ & or,\dfrac{x}{m}=1 \\ \end{align}
As we can see from the above calculation that gas absorbed per gram of charcoal is 1.
So, the correct answer is “Option A”.

Note: The Freundlich adsorption equation is not valid at very high pressure. If the plot of the isotherm plot is a straight line then only the Freundlich isotherm is valid. Another limitation of Freundlich adsorption isotherm is that it tells us only about the physical adsorption and nothing about the chemical adsorption.