Questions & Answers

Question

Answers

[A] 1.0

[B] 2.0

[C] 0.50

[D] 0.25

Answer
Verified

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.