Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the mass of the

Q: Why do we use the logarithmic form of the Freundlich isotherm?

The original equation is an exponential curve, which is hard to analyze visually. By taking the logarithm, we transform it into a straight line ($y = mx + c$). This makes it incredibly easy to determine the constants $K$ and $1/n$ directly from the intercept and slope of the graph!

Q: What does the intercept of this graph represent?

The y-intercept is the point where the line crosses the y-axis (where $\log p = 0$, or $p = 1$). In our equation, this corresponds to $\log K$. So, the intercept gives us the value of the empirical constant $K$, which depends on the nature of the gas and the adsorbent.

Q: Can the value of $1/n$ be greater than 1 or negative?

No! For the Freundlich isotherm, the value of $1/n$ strictly lies between 0 and 1. A value of 1 represents low pressure where adsorption is directly proportional to pressure. A value of 0 represents high pressure where the surface is saturated and adsorption is independent of pressure.

Q: How did we know the slope is exactly $2/4$?

The problem provides a right-angled triangle on the graph. The vertical side represents the change in the y-axis ($\Delta \log(x/m)$), which is 2 units. The horizontal side represents the change in the x-axis ($\Delta \log p$), which is 4 units. Slope is simply 'rise over run', giving us $2/4 = 1/2$.

Q: Does the Freundlich isotherm work at all pressures?

That's the catch! The Freundlich isotherm is an empirical model and it actually fails at very high pressures. At high pressures, the experimental graphs become completely horizontal (saturation), but the Freundlich equation doesn't perfectly predict this plateau. That's where the Langmuir isotherm comes in!

Question

Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, $x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent at pressure $p$. $\frac{x}{m}$ is proportional to

Options:

(A) $p^2$

(B) $p^{1/4}$

(C) $p^{1/2}$

(D) $p$

Sigma Percentile · Accepted Answer

### The Signature of Adsorption

Imagine you are observing a gas slowly clinging to the surface of a solid piece of charcoal. This fascinating phenomenon is called adsorption. To mathematically describe how much gas gets adsorbed at a given pressure, we rely on the **Freundlich Adsorption Isotherm**.

The fundamental equation that governs this behavior is given by:
$$ \frac{x}{m} = K p^{\frac{1}{n}} $$

Here, $x$ represents the mass of the gas adsorbed, $m$ is the mass of the solid adsorbent, and $p$ is the pressure of the gas. The terms $K$ and $n$ are empirical constants that depend on the specific nature of the gas and the solid surface at a given temperature.

### Decoding the Logarithmic Plot

While the exponential equation is powerful, it plots as a curve, which can be tricky to analyze visually. To make our lives easier, we use a brilliant mathematical trick: we take the logarithm of both sides!

$$ \log\left(\frac{x}{m}ight) = \log K + \frac{1}{n}\log p $$

I know this might look like just another equation, but let's take a breath and look closer. Does it remind you of something familiar from coordinate geometry? Yes! It perfectly mirrors the equation of a straight line:
$$ y = mx + c $$

In our transformed equation, the y-axis represents $\log\left(\frac{x}{m}ight)$, the x-axis represents $\log p$, the y-intercept $c$ is $\log K$, and most importantly, the **slope of the line** is $\frac{1}{n}$. This is exactly what the problem's graph is showing us!

### The Geometry of the Slope

Now, let's turn our attention to the visual clues provided in the graph. We are given a straight line with a right-angled triangle drawn on it. This triangle is the key to unlocking the value of our slope.

Remember, the slope of any straight line is simply the "rise over run", or the change in the y-coordinates divided by the change in the x-coordinates:
$$ 	ext{Slope} = \frac{\Delta y}{\Delta x} $$

Looking at the triangle, the vertical side (the rise) is given as **2 units**, and the horizontal side (the run) is given as **4 units**.

Let's plug these values in:
$$ 	ext{Slope} = \frac{2}{4} = \frac{1}{2} $$

### The Final Connection

We have now established two crucial facts. From our theoretical equation, the slope is $\frac{1}{n}$. From our geometric graph, the slope is $\frac{1}{2}$.

By equating these two findings, we get:
$$ \frac{1}{n} = \frac{1}{2} $$

Finally, let's substitute this value back into our original Freundlich proportionality:
$$ \frac{x}{m} \propto p^{\frac{1}{n}} $$
$$ \frac{x}{m} \propto p^{\frac{1}{2}} $$

And there we have it! The extent of adsorption is directly proportional to the square root of the pressure. This elegant result tells us that the adsorption is occurring in the intermediate pressure range, perfectly bridging the gap between abstract math and physical chemistry.

Animated Solution for Chemistry - Surface Chemistry: Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, $x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent at pressure $p$ . $\frac{x}{m}$ is proportional to

Select Answer:

Visualized Solution

\text{Analyzing the Graph}

\text{Freundlich Adsorption Isotherm}

\text{Logarithmic Transformation}

\text{Calculating the Slope}

\text{Equating the Slopes}

\text{Final Proportionality}

\text{Physical Significance}

The Sigma Insight: Adsorption

The Signature of Adsorption

Decoding the Logarithmic Plot

The Geometry of the Slope

The Final Connection

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