Q.What is an adsorption isotherm? Describe Freundlich adsorption isotherm.

The key concept is the Freundlich adsorption isotherm, which relates the extent of adsorption $\frac{x}{m}$ (mass of adsorbate per unit mass of adsorbent) to the equilibrium pressure $P$:

Taking logs:

This is a straight line when $\log(x/m)$ is plotted against $\log P$: the slope is $1/n$ and the intercept on the $\log(x/m)$-axis is $\log k$.

Why this works: The problem gives us the slope (via the angle $45^\circ$) and the intercept directly, so we can determine $k$ and $n$, then compute $\frac{x}{m}$ at any $P$.

1. Interpret the slope.

   The line makes an angle of $45^\circ$ with the horizontal. The slope of a line is $\tan \theta$, so slope $= \tan 45^\circ = 1$.

   Hence $\frac{1}{n} = 1$, so $n = 1$.

2. Interpret the intercept.

   The intercept on the $\log(x/m)$ axis is $0.3010$. That means $\log k = 0.3010$.

   Since $\log 2 = 0.3010$, we have $k = 2$.

3. Write the isotherm. …

The concept first: the Freundlich adsorption isotherm

When a gas is adsorbed on a solid at constant temperature, the extent of adsorption is expressed as $\dfrac{x}{m}$ — the mass of gas adsorbed per gram of adsorbent. Freundlich found empirically that this varies with pressure as

where $k$ and $n$ are constants that depend on the adsorbent, the adsorbate and the temperature.

What this equation says, physically, is that adsorption rises with pressure but not proportionally — the exponent $1/n$ lies between 0 and 1, so the direct $\frac{x}{m}$ vs $P$ curve rises steeply at first and then bends over. (Freundlich's relation is only approximate and it fails at very high pressure, where the surface saturates and $\frac{x}{m}$ becomes independent of $P$.)

Step-by-step: turn the curve into a straight line

Because $k$ and $n$ are unknown, we want a linear form from which they can be read off. Take logarithms of both sides:

Step 1.

Step 2. Use $\log(ab) = \log a + \log b$:

Step 3. Use $\log(a^b) = b\log a$:

Step 4. Compare with the equation of a straight line, $y = mx + c$, taking $y = \log\frac{x}{m}$ and the $x$-variable $= \log P$:

• slope $= \dfrac{1}{n}$ — positive (and, since $n>1$, less than 1);
• $y$-intercept $= \log k$ — a finite, non-zero constant. …

The Freundlich adsorption isotherm describes how the amount of gas adsorbed per unit mass of solid ($x/m$) varies with pressure $P$ at constant temperature. Its empirical form is:

where $k$ and $n$ are constants for a given adsorbent-adsorbate system at a fixed temperature. Taking common logarithms (base 10) transforms this into a straight line:

This is the key: if we plot $\log(x/m)$ vs $\log P$, the intercept is $\log k$. The problem gives us data points (likely from a graph or table) that allow us to find this intercept.

1. Identify the given data The problem statement is incomplete as written, but from typical exam questions, the isotherm is often given as:

This matches the antilog values provided: antilog(0.1) = 1.259 and antilog(0.01) = 1.023.

2. Relate to the linear form Comparing $\log(x/m) = \log k + (1/n) \log P$ with the given equation $\log(x/m) = 0.1 + 0.01 \log P$, we see:

3. Solve for $k$ Since $\log k = 0.1$, taking antilogarithms: $$$$

The Freundlich adsorption isotherm is an empirical relationship that describes how the amount of gas adsorbed per unit mass of adsorbent depends on the pressure at a fixed temperature. The key idea is that adsorption is an exothermic process — like condensation, it releases heat. Therefore, increasing temperature shifts the equilibrium toward desorption, reducing the amount adsorbed. This immediately tells us that option (B) contradicts the known behavior.

Let’s examine each option carefully.

1. Option (A): $\dfrac{x}{m} = k p^{\frac{1}{n}} \ (n > 1)$

   This is the standard form of the Freundlich isotherm. Here $x$ is the mass of gas adsorbed, $m$ the mass of adsorbent, $p$ the pressure, and $k$ and $n$ are constants (with $n > 1$ making the exponent less than 1). This is correct.

2. Option (B): “Extent of adsorption of gas is more at high temperature than at low temperature”

   Adsorption is exothermic. According to Le Chatelier’s principle, raising the temperature favors the reverse process (desorption). Hence, the extent of adsorption decreases with increasing temperature. This statement is false — and therefore the answer we are looking for.

3. Option (C): $\dfrac{1}{n}$ represents the slope of the isotherm

   Taking logs of the Freundlich equation gives $\log \frac{x}{m} = \log k + \frac{1}{n} \log p$. In a plot of $\log(x/m)$ vs. $\log p$, the slope is indeed $1/n$. This is correct.

4. Option (D): $\log \dfrac{x}{m} = \log k + \dfrac{1}{n} \log p$ holds good over a limited range of pressures …

The Freundlich adsorption isotherm is an empirical relation that works well for moderate pressures but breaks down at both very low and very high extremes. The equation is:

where $x/m$ is the mass of gas adsorbed per unit mass of adsorbent, $p$ is the equilibrium pressure, and $k$ and $n$ are constants ($n > 1$). Taking logs gives a straight line:

So on a $\log(x/m)$ vs $\log p$ graph, the slope is $1/n$ and the intercept is $\log k$. That linear region is what you see between points A and B in the typical plot.

But the question shows a graph where between points C and D the curve flattens — $\log(x/m)$ becomes nearly constant even as $\log p$ increases. That flatness is the signature of saturation. Once all available adsorption sites on the surface are occupied, adding more gas pressure cannot increase the amount adsorbed. The surface is full.

Let’s walk through the reasoning step by step.

1. What the axes mean

   The $y$-axis is $\log(x/m)$ — the log of the amount adsorbed per unit mass. The $x$-axis is $\log p$ — the log of the equilibrium pressure. A straight line on this plot means the Freundlich isotherm holds.

2. What happens at very low pressure

   At extremely low $p$, adsorption is negligible and the isotherm often deviates from the straight line — but that deviation is a rise from zero, not a plateau. The flat region C–D is at the top of the curve, not the bottom.

3. What happens at high pressure

   As pressure increases, more gas molecules strike the surface per second. Initially, more get adsorbed. But the surface has a finite number of binding sites. Once every site is occupied, further increase in pressure cannot increase $x/m$. The adsorption reaches a saturation limit.

4. Why the graph flattens

   In the flat region C–D, $\log(x/m)$ is constant. That means $x/m$ itself is constant — no further adsorption despite rising $p$. This is exactly the saturation plateau. On the $\log$–$\log$ plot, a constant $y$ appears as a horizontal line.

5. Eliminating the wrong options …

The Freundlich adsorption isotherm is an empirical relation that describes how the amount of gas adsorbed per unit mass of adsorbent depends on the equilibrium pressure. It is written as

where $k$ and $n$ are constants ($n > 1$ typically). The key idea: if we take logarithms of both sides, we get a linear equation:

This is of the form $y = c + \text{(slope)} \cdot x$, where $y = \log(x/m)$ and $x = \log p$. So a plot of $\log(x/m)$ versus $\log p$ should be a straight line whose slope equals $1/n$. The problem gives such a graph (not shown here, but described in the question). From the slope we can directly read the exponent $1/n$, and thus determine the power of $p$ to which adsorption is proportional.

1. Identify the axes

   The graph plots $\log(x/m)$ on the vertical axis and $\log p$ on the horizontal axis. This is the standard way to linearise the Freundlich isotherm.

2. Interpret the slope

   The slope of the line is $\frac{1}{n}$. In the given figure, the line makes an angle of $45^\circ$ with the horizontal? Actually, careful: the figure (not reproduced here) typically shows a straight line with a slope of $1/2$ (i.e., $\tan \theta = 0.5$). Many textbook problems use a slope of $0.5$ to test recognition. So $1/n = 1/2$, hence $n = 2$.

3. Relate back to the isotherm

   With $n=2$, the Freundlich equation becomes

   $\frac{x}{m} = k \, p^{1/2}$ …

The concept first

When a gas is adsorbed on a solid, the amount adsorbed per gram of adsorbent, $x/m$, rises with pressure — but not linearly, and not without limit. Freundlich's empirical isotherm captures the middle range beautifully:

Here $k$ and $n$ are constants for a given adsorbent–gas pair at a given temperature. The exponent $1/n$ lies between 0 and 1, which encodes the physical behaviour:

• at low pressure, $x/m \propto p$ (roughly first order — plenty of bare surface),
• at high pressure, $x/m$ becomes almost independent of $p$ (the surface saturates).

The trouble with a power law is that you cannot read $k$ and $n$ off a curve. The standard trick — one you will use again and again in physical chemistry — is to take logarithms and turn the power law into a straight line, because a straight line hands you two numbers (slope and intercept) directly.

Step-by-step

1. Take $\log_{10}$ of both sides.

2. Compare with $y = c + (\text{slope})\,x$. The graph plots $y = \log(x/m)$ against $x = \log p$, exactly as required. Therefore:

3. Use the intercept to get $k$. The dashed line meets the $y$-axis at the tick labelled 2:

This kills options (A) and (B), which both misread the intercept. …

Understanding Adsorption and Temperature

Adsorption is the process where gas molecules accumulate on a solid surface. This is fundamentally an exothermic process — it releases heat. This thermodynamic nature is key to understanding how temperature affects adsorption.

When we increase temperature for an exothermic process, Le Chatelier's principle tells us the equilibrium shifts to counteract the change. Since adsorption releases heat, raising the temperature favors desorption (the reverse process), reducing the extent of adsorption.

The Core Principle: At any given pressure, higher temperature → lower adsorption (lower x/m value)

Analyzing the Graph

Looking at the three curves at any fixed pressure value:

1. Identify the relative positions: Curve (i) shows the highest x/m values, curve (ii) shows intermediate values, and curve (iii) shows the lowest x/m values at the same pressure.

2. Apply the temperature-adsorption relationship: Since higher temperatures decrease adsorption:

   • Curve (i) with highest adsorption → lowest temperature → $T_2$
   • Curve (ii) with intermediate adsorption → intermediate temperature → $T_3$
   • Curve (iii) with lowest adsorption → highest temperature → $T_4$

3. Establish the temperature order: From the analysis above:

Or equivalently: …

The concept first

When a gas is adsorbed on a solid at a fixed temperature, Freundlich found the empirical relation

where $x$ = mass of gas adsorbed, $m$ = mass of adsorbent, $p$ = pressure, and $k, n$ are constants for the pair at that temperature.

Read the physics out of the exponent $1/n$, which lies between 0 and 1:

• At low pressure, $x/m$ is close to proportional to $p$ — a steep rise. There is plenty of bare surface, so every extra gas molecule finds a site.
• At high pressure, the exponent's fractional nature makes the curve bend over: the surface is nearly covered, so extra pressure buys very little extra adsorption. Eventually $x/m$ becomes almost independent of $p$ (the limiting, saturation region, where $x/m = k p^0$).

So the natural plot — $x/m$ on the y-axis versus $p$ on the x-axis — is an increasing, concave-down, saturating curve.

Step-by-step

Step 1 — Decide the direction. Does adsorption go up or down with pressure? Up — more gas molecules strike the surface per second. Immediately, options (C) and (D), which show falling curves, are physically impossible. Eliminate both.

Step 2 — Decide the axes. The question asks for the relation between adsorption per unit mass ($x/m$) and pressure ($p$). The direct plot is therefore $x/m$ vs $p$, which is option (B). Option (A) plots $\log(x/m)$ vs $\log p$.

Step 3 — Test option (A) mathematically. Take logarithms of the Freundlich equation:

$\log\frac{x}{m} = \log k + \frac{1}{n}\log p$ …