Which of the following correctly represents the graph between $\log\left(\frac{m}{x}\right)$ and $\log P$?
Show Hint
Freundlich isotherm becomes linear only after taking logarithm:
\[
\log\left(\frac{x}{m}\right) \text{ vs } \log P
\]
Slope = $\frac{1}{n}$ and intercept = $\log k$.
Concept:
This question is based on the Freundlich Adsorption Isotherm, which describes the empirical relationship between the amount of gas adsorbed onto a solid adsorbent surface and the pressure of the gas at a constant temperature.
Mathematical expression:
\[
\frac{x}{m} = k P^{\frac{1}{n}}
\]
where:
• $x$ is the mass of the gas (adsorbate) adsorbed.
• $m$ is the mass of the solid adsorbent.
• $\frac{x}{m}$ represents the extent of adsorption.
• $P$ is the equilibrium pressure of the gas.
• $k$ and $n$ are empirical constants ($n > 1$).
Step 1: Taking logarithm on both sides.
\[
\frac{x}{m} = k P^{\frac{1}{n}}
\]
Taking log:
\[
\log\left(\frac{x}{m}\right) = \log k + \frac{1}{n}\log P
\]
Rewriting:
\[
\log\left(\frac{x}{m}\right) = \frac{1}{n}\log P + \log k \quad (1)
\]
Step 2: Comparison with straight line equation.
Standard straight line form:
\[
y = mx + c
\]
Comparison:
• $y = \log\left(\frac{x}{m}\right)$
• $x = \log P$
• slope $= \frac{1}{n}$
• intercept $= \log k$
Since $k > 0$, $\log k \neq 0$, so the line does not pass through the origin.
Step 3: Graph identification.
• The relation is linear → straight line.
• Positive slope since $\frac{1}{n} > 0$.
• Non-zero intercept → does not pass through origin.
Hence, correct graph is Option (3).
