Problem 14 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The mass of gas adsorbed, \(x\), per unit mass of adsorbate, \(m\), was measured at various pressures, \(p .\) A graph between \(\log \frac{x}{m}\) and \(\log p\) gives a straight line with slope equal to 2 and the intercept equal to \(0.4771 .\) The value of \(\frac{x}{m}\) at a pressure of \(4 \mathrm{~atm}\) is : (Given \(\log 3=\) \(0.4771)\)

Step-by-Step Solution

Answer

\(\frac{x}{m} = 48\) at 4 atm.

Step 1: Understand the Relation

The relationship described in the problem is of the form \( \log \frac{x}{m} = n \log p + \log K \), where \( n \) is the slope and \( \log K \) is the intercept. Here, \( n = 2 \) and \( \log K = 0.4771 \).

Step 2: Formulate the Equation

Substitute the given slope and intercept into the equation: \( \log \frac{x}{m} = 2 \log p + 0.4771 \). This equation will be used to find \( \frac{x}{m} \) at a given pressure.

Step 3: Calculate the Logarithm of Pressure

Convert the given pressure of \( 4 \text{ atm} \) to its logarithmic form: \( \log 4 = \log (2^2) = 2 \log 2 \). Since \( \log 3 = 0.4771 \), use the approximation \( \log 2 \approx 0.3010 \). So, \( \log 4 \approx 2 \times 0.3010 = 0.6020 \).

Step 4: Substitute and Solve for \(\log \frac{x}{m}\)

Substitute \( \log 4 \approx 0.6020 \) into the formulated equation: \[ \log \frac{x}{m} = 2 \times 0.6020 + 0.4771 \] \[ \log \frac{x}{m} = 1.2040 + 0.4771 = 1.6811 \].

Step 5: Convert Back from Logarithm

To find \( \frac{x}{m} \), take the antilogarithm of \( 1.6811 \). \( \frac{x}{m} = 10^{1.6811} \). Using an antilogarithm table or calculator, \( 10^{1.6811} \approx 48 \).

Key Concepts

Freundlich isotherm

The Freundlich isotherm is an empirical model used to describe adsorption processes. Essentially, it helps to explain how molecules are adsorbed onto a surface. This applies particularly well across various pressures and concentrations, making it ideal for studying gas adsorption. The Freundlich isotherm can be represented mathematically by the equation:

• \( \frac{x}{m} = Kp^n \)

Where:

• \( \frac{x}{m} \) is the amount of gas adsorbed per unit mass of adsorbent.
• \( K \) is the adsorption coefficient indicating capacity of the adsorbent.
• \( p \) is the equilibrium pressure.
• \( n \) is the heterogeneity factor which indicates the intensity of the adsorption.

In our exercise, the relationship is linearized using logarithms, making it easier to interpret and solve. This model assumes that the surface of the adsorbent is heterogeneous, meaning different sites have different adsorption energies.

logarithmic transformations

Logarithmic transformations are very useful for turning a power-based relationship, like the one in the Freundlich isotherm, into a linear form. Linear forms are much easier to work with, especially when plotting data or solving variables. By applying a logarithm to both sides of the equation \( \frac{x}{m} = Kp^n \), we obtain:

• \( \log \frac{x}{m} = n \log p + \log K \)

This transformation allows us to use the slope-intercept form of a line, which relates the slope of the line \( n \) to the adsorption intensity and the intercept \( \log K \) to the adsorption capacity. This process simplifies calculations and helps visualize how changes in pressure affect adsorption. The logarithmic transformation also compresses large ranges into more manageable numbers, making calculations and graphical analysis more straightforward.

graphical analysis in chemistry

Graphical analysis in chemistry is a powerful tool that allows researchers to visualize relationships between different chemical variables. By plotting the logarithmic transformation of our isotherm on a log-log graph with \( \log \frac{x}{m} \) versus \( \log p \), the Freundlich equation becomes a straight line. This line has a slope equal to \( n \), indicating the intensity of the adsorption, and an intercept of \( \log K \), revealing insights about the adsorption capacity of the material.
Using graphs, chemists can quickly determine key properties of adsorbents and compare them visually. Additionally, data visualized graphically is often easier to interpret than tabulated data. It assists in identifying patterns, anomalies, or deviations from expected behavior faster. With the slope and intercept readily obtainable from a graph, we can back-calculate values like \( \frac{x}{m} \) for varying pressures, providing a clear and concise understanding of the adsorption process.