Problem 29 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases \(x\) times. The value of \(x\) is ____ .

Step-by-Step Solution

Answer

The diffusion coefficient increases 4 times, so \( x = 4 \).

Step 1: Understanding the Relationship

The diffusion coefficient \( D \) is given by the equation \( D = C \cdot \lambda \cdot \bar{v} \), where \( C \) is a constant, \( \lambda \) is the mean free path, and \( \bar{v} \) is the mean speed. Our task is to find the change in \( D \) when temperature \( T \) is increased 4 times and pressure \( P \) is increased 2 times.

Step 2: Mean Free Path Calculation

The mean free path \( \lambda \) for an ideal gas is given by \( \lambda \propto \frac{T}{P} \). Therefore, after the changes, \( \lambda' = \frac{4T}{2P} = 2 \times \text{original} \; \lambda \).

Step 3: Mean Speed Calculation

The mean speed \( \bar{v} \) is proportional to the square root of temperature \( \bar{v} \propto \sqrt{T} \). After the change, the new mean speed \( \bar{v}' = \sqrt{4T} = 2 \times \text{original} \; \bar{v} \).

Step 4: Diffusion Coefficient Calculation

The new diffusion coefficient \( D' = C \cdot \lambda' \cdot \bar{v}' = C \cdot (2\lambda) \cdot (2\bar{v}) = 4 \times C \cdot \lambda \cdot \bar{v} = 4D \). Thus, \( x = 4 \), meaning \( D \) increases 4 times.

Key Concepts

Mean Free Path

The mean free path, often denoted as \( \lambda \), is a critical concept in understanding the behavior of gases. It refers to the average distance a gas molecule travels before colliding with another molecule. This depends on several factors, including:

• The size of the molecules: Smaller molecules tend to have longer mean free paths.
• The density of the gas: Higher density means more frequent collisions, reducing the mean free path.

The mean free path in an ideal gas is related to temperature and pressure, given by the relationship:\[ \lambda \propto \frac{T}{P} \]where \( T \) is temperature and \( P \) is pressure. This formula tells us that increasing the temperature while holding pressure constant increases \( \lambda \), as molecules move faster and collide less often. Conversely, increasing pressure at constant temperature reduces \( \lambda \), as particles are packed closer together and collide more frequently.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics that describes how gases behave under different conditions. This law is expressed as:\[ PV = nRT \]where:

• \( P \) represents the pressure of the gas.
• \( V \) is the volume.
• \( n \) stands for the number of moles of gas.
• \( R \) is the ideal gas constant.
• \( T \) is the absolute temperature.

In the context of this exercise, we are examining changes in both temperature and pressure while considering how they affect other properties like mean free path and mean speed. The ideal gas law helps us understand how these changes affect the overall behavior of the gas molecules. For instance, increasing temperature leads to higher energy and speed of gas molecules, thus influencing mean speed and mean free path as derived from these relationship principles.

Temperature and Pressure Effects

Temperature and pressure are key factors that influence the properties of gases, including diffusive behavior. When you increase the temperature of an ideal gas:

• Molecules receive more energy, which increases their mean speed \( \bar{v} \) since \( \bar{v} \propto \sqrt{T} \).
• This increased speed usually results in a longer mean free path, as discussed earlier.

On the other hand, increasing the pressure has the opposite outcome. It tends to compact the gas:

• Higher pressure means molecules are closer together, reducing the mean free path \( \lambda \).

In the exercise, we see these temperature and pressure effects acting simultaneously. The temperature is increased fourfold, leading to significant increases in energy and speed, while the pressure is doubled, partially counteracting this by reducing the mean free path. Together, these changes result in a complex balance where the diffusion coefficient increases by a factor of four.