Problem 41 - A Complete Resource Book in Chemistry for JEE Main

Question

At constant volume and temperature conditions, the rates of diffusion \(\mathrm{D}_{A}\) and \(\mathrm{D}_{\mathrm{B}}\) of gases \(\mathrm{A}\) and \(\mathrm{B}\) having densities \(\rho_{A}\) and \(\rho_{B}\) are related by the expression (a) \(\mathrm{D}_{\mathrm{A}}=\left[\mathrm{D}_{\mathrm{B}} \rho_{\mathrm{A}} / \rho_{\mathrm{B}}\right]^{1 / 2}\) (b) \(\mathrm{D}_{\mathrm{A}}=\left[\mathrm{D}_{\mathrm{B}} \rho_{\mathrm{B}} / \rho_{\mathrm{A}}\right]^{1 / 2}\) (c) \(\mathrm{D}_{\mathrm{A}}=\mathrm{D}_{\mathrm{B}}\left[\rho_{\mathrm{A}} / \rho_{\mathrm{B}}\right]^{1 / 2}\) (d) \(\mathrm{D}_{\mathrm{A}}=\mathrm{D}_{\mathrm{B}}\left[\rho_{\mathrm{B}} / \rho_{\mathrm{A}}\right]^{1 / 2}\)

Step-by-Step Solution

Answer

The correct answer is (d): \( \mathrm{D}_{\mathrm{A}} = \mathrm{D}_{\mathrm{B}} \left(\frac{\rho_{\mathrm{B}}}{\rho_{\mathrm{A}}}\right)^{1/2} \).

Step 1: Identify the Principle

The problem relates to Fick's laws of diffusion, specifically dealing with the rates of diffusion for gases. At constant temperature and pressure, these rates can be derived from Graham's law of effusion, which states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass.

Step 2: Apply Graham's Law

According to Graham's law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass: \( \frac{\mathrm{D}_{\mathrm{A}}}{\mathrm{D}_{\mathrm{B}}} = \left(\frac{M_{\mathrm{B}}}{M_{\mathrm{A}}}\right)^{1/2} \). Here, \( \mathrm{D}_{\mathrm{A}} \) and \( \mathrm{D}_{\mathrm{B}} \) are the rates of diffusion for gases \( A \) and \( B \), and \( M_{\mathrm{A}} \) and \( M_{\mathrm{B}} \) are their molar masses.

Step 3: Relate Molar Mass and Density

The density of a gas \( \rho \) is related to its molar mass \( M \) and volume \( V \) through the ideal gas equation. At constant temperature and pressure, the densities \( \rho_{\mathrm{A}} \) and \( \rho_{\mathrm{B}} \) are proportional to their molar masses. Therefore, \( \frac{\rho_{\mathrm{B}}}{\rho_{\mathrm{A}}} = \frac{M_{\mathrm{B}}}{M_{\mathrm{A}}} \).

Step 4: Substitute Relation Into Graham's Law

Substitute the density relation into Graham's law: \( \frac{\mathrm{D}_{\mathrm{A}}}{\mathrm{D}_{\mathrm{B}}} = \left(\frac{\rho_{\mathrm{B}}}{\rho_{\mathrm{A}}}\right)^{1/2} \).Multiply both sides by \( \mathrm{D}_{\mathrm{B}} \) to express \( \mathrm{D}_{\mathrm{A}} \):\[ \mathrm{D}_{\mathrm{A}} = \mathrm{D}_{\mathrm{B}} \left(\frac{\rho_{\mathrm{B}}}{\rho_{\mathrm{A}}}\right)^{1/2} \]

Step 5: Identify the Correct Formula

Compare the derived formula \( \mathrm{D}_{\mathrm{A}} = \mathrm{D}_{\mathrm{B}} \left(\frac{\rho_{\mathrm{B}}}{\rho_{\mathrm{A}}}\right)^{1/2} \) with the options provided. This matches option (d).

Key Concepts

Fick's Laws of Diffusion

Fick's laws of diffusion give us a scientific way to understand how particles spread out over time. There are two main laws that explain this process:

1. **Fick's First Law** states that mass flux is proportional to the concentration gradient. Simply put, substances diffuse from areas of higher concentration to areas of lower concentration, and the rate at which this happens depends on how steeply the concentration drops.
2. **Fick's Second Law** describes how concentration changes over time. It takes into account not just the concentration gradient but also how it varies as diffusion occurs, predicting how substance concentration at a given location will change with time.

These two laws together help scientists and engineers understand and predict how gases and liquids move in various environments. In practical terms, they are the reason why food coloring spreads in water and why perfume scents drift across a room. When applied to gases, like in the exercise above, they help us calculate how quickly one type of gas might spread into another, using additional tools like Graham's Law to fine-tune our predictions.

Ideal Gas Equation

The Ideal Gas Equation is a handy tool that science uses to describe how gases behave under different conditions. Written as \( PV = nRT \), each letter stands for something specific:

- \( P \): Pressure
- \( V \): Volume
- \( n \): Number of moles
- \( R \): Ideal gas constant
- \( T \): Temperature

This equation helps us determine the relationship between these variables. For example, if you increase the temperature while keeping the volume constant, the pressure will rise too. Or if the pressure and temperature are constant, the number of moles (the amount of gas) can help us determine the volume.

The Ideal Gas Equation is key when we talk about density and molar mass, especially in the context of gases. It assumes gases are made of particles that don't interact, allowing us to relate volume and density to pressure and temperature. This equation is at the heart of many calculations in chemistry and physics, including those involving diffusion, as in the exercise discussed.

Density and Molar Mass Relationship

To understand how density and molar mass relate, remember that density is the mass of an object divided by its volume. For gases, we often refer to the molar mass, which is the mass of a given substance (the gas, in this case) divided by the number of moles.

- **Molar Mass** is often expressed in grams per mole, and it tells us how much a mole of each substance weighs.
- **Density** is typically given in grams per liter for gases and indicates how much mass there is in a given volume.

In terms of gases, particularly under conditions of constant temperature and pressure, their densities are proportional to their molar masses. Essentially, a heavier gas (higher molar mass) will have a higher density if the conditions are constant, assuming the gases behave like ideal gases.

This relationship plays a critical role in explaining diffusion behaviors using Graham's Law, as we saw in the provided exercise. When we know the densities of the gases involved, we can use these values with Graham's Law to determine the rate of diffusion for each.