Problem 79 - A Complete Resource Book in Chemistry for JEE Main

Question

If Vrms of \(\mathrm{H}_{2}\) at \(300 \mathrm{~K}\) is \(1.9 \times 10^{3} \mathrm{~m} / \mathrm{s}\). What is the value of Vrms of \(\mathrm{O}_{2}\) at \(1200 \mathrm{~K} ?\) (a) \(1.9 \times 10^{3}\) (b) \(3.8 \times 10^{3} \mathrm{~m} / \mathrm{s}\) (c) \(0.475 \times 10^{3} \mathrm{~m} / \mathrm{s}\) (d) \(0.95 \times 10^{3} \mathrm{~m} / \mathrm{s}\)

Step-by-Step Solution

Answer

The Vrms of \( \text{O}_2 \) at 1200 K is \( 0.95 \times 10^3 \) m/s (option d).

Step 1: Understand Vrms Formula

The root mean square velocity, \( V_{rms} \), is given by the formula \( V_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is the Boltzmann constant, \( T \) is the temperature in Kelvin, and \( m \) is the molar mass of the gas molecule.

Step 2: Compare Vrms of Different Gases

For two different gases, the ratio of their \( V_{rms} \) is given by \( \frac{V_{rms,1}}{V_{rms,2}} = \sqrt{\frac{T_1}{T_2} \cdot \frac{M_2}{M_1}} \), where \( M \) is the molar mass of the gases.

Step 3: Plug Known Values into the Formula

Given that \( V_{rms} \) of \( \text{H}_2 \) at 300 K is \( 1.9 \times 10^3 \) m/s, the molar mass of \( \text{H}_2 \) is 2 g/mol, and the molar mass of \( \text{O}_2 \) is 32 g/mol, we now find \( V_{rms} \) of \( \text{O}_2 \) at 1200 K using: \[ \frac{V_{rms, O_2}}{1.9 \times 10^3} = \sqrt{\frac{1200}{300} \cdot \frac{2}{32}} \].

Step 4: Simplify and Calculate

First, simplify the expression: \( \frac{1200}{300} = 4 \) and \( \frac{2}{32} = \frac{1}{16} \). This gives \( \sqrt{4 \cdot \frac{1}{16}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \). Therefore, \( \frac{V_{rms, O_2}}{1.9 \times 10^3} = \frac{1}{2} \).

Step 5: Final Calculation

Multiply both sides by \( 1.9 \times 10^3 \) to find \( V_{rms, O_2} \): \( V_{rms, O_2} = \frac{1}{2} \times 1.9 \times 10^3 = 0.95 \times 10^3 \) m/s.

Key Concepts

Root Mean Square Velocity (Vrms)

The root mean square velocity, often abbreviated as \( V_{rms} \), is a measure of the velocity of gas particles in a substance. It provides an average speed at which molecules in a gas move. The formula to calculate \( V_{rms} \) is given by:

• \( V_{rms} = \sqrt{\frac{3kT}{m}} \)

Here, \( k \) is the Boltzmann constant, \( T \) is the temperature in Kelvin, and \( m \) is the molar mass of the particle. This formula shows that \( V_{rms} \) depends on both the temperature of the gas and its molar mass.
Understanding \( V_{rms} \) is crucial for diving into concepts such as kinetic energy, diffusion, and gas dynamics.

Molar Mass of Gases

The molar mass of a gas is a fundamental property that affects its behavior in various conditions. Molar mass, denoted as \( M \), is the mass of one mole of a substance measured in grams per mole. Different gases have different molar masses due to their molecular structure. For instance:

• Hydrogen (\( \text{H}_2 \)) has a molar mass of 2 g/mol.
• Oxygen (\( \text{O}_2 \)) has a molar mass of 32 g/mol.

Understanding molar mass is important because it influences the speed of the gas particles and how they react to temperature changes. With a higher molar mass, gases generally move slower since heavier particles take more energy to move.

Temperature and Vrms Relationship

Temperature has a significant impact on the root mean square velocity of gas particles. The relationship between temperature and \( V_{rms} \) can be understood through the formula \( V_{rms} = \sqrt{\frac{3kT}{m}} \). As temperature \( T \) increases, so does \( V_{rms} \), which means higher temperatures make gas particles move faster.
This relationship is vital when comparing gases at different temperatures:

• A higher temperature typically results in an increase in \( V_{rms} \).
• This increase affects how gases behave, making them expand more and move faster.
• Comparing \( V_{rms} \) at different temperatures involves factoring in the change in both temperature and molar mass.

Thus, when considering the effect of temperature changes, it is crucial to adjust for these parameters to accurately determine particle speed.