Problem 138 - A Complete Resource Book in Chemistry for JEE Main

Question

The ration between the r.m.s velocity of \(\mathrm{H}_{2}\) at \(50 \mathrm{~K}\) ad that of \(\mathrm{O}_{2}\) at \(800 \mathrm{~K}\) is

Step-by-Step Solution

Answer

The ratio of the r.m.s velocities is 1.

Step 1: Understanding r.m.s. velocity

The root mean square (r.m.s.) velocity of a gas 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 mass of one molecule of the gas.

Step 2: r.m.s. Velocity Formula Ratio

To find the ratio between the r.m.s velocities of \( \text{H}_2 \) and \( \text{O}_2 \), we use the formula: \( \frac{v_{rms, ext{H}_2}}{v_{rms, ext{O}_2}} = \sqrt{\frac{3kT_{ ext{H}_2}/m_{ ext{H}_2}}{3kT_{ ext{O}_2}/m_{ ext{O}_2}}} \). This simplifies to \( \sqrt{\frac{T_{ ext{H}_2}m_{ ext{O}_2}}{T_{ ext{O}_2}m_{ ext{H}_2}}} \), as the \( 3k \) terms cancel out.

Step 3: Substituting Values

Substituting in the given temperatures \( T_{ ext{H}_2} = 50 \ \text{K} \) and \( T_{ ext{O}_2} = 800 \ \text{K} \), and using the molar masses \( m_{ ext{H}_2} = 2 \ \text{g/mol} \) and \( m_{ ext{O}_2} = 32 \ \text{g/mol} \) (assuming molar mass in unified atomic mass units), we get: \( \frac{v_{rms, ext{H}_2}}{v_{rms, ext{O}_2}} = \sqrt{\frac{50 \times 32}{800 \times 2}} \).

Step 4: Simplifying the Expression

Simplifying the expression further: \[ \frac{v_{rms, ext{H}_2}}{v_{rms, ext{O}_2}} = \sqrt{\frac{1600}{1600}} = \sqrt{1} = 1. \] Thus, the r.m.s velocity of \( \text{H}_2 \) at 50 K is equal to that of \( \text{O}_2 \) at 800 K.

Key Concepts

The Kinetic Theory of Gases

The kinetic theory of gases is a fundamental scientific theory that explains the behavior of gases in terms of the motion of their particles. According to this theory, gases are composed of a large number of small particles, usually molecules, that are in constant random motion. This motion is what accounts for the pressure, temperature, and volume relationships in gases as explained by the ideal gas law.
The essence of this theory is that the particles themselves have very little volume compared to the volume of the container they occupy, allowing gases to be highly compressible. Importantly, this model assumes that these gas particles do not exert forces on each other except during collisions.

• The particles are considered point masses, meaning they are assumed to have negligible size compared to the space they move in.
• It also assumes collisions between particles are perfectly elastic, meaning they do not lose kinetic energy from one collision to the next.
• The kinetic energy of the gas is proportional to its temperature measured in Kelvin.

Understanding the principles of the kinetic theory forms the basis for calculating rms velocity and analyzing how gas behaves under different conditions.

Temperature Effect on Gas Particles

Temperature has a direct effect on the motion of gas particles. When the temperature of a gas increases, the particles move more quickly because they have more kinetic energy. Conversely, at lower temperatures, the speed of gas particles decreases.
This relationship is reflected in the root mean square (rms) velocity equation: \[ v_{rms} = \sqrt{\frac{3kT}{m}} \]Here, \( T \) is the temperature in Kelvin. As \( T \) increases, so does the numerator, resulting in higher rms velocity, indicating that gas particles are moving faster.

• At higher temperatures, gas expands because particles move faster and collide with the walls of the container with greater energy.
• At absolute zero (0 K), the theoretical lowest temperature possible, particles have minimal kinetic energy and thus do not move at all.

This interplay between temperature and particle velocity is critical for understanding how gases behave under various thermal conditions.

Molecular Mass Impact on Velocity

The molecular mass or molecular weight of a gas also impacts the speed at which its particles move. Specifically, lighter gas molecules will move faster than heavier ones at the same temperature.
The equation for rms velocity, \( v_{rms} = \sqrt{\frac{3kT}{m}} \), indicates this relationship clearly where \( m \) is the mass of a single gas molecule. As \( m \) decreases, \( v_{rms} \) increases, demonstrating that lighter gases have higher velocities.
This is why gases such as hydrogen () have higher rms speeds compared to heavier gases like oxygen ().

• The mass of the molecules determines the effect a given temperature will have on the speed of gas particles.
• Heavier gases may take longer to diffuse through a space compared to lighter gases because of their lower velocity.

Understanding how molecular mass impacts velocity is essential for predicting the behavior of different gases under identical conditions.