Question

The ratio between the RMS velocity of \(\mathrm{H}_{2}\) at \(50 \mathrm{~K}\) and that of \(\mathrm{O}_{2} 800 \mathrm{~K}\) is (1) 4 (2) 2 (3) 1 (4) \(1 / 4\)

Short answer

The ratio is 1. Option (3) is correct.

Step-by-step solution

- Understand the formula for RMS Velocity

The root mean square (RMS) 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 molar mass of the gas.

- Set up the ratio of RMS velocities

We need to find the ratio of the RMS velocities of \( \mathrm{H}_{2} \) at \( 50 \) K and \( \mathrm{O}_{2} \) at \( 800 \) K. Using the formula for \( v_{rms} \), the ratio can be written as \[\frac{v_{rms}(\mathrm{H}_{2}~50K)}{v_{rms}(\mathrm{O}_{2}~800K)} = \frac{\sqrt{\frac{3k \cdot 50}{m_{H2}}}}{\sqrt{\frac{3k \cdot 800}{m_{O2}}}} = \sqrt{ \frac{50}{m_{H2}} \cdot \frac{m_{O2}}{800} } \].

- Use molar masses of the gases

The molar mass of \ \mathrm{H}_{2} \ is approximately \( 2 \) g/mol, and the molar mass of \ \mathrm{O}_{2} \ is approximately \( 32 \) g/mol. Substitute these values into the expression from Step 2: \[\frac{v_{rms}(\mathrm{H}_{2}~50K)}{v_{rms}(\mathrm{O}_{2}~800K)} = \sqrt{ \frac{50}{2} \cdot \frac{32}{800} } = \sqrt{ 25 \cdot \frac{1}{25} } = \sqrt{ 1 } = 1 \].

Key concepts

Root Mean Square Velocity

The root mean square (RMS) velocity is a measure of the speed of particles in a gas. Imagine all the molecules bumping into each other at different speeds. The RMS velocity is like finding the average speed but considering the squares of the velocities first, then taking a square root. This is given by \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is the Boltzmann constant, \( T \) is temperature, and \( m \) is the molar mass.
This value is important in understanding how fast particles are moving on average, which affects things like pressure and temperature.

• Higher RMS velocity means the gas particles are moving faster.
• The RMS velocity increases with temperature as warmer particles move more energetically.

It's a useful property in many areas of physics and chemistry, especially when dealing with the kinetic theory of gases.

Temperature Dependence

Temperature is a key factor in determining the RMS velocity of gas particles. At higher temperatures, gas particles move faster, thus increasing their RMS velocity. This relationship is directly proportional, shown by the formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \)
Here’s how it works:

• As temperature \( T \) increases, the term \( \frac{3kT}{m} \) also increases.
• As we take the square root of this, the RMS velocity increases.

If we compare hydrogen gas (\( \mathrm{H}_{2} \)) at two different temperatures, say 50 K and 800 K for oxygen gas (\( \mathrm{O}_{2} \)), we'll find the particles at the higher temperature (oxygen in this case) move significantly faster than those at the lower temperature.
This concept helps explain why gases expand when heated, as faster-moving particles need more room.

Molar Mass

Molar mass is another crucial factor in computing RMS velocity. It refers to the mass of one mole of a substance, typically measured in grams per mole (g/mol).
In the formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \), \( m \) represents the molar mass, and it affects the RMS velocity inversely. Here’s a breakdown:

• Lighter gases (small molar mass) move faster, resulting in higher RMS velocity.
• Heavier gases (large molar mass) move slower, leading to lower RMS velocity.

In the exercise, hydrogen (\( \mathrm{H}_{2} \)) at 50 K has a molar mass of approximately 2 g/mol, while oxygen (\( \mathrm{O}_{2} \)) at 800 K has a molar mass of about 32 g/mol.
When calculating their RMS velocities, we see that despite the higher temperature, the higher molar mass of oxygen results in a lower RMS velocity compared to hydrogen. This inverse relationship is vital in understanding the behavior of different gases in various conditions.