Question

The ratio between the root mean square velocity of \(\mathrm{H}_{2}\) at \(50 \mathrm{~K}\) and that of \(\mathrm{O}_{2}\) at \(800 \mathrm{~K}\) is: (a) 4 (b) 2 (c) 1 (d) \(\frac{1}{4}\)

Short answer

The ratio of the root mean square velocities is approximately 1, option (c).

Step-by-step solution

Understanding Root Mean Square Velocity

The root mean square (RMS) velocity 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 particles in kilograms per mole.

Expression for RMS velocities

For \(\mathrm{H}_{2}\), the RMS velocity is \( v_{H_2} = \sqrt{\frac{3k \cdot 50}{m_{H_2}}} \). For \(\mathrm{O}_{2}\), the RMS velocity is \( v_{O_2} = \sqrt{\frac{3k \cdot 800}{m_{O_2}}} \).

Using Molar Masses

The molar mass \( m_{H_2} \) for hydrogen is approximately \(2 \times 1.008 = 2.016\) g/mol which converts to \(0.002016\) kg/mol. Similarly, \( m_{O_2} \) for oxygen is \(32\) g/mol which converts to \(0.032\) kg/mol.

Substituting Values into RMS Formula

Plug in the molar masses into the RMS velocity formulas: \[ v_{H_2} = \sqrt{\frac{3k \cdot 50}{0.002016}} \]and\[ v_{O_2} = \sqrt{\frac{3k \cdot 800}{0.032}} \].

Ratio of RMS Velocities

Compute the ratio of the RMS velocities:\[\frac{v_{H_2}}{v_{O_2}} = \frac{\sqrt{\frac{3k \cdot 50}{0.002016}}}{\sqrt{\frac{3k \cdot 800}{0.032}}}\] which simplifies to \[\sqrt{\frac{50 \cdot 0.032}{800 \cdot 0.002016}}\].

Simplifying the Ratio

Simplify the expression inside the square root:\[\frac{50 \times 0.032}{800 \times 0.002016} = \frac{1.6}{1.6128} \approx 0.9926\]Therefore, \[\sqrt{0.9926} \approx 1\].

Conclusion

The calculated ratio indicates that the RMS velocities are approximately equal, which suggests the ratio is 1.

Key concepts

Gas Laws

Gas laws are fundamental in understanding how gases behave under various conditions. They describe the relationship between pressure, volume, temperature, and the number of gas particles in a container. These laws are mainly derived from the Ideal Gas Law, expressed as:

• The Ideal Gas Law: \( PV = nRT \), where \( P \) stands for pressure, \( V \) for volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is temperature in Kelvin.
• Boyle’s Law: It states that the pressure of a gas is inversely proportional to its volume, when temperature is held constant. This can be stated mathematically as \( P \propto \frac{1}{V} \) or \( PV = k \).
• Charles’s Law: This law outlines that the volume of a gas is directly proportional to its temperature, keeping pressure constant. It can be expressed as \( V \propto T \) or \( \frac{V}{T} = k \).

For our discussion of root mean square velocity (RMS velocity), we relate these laws by understanding how temperature influences the speed of gas particles. Temperature, as per the kinetic molecular theory, directly affects the kinetic energy, changing how fast particles move, which is represented by RMS velocity.

Kinetic Molecular Theory

The Kinetic Molecular Theory is an essential framework for explaining the behavior of gases at a molecular level. It provides insights into why gases obey the gas laws. The theory is based on several key postulates:

• Gas particles are in constant, random motion, colliding with each other and container walls.
• These particles are considered point masses with no volume, meaning individual gas particle volume is negligible compared to the volume of their container.
• Collisions between gas particles are perfectly elastic, so there is no net loss in kinetic energy during these collisions.
• The average kinetic energy of gas particles is proportional to the temperature, given by \( KE = \frac{3}{2} kT \) where \( k \) is the Boltzmann constant.

The theory ties directly into the root mean square velocity by explaining that RMS velocity \( v_{rms} = \sqrt{\frac{3kT}{m}} \) is derived from the measure of the average kinetic energy of gas molecules. Thus, the RMS velocity is influenced by both the temperature and mass of particles, revealing why different gases at different temperatures have different speeds.

Molar Mass Calculations

Molar mass plays a crucial role in understanding gas behavior, particularly when examining root mean square velocities. Molar mass is essentially the mass of one mole of a substance. It is an aggregate value derived by summing the atomic masses of all atoms constituting a molecule, often measured in grams per mole (g/mol).

• The molar mass of \( H_2 \), for example, is calculated as \( 2 \times 1.008 \text{ g/mol} = 2.016 \text{ g/mol} \), which translates to \( 0.002016 \text{ kg/mol} \).
• For \( O_2 \), the molar mass is \( 32 \text{ g/mol} \) or \( 0.032 \text{ kg/mol} \).

When calculating RMS velocity, the molar mass must be converted into kilograms per mole (as seen in the equation \( v_{rms} = \sqrt{\frac{3kT}{m}} \)) to appropriately match units with the other constants in the equation. Therefore, understanding how to manipulate molar mass values is critical in solving not just RMS velocity problems, but a plethora of calculations involving gases.