Question

At what temperature does the rms speed of \(\mathrm{O}_{2}\) molecules equal \(475 . \mathrm{m} / \mathrm{s}\) ?

Short answer

The temperature is approximately 269 K.

Step-by-step solution

Understanding the Relationship

The root mean square (rms) speed of gas molecules depends on the temperature and the mass of the molecules. The formula for rms speed is \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( v_{rms} \) is the rms speed, \( k \) is Boltzmann's constant, \( T \) is the temperature in Kelvin, and \( m \) is the mass of one molecule of the gas.

Rearrange the Formula

We need to determine the temperature \( T \). Rearrange the rms speed formula to solve for \( T \): \[ T = \frac{m v_{rms}^2}{3k} \]

Identify Known Values

We are given \( v_{rms} = 475 \ \text{m/s} \). The mass of one \( \mathrm{O}_2 \) molecule \( m \) can be found using its molar mass. \( \mathrm{O}_2 \) has a molar mass of approximately 32.00 g/mol. Convert it to kg per molecule: \[ m = \frac{32.00 \times 10^{-3}}{6.022 \times 10^{23}} \text{ kg/molecule} \]Boltzmann's constant \( k = 1.38 \times 10^{-23} \ \text{J/K} \).

Perform Calculations

Calculate the mass per molecule: \[ m = \frac{32.00 \times 10^{-3} \text{ kg/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 5.32 \times 10^{-26} \text{ kg/molecule} \]Substitute known values into the rearranged formula:\[ T = \frac{5.32 \times 10^{-26} \cdot (475)^2}{3 \times 1.38 \times 10^{-23}} \]Compute \( T \):\[ T \approx 269 \text{ K} \]

Conclusion

The temperature at which the rms speed of \( \mathrm{O}_2 \) molecules is 475 m/s is approximately 269 K.

Key concepts

Root Mean Square Speed

Root mean square (rms) speed is an essential concept in understanding how gas molecules move. It represents an average speed of gas molecules and is determined using the formula: \[ v_{rms} = \sqrt{\frac{3kT}{m}} \] where:

• \( v_{rms} \) is the root mean square speed,
• \( k \) is Boltzmann's constant,
• \( T \) is the temperature in Kelvin,
• and \( m \) is the mass of one molecule of the gas.

The rms speed gives you an idea of the energy of the molecules at a particular temperature. It differs from average speed as it squares all the individual speeds first before averaging them, so it accounts better for variations in speed among molecules. Understanding rms speed is crucial because it relates directly to multiple thermodynamic properties, such as temperature and pressure.

Kinetic Theory of Gases

The kinetic theory of gases provides a mathematical framework to describe the behavior of gases. It assumes that gas consists of many small particles (atoms or molecules) in constant, random motion. This theory helps to relate temperature, pressure, volume, and number of particles or moles. Some core assumptions made include:

• Particles have negligible volume compared to the volume of their container.
• The collisions between particles are perfectly elastic, meaning no kinetic energy is lost.
• There are no intermolecular forces between the particles, except during collisions.

These fundamental ideas allow us to derive the formula for rms speed, helping us connect the microscopic motion of gas particles to macroscopic observable properties like temperature.

Boltzmann Constant

Boltzmann's constant, denoted as \( k \), acts as a bridge between macroscopic and microscopic physics. It plays a crucial role in statistical mechanics and thermodynamics. The constant is approximately equal to:\[ k = 1.38 \times 10^{-23} \, \text{J/K} \] This value links temperature to kinetic energy, helping to convert between energy and temperature scales. When used in the rms speed formula, Boltzmann's constant allows us to calculate the rms speed of individual gas molecules based on the temperature. Hence, it's vital for understanding how energy at the atomic scale translates to observable phenomena like heat and gas pressure.

Temperature Calculations

Temperature is a measure of the average kinetic energy of the particles in a substance. In solving the exercise, we transformed the rms speed formula to find the temperature at which the rms speed of \( \mathrm{O}_2 \) molecules equals 475 m/s. We rearranged the formula:\[ T = \frac{m v_{rms}^2}{3k} \]where you input:

• mass \( m \) of one \( \mathrm{O}_2 \) molecule,
• given rms speed \( v_{rms} \),
• and Boltzmann's constant \( k \).

After calculations, we found that the temperature is approximately 269 K. This example underscores the connection between temperature and the movement of molecules, as kinetic energy from motion is what we measure as temperature.