Question

The average translational energy and \(\mathrm{rms}\) speed of molecules in sample of oxygen gas at \(300 \mathrm{~K}\) are \(6.21 \times 10^{-21} \mathrm{~J}\) and \(484 \mathrm{~m} / \mathrm{s}\) respectively. The corresponding values at \(600 \mathrm{~K}\) are nearly (assuming ideal gas behavior) (A) \(6.21 \times 10^{-21} \mathrm{~J}, 968 \mathrm{~m} / \mathrm{s}\) (B) \(12.42 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~m} / \mathrm{s}\) (C) \(12.42 \times 10^{-21} \mathrm{~J}, 968 \mathrm{~m} / \mathrm{s}\) (D) \(8.78 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~m} / \mathrm{s}\)

Short answer

The corresponding average translational energy and rms speed of oxygen gas molecules at 600 K are nearly \(12.42 \times 10^{-21} \mathrm{J}\) and \(684 \mathrm{m/s}\), making the correct answer option (B).

Step-by-step solution

Calculate the new average translational energy

For an ideal gas, the average translational energy is given by the formula: \(\bar{E_k} = \frac{3}{2} k_B T\) Where: \(\bar{E_k}\) = average translational energy, \(k_B\) = Boltzmann constant = \(1.38 \times 10^{-23} \mathrm{J} / \mathrm{K}\), \(T\) = temperature. We are given the relationship: At 300 K, \(\bar{E_k} = 6.21 \times 10^{-21} \mathrm{J}\) At 600 K, we need to find the new average translational energy: \(\bar{E_k}_{new} = \frac{3}{2} k_B T_{new}\) Since the temperature is doubled, the new translational energy will also double: \(\bar{E_k}_{new} = 2 \times \bar{E_k} = 2 \times 6.21 \times 10^{-21} \mathrm{J} = 12.42 \times 10^{-21} \mathrm{J}\)

Calculate the new speed

The rms speed is given by the following equation: \(v_{rms} = \sqrt{\frac{3RT}{M}}\) Where: \(v_{rms}\) = rms speed, \(R\) = gas constant = \(8.314 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\), \(T\) = temperature, \(M\) = molar mass of gas (For O2: M = 32 g/mol = \(0.032 \mathrm{kg} / \mathrm{mol}\)). Given the relationship at 300 K: \(v_{rms} = 484 \mathrm{m/s}\). For the new rms speed, we need to look at how the temperature affects the rms speed: \(v_{rms,new} = \sqrt{\frac{3R(T_{new})}{M}}\) Since the equation involves the square root of the temperature and the temperature doubles, the new rms speed will be multiplied by the square root of 2: \(v_{rms,new} = 484 \mathrm{m/s} \times \sqrt{2} = 684 \mathrm{m/s}\)

Conclusion

Thus, when the temperature is increased to 600K, the average translational energy is \(12.42 \times 10^{-21} \mathrm{J}\) and the rms speed is 684 m/s. The correct answer is option (B) \(12.42 \times 10^{-21} \mathrm{~J}, 684 \mathrm{~m} / \mathrm{s}\).

Key concepts

Average Translational Energy

The average translational energy of gas molecules is an essential concept in the kinetic theory of gases. It represents the energy associated with the random motion of molecules in an ideal gas. The formula to calculate this is \( \bar{E_k} = \frac{3}{2} k_B T \), where \( k_B \) is the Boltzmann constant, and \( T \) is the temperature in Kelvin.

At a microscopic level, molecules in a gas move at various speeds and directions, their kinetic energy being proportional to the temperature of the gas. As the temperature increases, so does the average translational energy. For instance, if the temperature of a gas sample is doubled, as we see from the problem, the average translational energy doubles as well. This means if it's \( 6.21 \times 10^{-21} \mathrm{~J} \) at 300 K, it'll be \( 12.42 \times 10^{-21} \mathrm{~J} \) at 600 K.

This relationship shows why temperature is a critical factor in determining the energy of gas molecules and helps in understanding phenomena like thermal expansion and pressure changes in gases.

RMS Speed

The root mean square (rms) speed is a statistical measure of the speed of particles in a gas. It's crucial for understanding how molecules move in a gas, and it can be calculated with the formula \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas.

This formula tells us that the rms speed increases with the square root of the absolute temperature. Hence, if the temperature is doubled, like in our example from 300 K to 600 K, the rms speed increases by a factor of \( \sqrt{2} \). This was calculated as \( 684 \mathrm{m/s} \), starting from \( 484 \mathrm{m/s} \).

The concept of rms speed is important in applications involving gas flow, diffusion, and effusion, as it ties directly into how fast molecules are moving on average, affecting the behavior of the gas as a whole.

Ideal Gas Behavior

Ideal gas behavior is a theoretical framework to understand the properties of gases. In the ideal gas model, the gas particles are considered point masses with no volume, experiencing elastic collisions and no intermolecular forces. This model simplifies the study of gases using the ideal gas law, \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature.

The problem assumes ideal gas behavior, meaning the calculations use mathematical models assuming no interactions between molecules besides collisions. This assumption holds well under many normal conditions.

Boltzmann Constant

The Boltzmann constant \( (k_B) \) is a fundamental physical constant that relates the average kinetic energy of particles in a gas with the temperature of the gas. It has a value of \( 1.38 \times 10^{-23} \mathrm{J} / \mathrm{K} \).

The role of the Boltzmann constant is pivotal in the formula \( \bar{E_k} = \frac{3}{2} k_B T \) to calculate the average translational energy of gas molecules. It essentially bridges the micro and macro world, linking the microscopic kinetic energy of individual particles to the macroscopic temperature of the system.

Understanding \( k_B \) aids in comprehending why energy is quantized at microscopic levels and supports the statistical mechanics that describe the behavior of gases.