Question

The thermal conductivity of Kr is roughly half that of Ar under identical pressure and temperature conditions. Both gases are monatomic such that \(C_{V, m}=3 / 2 R\) a. Why would one expect the thermal conductivity of Kr to be less than that of Ar? b. Determine the ratio of collisional cross sections for Ar relative to Kr assuming identical pressure and temperature conditions. c. For \(\mathrm{Kr}\) at \(273 \mathrm{K}\) at \(1 \mathrm{atm}, \kappa=0.0087 \mathrm{J} \mathrm{K}^{-1} \mathrm{m}^{-1} \mathrm{s}^{-1}\) Determine the collisional cross section of Kr.

Short answer

The thermal conductivity of Kr is less than that of Ar due to Kr having a larger atomic mass, leading to a lower tendency to transfer heat. The ratio of collisional cross sections for Ar relative to Kr under identical pressure and temperature conditions is 1:2. The collisional cross section of Kr is approximately \( 2.682 \times 10^{-19} \, \mathrm{m^2} \).

Step-by-step solution

Part a: Reason for the different thermal conductivities of Kr and Ar.

The thermal conductivity of any gas depends on several factors, such as the size of the particles, mass, and collisions between particles. Larger and heavier particles tend to have more difficulty in transferring heat, and their collisions may also be more effective in preventing heat transfer. In our case, Kr has a larger atomic mass (83.80 u) than Ar (39.95 u), which means that the atoms of Kr are heavier. This difference in mass leads to a lower tendency of Kr atoms to transfer heat in comparison to the lighter Ar atoms. This explains why the thermal conductivity of Kr is less than that of Ar.

Part b: Finding the ratio of collisional cross sections.

The formula for thermal conductivity for ideal gases can be approximated as: \[ \kappa = \frac{1}{3} C_{V, m} n v L \] Where: - C_{V, m} is the molar heat capacity at constant volume (\(3/2R\)) - n is the number density (ratio of particles in a volume) - v is the average velocity of particles - L is the mean free path between collisions Under identical pressure and temperature conditions, the number densities and the average velocities of the particles are the same for both Kr and Ar. The ratio of the thermal conductivities can then be written as: \[ \frac{\kappa_\mathrm{Kr}}{\kappa_\mathrm{Ar}} = \frac{L_\mathrm{Kr}}{L_\mathrm{Ar}} = \frac{1}{2} \] The mean free path (L) is inversely proportional to the collisional cross section (σ), so the ratio of cross sections can be expressed as: \[ \frac{\sigma_\mathrm{Ar}}{\sigma_\mathrm{Kr}} = \frac{L_\mathrm{Kr}}{L_\mathrm{Ar}} \] \[ \frac{\sigma_\mathrm{Ar}}{\sigma_\mathrm{Kr}} = \frac{1}{2} \] Therefore, the ratio of collisional cross sections for Ar relative to Kr is 1:2 under identical pressure and temperature conditions.

Part c: Finding the collisional cross section of Kr.

To determine the collisional cross section (σ) of Kr, we will use the values provided in the exercise: - Temperature (T) = 273 K - Pressure (P) = 1 atm - Thermal conductivity (κ) = 0.0087 J∙K⁻¹∙m⁻¹∙s⁻¹ First, we must find the molar concentration (nc) expressed in mol/m³ as follows: \[ nc = \frac{P}{R \times T}\] Where: - R is the gas constant, equal to 8.314 J∙mol⁻¹∙K⁻¹ \[ nc = \frac{1 \times 10^5 \mathrm{Pa}}{8.314 \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 273 \mathrm{K}} = 4.466 \times 10^{25} \mathrm{mol/m^3} \] Now, we rewrite the formula for thermal conductivity, isolating the mean free path (L): \[ L_\mathrm{Kr} = \frac{3 \kappa_\mathrm{Kr} }{ C_{V, m} n v } \] To find the average velocity of the particles (v), we can use the relationship between temperature and kinetic energy: \[ v = \sqrt{\frac{2RT}{M}} \] Where: - M is the molar mass of Kr (0.0838 kg/mol) \[ v = \sqrt{\frac{2 \times 8.314 \mathrm{J/mol \cdot K \times} 273 \mathrm{K}}{0.0838 \mathrm{kg/mol}}} = 244.4 \mathrm{\,m/s} \] Now, we can find the mean free path for Kr: \[ L_\mathrm{Kr} = \frac{3 \times 0.0087 \mathrm{J \cdot K^{-1} \cdot m^{-1} \cdot s^{-1}}}{ \frac{3}{2} \times 8.314 \mathrm{J \cdot mol^{-1} \cdot K^{-1}} \times 4.466 \times 10^{25}\,\mathrm{mol/m^3} \times 244.4\,\mathrm{m/s}} = 8.215 \times 10^{-8}\,\mathrm{m} \] Finally, knowing the mean free path (L) of Kr, we can find the collisional cross-section (σ) using the formula: \[ \sigma_\mathrm{Kr} = \frac{1}{n \times L_\mathrm{Kr}} \] \[ \sigma_\mathrm{Kr} = \frac{1}{4.466 \times 10^{25} \mathrm{mol/m^3} \times 8.215 \times 10^{-8}\,\mathrm{m}} = 2.682 \times 10^{-19}\,\mathrm{m^2} \] Thus, the collisional cross section of Kr is approximately \( 2.682 \times 10^{-19} \, \mathrm{m^2} \).

Key concepts

Collisional Cross Section

The concept of collisional cross section is vital when studying the behavior of gas particles. In essence, it represents the effective area that one particle presents as a target for collision with another particle.
This can help us understand how often collisions occur, particularly in gases, and is a key parameter in analyzing gas interactions.
The collisional cross section is typically measured in square meters. It depends on the size of the gas particles and their spacing when at a given temperature and pressure. Some important points to understand about collisional cross section include:

• It varies inversely with the mean free path. That means, as the collisional cross section increases, the mean free path decreases.
• Larger particles tend to have larger collisional cross sections, leading to more frequent collisions.
• Temperature and pressure conditions can impact the calculated value.

In the case of Ar and Kr, despite having a bigger size, Kr particles demonstrate a significantly larger collisional cross section compared to Ar, correlating with its lower thermal conductivity. The wider target area means collisions occur more frequently, impeding efficient heat transfer.

Mean Free Path

The mean free path is a fundamental concept in understanding gas behavior. It describes the average distance a particle travels before it collides with another particle. Key insights about the mean free path include:

• It is inversely proportional to the collisional cross section. Smaller cross sections lead to longer mean free paths, as particles have a reduced likelihood of encountering each other.
• The mean free path can be used to derive other properties, like viscosity and thermal conductivity.
• Factors such as temperature, pressure, and the type of molecules influence the mean free path.

For a given temperature and pressure, the mean free path helps predict how a gas will behave under kinetic theory. When solving for Kr's mean free path in the provided problem, the calculated value was key to determining its collisional cross section. A shorter mean free path in Kr compared to Ar explains its relatively lower heat transfer rate.

Kinetic Theory of Gases

The kinetic theory of gases provides a framework for understanding the macroscopic properties of gases in terms of their molecular composition and motion. This theory assumes that gases consist of a large number of small particles that are in constant, random motion. The primary tenets of the kinetic theory include:

• Gas particles are point-like with no volume, and they move in straight lines until they collide with one another or the walls of the container.
• Collisions between gas particles or with a container are perfectly elastic, meaning there is no loss of kinetic energy.
• The average kinetic energy of gas particles is directly related to the temperature of the gas. This explains why gas properties like speed change with temperature.

Understanding these principles helps explain why heavier and larger particles, like Kr compared to Ar, tend to have lower velocities at the same temperature and thus lower thermal conductivity. In the case of Kr and Ar, even though both are monatomic and behave ideally, Kr's larger mass leads to different thermal conductivity outcomes due to the mechanics described by kinetic theory.