Problem 38 - University Physics with Modern Physics

Question

Calculate the mean free path of air molecules at a pressure of \(3.50 \times 10^{-13}\) atm and a temperature of 300 \(\mathrm{K}\) . (This pressure is readily attainable in the laboratory; see Exercise \(18.24 .\) ) As in Example \(18.8,\) model the air molecules as spheres of radius \(2.0 \times 10^{-10} \mathrm{m} .\)

Step-by-Step Solution

Answer

The mean free path of air molecules is approximately \(5.82 \times 10^7\) meters.

Step 1: Understanding the Mean Free Path Formula

The mean free path \( \lambda \) is the average distance a particle travels between collisions. It can be calculated using the formula: \[ \lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P} \] where \( k_B \) is the Boltzmann constant \( (1.38 \times 10^{-23} \text{ J/K}) \), \( T \) is the temperature in Kelvin, \( d \) is the diameter of the molecules, and \( P \) is the pressure in Pascals.

Step 2: Convert Pressure to Pascals

The given pressure is \(3.50 \times 10^{-13}\) atm. To use it in the formula, convert it to Pascals using: \[ P = 3.50 \times 10^{-13} \times 1.013 \times 10^5 = 3.55 \times 10^{-8} \text{ Pa} \] since 1 atm = 1.013 \times 10^5 Pa.

Step 3: Determine the Molecule Diameter

The radius \( r \) of the air molecule is given as \( 2.0 \times 10^{-10} \text{ m} \). Therefore, the diameter \( d \) is twice the radius: \[ d = 2 \times 2.0 \times 10^{-10} = 4.0 \times 10^{-10} \text{ m} \]

Step 4: Calculate the Mean Free Path

Substitute the values into the mean free path formula: \[ \lambda = \frac{1.38 \times 10^{-23} \times 300}{\sqrt{2} \pi (4.0 \times 10^{-10})^2 \times 3.55 \times 10^{-8}} \] Calculate the value step-by-step.

Step 5: Complete the Calculation

Evaluating each part, we find: - Numerator: \( 1.38 \times 10^{-23} \times 300 = 4.14 \times 10^{-21} \)- Denominator: \[ \sqrt{2} \pi (4.0 \times 10^{-10})^2 \times 3.55 \times 10^{-8} = 7.11 \times 10^{-29} \]Thus, the mean free path \( \lambda \approx \frac{4.14 \times 10^{-21}}{7.11 \times 10^{-29}} \approx 5.82 \times 10^7 \text{ m} \)

Key Concepts

Kinetic Theory

Kinetic Theory is a vital component of understanding gases and their behaviors. It posits that gases consist of a large number of minuscule particles, which are in constant, random motion. The theory explains how these particles collide with the walls of their container, leading to pressure. These collisions are perfectly elastic, meaning there is no net loss of energy. By examining the motion and interactions of these particles, Kinetic Theory provides insights into the properties of gases, such as temperature and pressure. It helps us understand phenomena like diffusion, viscosity, and thermal conductivity. In essence, it describes how physical conditions affect particle motion, crucially informing calculations like those used to determine the mean free path. Mean free path describes how far a molecule travels before colliding with another, integrating concepts from Kinetic Theory to offer a measure of how molecular interactions impact gas behavior.

Pressure Conversion

Pressure Conversion is an essential skill when working with equations in physics and chemistry. Different units are used to express pressure, and being able to convert between these units allows us to apply consistent calculations. Common units include atmospheres (atm), Pascals (Pa), and millimeters of mercury (mmHg). In the context of our exercise, we converted pressure from atmospheres to Pascals. This conversion is critical as the formula for mean free path natively uses Pascals. By using the conversion factor of 1 atm = 1.013 x 10^5 Pa, we ensured that the input values are congruent with the required units in the equation. Understanding this conversion helps create consistency in scientific measurements, making it easier to share, compare, and compute data accurately.

Molecular Diameter

Molecular Diameter is a critical parameter when modeling molecules as spheres to calculate their interactions and movement. The diameter is particularly significant in computations that involve the geometry of molecules, such as determining the mean free path. To find the molecular diameter, you often start with the molecular radius, as in our exercise where the radius was given as 2.0 x 10^-10 m. The diameter, being twice the radius, allows for a more comprehensive understanding of the size of the molecule. Knowing the diameter helps in calculating distances that particles cover in their movement and the surface area available for collisions. This knowledge is useful in applying Kinetic Theory and in understanding how particle size affects gas properties.

Ideal Gas Law

The Ideal Gas Law is a cornerstone of chemical and physical sciences, relating the pressure, volume, temperature, and amount of gas. Its mathematical expression is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is temperature in Kelvin.This law assumes an ideal gas where particles do not interact except for elastic collisions, and occupy no volume. It provides a good approximation for the behavior of real gases under many conditions.In our exercise, while the Ideal Gas Law wasn't directly used for the mean free path, it provides a broader framework to understand gas behaviors and complements the foundational concepts like temperature and pressure used in the calculations. It's a powerful tool for scientists to predict how changing one condition affects others within a gaseous system.