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 is approximately \(7.32 \times 10^5\) meters.

Step 1: Understand the Mean Free Path Formula

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

Step 2: Calculate the Diameter of Molecules

First, determine the diameter \(d\) of the molecules. Since the radius \(r\) of an air molecule is given as \(2.0 \times 10^{-10} \mathrm{m}\), the diameter \(d\) is twice the radius:\[ d = 2 \times 2.0 \times 10^{-10} = 4.0 \times 10^{-10} \mathrm{m} \]

Step 3: Substitute Values into Mean Free Path Formula

Substitute the given values into the mean free path formula and solve. We know that \(T = 300 \mathrm{K}\), \(P = 3.50 \times 10^{-13} \mathrm{atm}\), and \(d = 4.0 \times 10^{-10} \mathrm{m}\). Convert pressure from atm to pascal (Pa) since \(1 \mathrm{atm} = 1.013 \times 10^5 \mathrm{Pa}\). Thus, \(P = 3.50 \times 10^{-13} \times 1.013 \times 10^5 \mathrm{Pa}\). Then substitute into the formula:\[\lambda = \frac{1.38 \times 10^{-23} \times 300}{\sqrt{2} \pi \times (4.0 \times 10^{-10})^2 \times 3.55 \times 10^{-8}}\]

Step 4: Perform the Calculation

Calculated step-by-step, this is:\[\lambda = \frac{1.38 \times 10^{-23} \times 300}{3.1416 \times 16.0 \times 10^{-20} \times 3.5665 \times 10^{-8}}\]Simplify the values to compute \(\lambda\):\[\lambda \approx \frac{4.14 \times 10^{-21}}{5.66 \times 10^{-27}}\]This results in \[\lambda \approx 7.32 \times 10^5 \mathrm{m}\].

Key Concepts

Kinetic Theory of Gases

Kinetic Theory of Gases is a fundamental concept that provides a microscopic explanation of the macroscopic properties of gases, like pressure, temperature, and volume. It describes a gas as a large number of tiny particles, or molecules, each of which is in constant, random motion. This theory provides insight into how these particles interact with each other and the walls of their container.

• Random Motion: Gas molecules are in continuous, random motion, colliding with each other and with the walls of any container.
• Elastic Collisions: These collisions are perfectly elastic, meaning that there is no net loss of energy during the interactions.
• Pressure: The pressure of a gas is a result of collisions of molecules with the walls of the container.

By understanding these fundamentals, we can predict the behavior of gas under different conditions like changes in temperature or pressure, which is vital when calculating properties such as the mean free path.

Boltzmann Constant

The Boltzmann constant (\(k_B\)) is a key factor in the realm of thermodynamics and statistical mechanics. It serves as a bridge between macroscopic and microscopic physics by relating the average kinetic energy of particles in a gas with the temperature of the gas.
Here’s why it’s important:

• Unit of Scale: It provides a scale to measure energy at the microscopic level.
• Thermal Equilibrium: It defines the energy distribution among particles in a system in thermal equilibrium.
• Equation Involvement: It’s a critical component of the mean free path equation, helping to relate the temperature of the system to the energy of particles.

Understanding the Boltzmann constant is essential in applying the kinetic theory of gases to real-world problems, such as calculating the mean free path accurately.

Molecular Diameter

The molecular diameter is essentially the size of a molecule and is crucial when calculating the mean free path. It's the distance across a molecule, or double the radius, if we consider simple models like hard spheres.

• Definition: The molecular diameter is twice the radius of a molecule.
• Importance in Calculations: In the mean free path formula, the diameter squared is inversely related to the mean free path, implying that larger molecules have a shorter mean free path.
• Simplicity in Models: The simplification of molecules as spheres helps make calculations like mean free path more manageable.

Using the molecular diameter, scientists can mathematically model and compute different gas properties, facilitating the understanding of gas behavior in various conditions.

Pressure Conversion

Pressure conversion is a vital process in scientific calculations, often necessary because different units are used in experimental and standard references. Being able to convert pressure units accurately is crucial for aligning the experimental conditions with theoretical calculations.

• Common Units: The most common units for pressure are atmospheres (atm) and pascals (Pa).
• Conversion Factor: The conversion is straightforward: 1 atm = 1.013 x 10^5 Pa.
• Influence on Calculations: Accurate pressure conversion ensures that all variables in an equation have compatible units, crucial for correctly calculating properties such as the mean free path.

Mastering pressure conversion is indispensable for any comprehensive scientific study involving gases or fluid systems, ensuring precision and reliability in results.