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Chapter 18: Problem 34

Calculate the mean free path of air molecules at 3.50 \(\times\) 10\({^-}{^1}{^3}\) atm and 300 K. (This pressure is readily attainable in the laboratory; see Exercise 18.23.) As in Example 18.8, model the air molecules as spheres of radius 2.0 \(\times\) 10\({^-}{^1}{^0}\) m.

Short Answer

Expert verified
The mean free path of air molecules is approximately \( 1.86 \times 10^{-6} \text{ m} \).

Step by step solution

01

Recall the Formula for Mean Free Path

The mean free path, \( \lambda \), is given by the formula: \( \lambda = \frac{kT}{\sqrt{2} \pi d^2 P} \), where \( k \) 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.
02

Convert Radius to Diameter

Given the radius of the air molecules is \( 2.0 \times 10^{-10} \text{ m} \). The diameter \( d \) will be twice the radius: \( d = 2 \times 2.0 \times 10^{-10} \text{ m} = 4.0 \times 10^{-10} \text{ m} \).
03

Insert Known Values into the Formula

Using \( T = 300 \text{ K} \), \( P = 3.50 \times 10^{-13} \text{ atm} \), and converting pressure from atm to pascals (1 atm \( = 1.013 \times 10^5 \text{ Pa}\)), we find \( P = 3.50 \times 10^{-13} \times 1.013 \times 10^5 \text{ Pa} = 3.5465 \times 10^{-8} \text{ Pa} \). Insert the values into the formula: \[ \lambda = \frac{(1.38 \times 10^{-23} \text{ J/K}) \times 300 \text{ K}}{\sqrt{2} \pi (4.0 \times 10^{-10} \text{ m})^2 \times 3.5465 \times 10^{-8} \text{ Pa}} \].
04

Calculate the Mean Free Path

Simplify the expression: \[ \lambda = \frac{4.14 \times 10^{-21} \text{ J}}{2.2212 \times 10^{-27} \text{ m}^2 \text{ Pa}} \]. This simplifies to \( \lambda \approx 1.86 \times 10^{-6} \text{ m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Theory of Gases
The kinetic theory of gases provides a fundamental understanding of how gases behave on a molecular level. It explains their macroscopic properties in terms of the motion of individual molecules.
  • Basic Premises: According to this theory, a gas is composed of a large number of tiny particles (atoms or molecules) that are in continuous random motion.
  • Interactions: These gas particles are considered to be point masses without volume, implying that the interactions between them occur only during collisions.
  • Collisions: The collisions between particles, as well as with the walls of the container, are perfectly elastic. This means kinetic energy is conserved, leading to the pressure exerted by the gas.
  • Pressure and Temperature: The pressure exerted by a gas results from the collisions of the molecules with the walls of the container, while temperature is a measure of the average kinetic energy of these particles.
By understanding these properties, the kinetic theory helps us predict the behavior of gases under different conditions, such as varying pressure and temperature levels. It sets a foundation for calculating elements like the mean free path, which measures the average distance traveled by a molecule between collisions.
Molecular Diameter
Molecular diameter is a key factor when calculating the mean free path of gas molecules. It represents the effective size of a molecule, which plays a significant role in how often molecules collide in a gas.
  • Definition: The molecular diameter can be thought of as the "size" of the molecule, calculated as twice its radius.
  • Importance in Calculations: When determining mean free path, the diameter directly influences how frequently molecules collide. This is because a larger diameter suggests that molecules will have more frequent interactions, reducing the mean free path.
  • Spherical Assumption: In practice, molecules are often considered spherical for simplicity in calculations. This assumption allows the use of straightforward geometric formulas to describe molecular interactions.
Understanding the molecular diameter helps when using equations like the mean free path formula, which incorporates this dimension to determine the average distance a molecule travels before a collision.
Boltzmann Constant
The Boltzmann constant is a fundamental factor in many physical formulas, including those used in the kinetic theory of gases. It links macroscopic and microscopic physical quantities.
  • Definition: The Boltzmann constant, denoted by the symbol \( k \), is a physical constant that relates the average kinetic energy of particles in a gas with the temperature of the gas, having a value of \( 1.38 \times 10^{-23} \text{ J/K} \).
  • Role in Mean Free Path: In the mean free path equation \( \lambda = \frac{kT}{\sqrt{2} \pi d^2 P} \), the Boltzmann constant \( (k) \) allows us to account for the temperature effects on the kinetic energy and movement of molecules. This is vital in determining how frequently molecules collide.
  • Fundamental in Thermodynamics: The constant plays a critical role in statistical mechanics and thermodynamics, bridging the statistical properties at atomic scales to thermodynamic properties at macroscopic levels.
The Boltzmann constant is indispensable for calculations describing gas behavior and energy distributions, enhancing our understanding of thermal dynamics and molecular motion.

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Most popular questions from this chapter

The gas inside a balloon will always have a pressure nearly equal to atmospheric pressure, since that is the pressure applied to the outside of the balloon. You fill a balloon with helium (a nearly ideal gas) to a volume of 0.600 L at 19.0\(^\circ\)C. What is the volume of the balloon if you cool it to the boiling point of liquid nitrogen (77.3 K)?

If deep-sea divers rise to the surface too quickly, nitrogen bubbles in their blood can expand and prove fatal. This phenomenon is known as the \(bends\). If a scuba diver rises quickly from a depth of 25 m in Lake Michigan (which is fresh water), what will be the volume at the surface of an N\(_2\) bubble that occupied 1.0 mm\(^3\) in his blood at the lower depth? Does it seem that this difference is large enough to be a problem? (Assume that the pressure difference is due to only the changing water pressure, not to any temperature difference. This assumption is reasonable, since we are warm-blooded creatures.)

If a certain amount of ideal gas occupies a volume \(V\) at STP on earth, what would be its volume (in terms of \(V\)) on Venus, where the temperature is 1003\(^\circ\)C and the pressure is 92 atm?

Helium gas with a volume of 3.20 L, under a pressure of 0.180 atm and at 41.0\(^\circ\)C, is warmed until both pressure and volume are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is 4.00 g/mol.

A physics lecture room at 1.00 atm and 27.0\(^\circ\)C has a volume of 216 m\(^3\). (a) Use the ideal-gas law to estimate the number of air molecules in the room. Assume that all of the air is N\(_2\). Calculate (b) the particle density-that is, the number of N\(_2 \) molecules per cubic centimeter-and (c) the mass of the air in the room.
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