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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

During a test dive in 1939, prior to being accepted by the U.S. Navy, the submarine \(Squalus\) sank at a point where the depth of water was 73.0 m. The temperature was 27.0\(^\circ\)C at the surface and 7.0\(^\circ\)C at the bottom. The density of seawater is 1030 kg/m\(^3\). (a) A diving bell was used to rescue 33 trapped crewmen from the \(Squalus\). The diving bell was in the form of a circular cylinder 2.30 m high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: Ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell.) (b) At what gauge pressure must compressed air have been supplied to the bell while on the bottom to expel all the water from it?

A 20.0-L tank contains \(4.86 \times 10{^-}{^4}\) kg of helium at 18.0\(^\circ\)C. The molar mass of helium is 4.00 g/mol. (a) How many moles of helium are in the tank? (b) What is the pressure in the tank, in pascals and in atmospheres?

Modern vacuum pumps make it easy to attain pressures of the order of 10\({^-}{^1}{^3}\) atm in the laboratory. Consider a volume of air and treat the air as an ideal gas. (a) At a pressure of 9.00\(\times\) 10\({^-}{^1}{^4}\) atm and an ordinary temperature of 300.0 K, how many molecules are present in a volume of 1.00 cm\(^3\)? (b) How many molecules would be present at the same temperature but at 1.00 atm instead?

An automobile tire has a volume of 0.0150 m\(^3\) on a cold day when the temperature of the air in the tire is 5.0\(^\circ\)C and atmospheric pressure is 1.02 atm. Under these conditions the gauge pressure is measured to be 1.70 atm (about 25 lb/in.\(^2\)). After the car is driven on the highway for 30 min, the temperature of the air in the tires has risen to 45.0\(^\circ\)C and the volume has risen to 0.0159 m\(^3\). What then is the gauge pressure?

A large cylindrical tank contains 0.750 m\(^3\) of nitrogen gas at 27\(^\circ\)C and 7.50 \(\times\) 10\({^3}\)Pa (absolute pressure). The tank has a tight-fitting piston that allows the volume to be changed. What will be the pressure if the volume is decreased to 0.410 m\(^3\) and the temperature is increased to 157\(^\circ\)C?

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