/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Calculate the mean free path of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 38

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 \(2.0 \times 10^{-10} \mathrm{m}\).

Short Answer

Expert verified
The mean free path is approximately \(7.32 \times 10^5\) meters.

Step by step solution

01

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.
02

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:\[ d = 2 \times 2.0 \times 10^{-10} = 4.0 \times 10^{-10} \mathrm{m} \]
03

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:\[\lambda = \frac{1.38 \times 10^{-23} \times 300}{\sqrt{2} \pi \times (4.0 \times 10^{-10})^2 \times 3.55 \times 10^{-8}}\]
04

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 \[\lambda \approx 7.32 \times 10^5 \mathrm{m}\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

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鈥檚 why it鈥檚 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鈥檚 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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The conditions of standard temperature and pressure (STP) are a temperature of \(0.00^{\circ} \mathrm{C}\) and a pressure of 1.00 \(\mathrm{atm}\) . (a) How many liters does 1.00 \(\mathrm{mol}\) of any ideal gas occupy at STP? (b) For a scientist on Venus, an absolute pressure of 1 Venusian-atmosphere is 92 Earth- atmospheres. Of course she would use the Venusian-atmosphere to define STP. Assuming she kept the same temperature, how many liters would 1 mole of ideal gas occupy on Venus?

A cylinder 1.00 \(\mathrm{m}\) tall with inside diameter 0.120 \(\mathrm{m}\) is used to hold propane gas (molar mass 44.1 \(\mathrm{g} / \mathrm{mol}\) ) for use in a barbecue. It is initially filled with gas until the gauge pressure is \(1.30 \times 10^{6} \mathrm{Pa}\) and the temperature is \(22.0^{\circ} \mathrm{C} .\) The temperature of the gas remains constant as it is partially emptied out of the tank, until the gauge pressure is \(2.50 \times 10^{5}\) Pa. Calculate the mass of propane that has been used.

A canister of 1.20 mol of nitrogen gas \((28.0 \mathrm{g} / \mathrm{mol})\) at \(25.0^{\circ} \mathrm{C}\) is left on Jupiter's satellite after completion of a future space mission. Europa has no appreciable atmosphere, and the acceleration due to gravity at its surface is 1.30 \(\mathrm{m} / \mathrm{s}^{2}\) . After some time, the canister springs a small leak, allowing molecules to escape through a small hole. What is the maximum height (in \(\mathrm{km}\) ) above Europa's surface that a \(\mathrm{N}_{2}\) molecule having speed equal to the rms speed will reach if it is shot straight up out of the hole in the canister? Ignore the variation in \(g\) with altitude,

The atmosphere of Mars is mostly \(\mathrm{CO}_{2}\) (molar mass 44.0 \(\mathrm{g} / \mathrm{mol} )\) under a pressure of 650 \(\mathrm{Pa}\) , which we shall assume remains constant. In many places the temperature varies from \(0.0^{\circ} \mathrm{C}\) in summer to \(-100^{\circ} \mathrm{C}\) in winter. Over the course of a martian year, what are the ranges of \((\mathrm{a})\) the rms speeds of the \(\mathrm{CO}_{2}\) molecules, and (b) the density (in mollm') of the atmosphere?

(a) Compute the specific heat capacity at constant volume of nitrogen \(\left(\mathrm{N}_{2}\right)\) gas, and compare with the specific heat capacity of liquid water. The molar mass of \(\mathrm{N}_{2}\) is 28.0 \(\mathrm{g} / \mathrm{mol}\) . (b) You warm 1.00 \(\mathrm{kg}\) of water at a constant volume of 1.00 \(\mathrm{L}\) from \(20.0^{\circ} \mathrm{C}\) to \(30.0^{\circ} \mathrm{C}\) in a kettle. For the same amount of beat, how many kilograms of \(20.0^{\circ} \mathrm{C}\) air would you be able to warm to \(30.0^{\circ} \mathrm{C} ?\) What volume (in liters) would this air occupy at \(20.0^{\circ} \mathrm{C}\) and a pressure of 1.00 atm? Make the simplifying assumption that air is 100\(\% \mathrm{N}_{2}\) .
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Quantum Physics

Read Explanation

Astrophysics

Read Explanation

Electricity and Magnetism

Read Explanation

Radiation

Read Explanation

Electromagnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.