/*! 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 46 In deep space the number density... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 46

In deep space the number density of particles can be one particle per cubic meter. Using the average temperature of \(3.00 \mathrm{K}\) and assuming the particle is \(\mathrm{H}_{2}\) with a diameter of \(0.200 \mathrm{nm},\) (a) determine the mean free path of the particle and the average time between collisions. (b) What If? Repeat part (a) assuming a density of one particle per cubic centimeter.

Short Answer

Expert verified
In part (a), the mean free path is approximately \(4.0 \times 10^{11} m\) (which is several times the distance from the Earth to the Sun) and the average time between collisions is approximately \(3.7 \times 10^{11} s\) (which is roughly 11000 years). In part (b), the mean free path reduces to about \(4.0 \times 10^{5} m\) (roughly the size of a city) and the average time between collisions drops to about \(3.7 \times 10^{5} s\) (or about 4 days).

Step by step solution

01

Determine the properties for \(\mathrm{H}_{2}\)

The diameter of an \(\mathrm{H}_{2}\) molecule is given as 0.200 nm or \(0.200 \times 10^{-9} m.\) The radius \(r\) is half of the diameter, therefore \(r = 0.100 \times 10^{-9} m.\) The volume for one molecule \(v\) is thus calculated using the formula for the volume of a sphere \(\frac{4}{3}\pi r^3.\)
02

Calculate the Mean Free Path (位)

The mean free path is given by the formula \(\frac{1}{n\pi d^2}\), where \(n\) is the number density and \(d\) is the diameter of the molecule. Substituting the given \(n = 1\) particle/m鲁 and \(d = 0.200 \times 10^{-9}\) m into the formula, calculate the mean free path.
03

Calculate the Average Time Between Collisions

The average time between collisions (蟿) is given by \(\frac{\lambda}{v}\), where \(v\) is the average speed of the molecule. The average speed of a molecule in a gas is given by \(\sqrt{\frac{8kT}{\pi m}}\), where \(k\) is the Boltzmann constant, \(T\) is the temperature, and \(m\) is the mass of the molecule. Use \(T = 3.00 K\), \(k = 1.38 \times 10^{-23} J/K\), and \(m = 2 \times 1.67 \times 10^{-27} kg\) (since \(\mathrm{H}_2\) contains two hydrogen atoms). Calculate \(v\) first, and then 蟿.
04

Redo the calculations with new density

Repeat Steps 2 and 3, but this time use \(n = 1\) particle/cm鲁 or \(1 \times 10^{6}\) particles/m鲁 (since 1 cm鲁 = \(1 \times 10^{-6}\) m鲁). Calculate the new mean free path and average time between collisions.

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
The kinetic theory of gases plays a central role in understanding how gases behave at the molecular level. This theory assumes that a gas is composed of small particles that are in constant, random motion. These particles are mostly far apart from each other, with interactions occurring only when they collide.
  • Each gas particle moves independently of the others unless a collision occurs.
  • The collisions between particles, and with the walls of their container, are perfectly elastic, meaning no energy is lost during these interactions.
  • By considering properties such as pressure, temperature, and volume, the kinetic theory provides valuable insights into gas behaviors.
Understanding these principles is crucial when calculating the mean free path or predicting collision frequencies in a given environment, like deep space in our exercise. Here, parameters such as particle size and temperature will directly influence these calculations.
Particle Density
Particle density, often referred to as number density, is a measure of how many particles occupy a unit volume, like cubic meter or cubic centimeter, of a substance. It's a vital concept in understanding how frequently particles may collide in a gas.
  • The higher the density, the more likely particles will collide, affecting both the mean free path and collision frequency.
  • In the context of the exercise, a particle density of one particle per cubic meter or cubic centimeter helps illustrate the differences in spatial arrangements and collision likelihoods.
  • Changes in density significantly affect the mean free path, shown by the different calculations when using one particle per cubic meter versus one per cubic centimeter.
This concept, when related to kinetic theory, helps us grasp why particles behave differently under diverse environmental conditions.
Collision Frequency
Collision frequency describes how often particles in a gas will collide over a specific period. It's fundamentally related to the particle density and mean free path, and is influenced by the temperature and size of the particles.
  • As particles move erratically and collide, the frequency of these collisions impacts the gas's behavior, such as pressure and temperature changes.
  • Using the average speed and particle density, we can calculate the average time between these collisions, which provides insights into the gas's kinetic energy distribution.
  • In our exercise, understanding collision frequency helps determine how often hydrogen molecules interact, an important factor in astrophysical studies or gas dynamics.
Understanding these interactions is crucial for applications ranging from chemistry to astronomy.
Boltzmann Constant
The Boltzmann constant (k) is a key parameter in the kinetic theory of gases, linking the microscopic motion of particles to macroscopic measurable quantities like temperature. It's named after the physicist Ludwig Boltzmann and has a value of approximately \(1.38 \times 10^{-23} \text{J/K}\).
  • This constant enables the calculation of average kinetic energy of particles within a substance.
  • For gases, it is used in deriving expressions that relate temperature and molecular speed.
  • For instance, in calculating the average speed of \(\mathrm{H}_2\) molecules in the exercise, it plays a crucial role in determining thermal behavior at 3 K.
This constant represents the bridge between microscopic physics and macroscopic observations, essential for calculations involving energy distributions and thermodynamic behavior.

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 largest bottle ever made by blowing glass has a volume of about \(0.720 \mathrm{m}^{3} .\) Imagine that this bottle is filled with air that behaves as an ideal diatomic gas. The bottle is held with its opening at the bottom and rapidly submerged into the ocean. No air escapes or mixes with the water. No energy is exchanged with the ocean by heat. (a) If the final volume of the air is \(0.240 \mathrm{m}^{3},\) by what factor does the internal energy of the air increase? (b) If the bottle is submerged so that the air temperature doubles, how much volume is occupied by air?

The dimensions of a room are \(4.20 \mathrm{m} \times 3.00 \mathrm{m} \times 2.50 \mathrm{m}\) (a) Find the number of molecules of air in the room at atmospheric pressure and \(20.0^{\circ} \mathrm{C} .\) (b) Find the mass of this air, assuming that the air consists of diatomic molecules with molar mass \(28.9 \mathrm{g} / \mathrm{mol} .\) (c) Find the average kinetic energy of one molecule. (d) Find the root-mean-square molecular speed. (e) On the assumption that the molar specific heat is a constant independent of temperature, we have \(F_{\text {int }}=5 n R T / 2 .\) Find the internal energy in the air. (f) What If? Find the internal energy of the air in the room at \(25.0^{\circ} \mathrm{C}\)

A vertical cylinder with a heavy piston contains air at a temperature of \(300 \mathrm{K}\). The initial pressure is \(200 \mathrm{kPa},\) and the initial volume is \(0.350 \mathrm{m}^{3} .\) Take the molar mass of air as \(28.9 \mathrm{g} / \mathrm{mol}\) and assume that \(C_{V}=5 R / 2 .\) (a) Find the specific heat of air at constant volume in units of \(\mathrm{J} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) A vertical cylinder with a heavy piston contains air at a temperature of \(300 \mathrm{K}\). The initial pressure is \(200 \mathrm{kPa},\) and the initial volume is \(0.350 \mathrm{m}^{3} .\) Take the molar mass of air as \(28.9 \mathrm{g} / \mathrm{mol}\) and assume that \(C_{V}=5 R / 2 .\) (a) Find the specific heat of air at constant volume in units of \(\mathrm{J} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\)

(a) If it has enough kinetic energy, a molecule at the surface of the Earth can "escape the Earth's gravitation," in the sense that it can continue to move away from the Earth forever, as discussed in Section \(13.7 .\) Using the principle of conservation of energy, show that the minimum kinetic energy needed for "escape" is \(m g R_{E},\) where \(m\) is the mass of the molecule, \(g\) is the free-fall acceleration at the surface, and \(R_{E}\) is the radius of the Earth. (b) Calculate the temperature for which the minimum escape kinetic energy is ten times the average kinetic energy of an oxygen molecule.

At what temperature would the average speed of helium atoms equal (a) the escape speed from Earth, \(1.12 \times 10^{4} \mathrm{m} / \mathrm{s}\) and \((\mathrm{b})\) the escape speed from the Moon, \(2.37 \times 10^{3} \mathrm{m} / \mathrm{s} ?\) (See Chapter 13 for a discussion of escape speed, and note that the mass of a helium atom is \(\left.6.64 \times 10^{-27} \mathrm{kg} .\right)\)
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Electricity

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Magnetism

Read Explanation

Force

Read Explanation

Atoms and Radioactivity

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.