/*! 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 40 Smoke particles in the air typic... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Smoke particles in the air typically have masses of the order of \(10^{-16} \mathrm{kg} .\) The Brownian motion (rapid, irregular movement) of these particles, resulting from collisions with air molecules, can be observed with a microscope. (a) Find the root-mean-square speed of Brownian motion for a particle with a mass of \(3.00 \times 10^{-16} \mathrm{kg}\) in air at 300 \(\mathrm{K}\) . (b) Would the root-mean-square speed be different if the particle were in hydrogen gas at the same temperature? Explain.

Short Answer

Expert verified
(a) \(v_{rms} \approx 6.44 \times 10^{-3} \text{ m/s}\). (b) No, it stays the same.

Step by step solution

01

Understand the Formula for Root-Mean-Square Speed

The root-mean-square speed \(v_{rms}\) of a particle can be calculated using the formula: \[v_{rms} = \sqrt{\frac{3kT}{m}}\] where \(k\) is the Boltzmann constant \((1.38 \times 10^{-23} \text{ J/K})\), \(T\) is the absolute temperature in Kelvin, and \(m\) is the mass of the particle.
02

Substitute the Given Values

For a particle with mass \(m = 3.00 \times 10^{-16} \text{ kg}\) at a temperature \(T = 300 \text{ K}\), substitute these values into the formula. Also use the Boltzmann constant \(k = 1.38 \times 10^{-23} \text{ J/K}\).
03

Calculate the Root-Mean-Square Speed

Now, perform the calculation:\[v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{3.00 \times 10^{-16}}}\]Simplify the expression:\[v_{rms} = \sqrt{\frac{1.242 \times 10^{-20}}{3.00 \times 10^{-16}}} = \sqrt{4.14 \times 10^{-5}}\]\[v_{rms} \approx 6.44 \times 10^{-3} \text{ m/s}\]
04

Analyze if Different Gas Affects Speed

The root-mean-square speed depends on the mass of the particle and the temperature, not on the type of gas surrounding the particle. Since the mass of the particle and the temperature remain unchanged, the root-mean-square speed of the particle in hydrogen gas would be the same.

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 91Ó°ÊÓ!

Key Concepts

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

Root-Mean-Square Speed
The root-mean-square speed, often abbreviated as \( v_{rms} \), is a measure used in physics to quantify the speed of particles in a gas. It provides an average speed, but specifically weighted so that it is more attuned to the energies involved.
To calculate \( v_{rms} \), the formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \) is used, where \( k \) is the Boltzmann constant, \( T \) is the absolute temperature, and \( m \) is the mass of a particle.
Understanding \( v_{rms} \) is crucial because it reflects the energy distribution of the molecules within a gas.
  • \( v_{rms} \) is higher when temperature \( T \) increases as it relates positively to kinetic energy.
  • Heavier particles mean lower \( v_{rms} \) as mass appears in the denominator in the formula.
  • \( v_{rms} \) is independent of the type of gas because it depends only on \( m \), \( T \), and \( k \).
This concept is related to Brownian motion by showing how gas molecules collide with particles, leading them to move faster as \( v_{rms} \) increases with temperature.
Kinetic Theory of Gases
The kinetic theory of gases is a pivotal concept that explains how gases behave on a molecular level. This theory underpins how temperature, pressure, and volume are interrelated through the movements of gas particles.
It's based on several key assumptions:
  • Gas consists of a large number of small particles, either molecules or atoms.
  • These particles are constantly in random, straight-line motion.
  • Collisions between particles are perfectly elastic, meaning there's no net loss of energy.
  • There's a vast amount of space between particles compared to their size, illustrating why gases are compressible.
In relation to Brownian motion, the kinetic theory helps describe why smoke or pollen particles suspended in a fluid (like air) exhibit random movements.
This erratic movement results from collisions with the various particles of gas, reinforcing the idea that temperature and particle size significantly influence \( v_{rms} \). As molecules collide, energy is transferred, causing visible effects on larger "floating" particles.
Boltzmann Constant
The Boltzmann constant, represented by \( k \), is a fundamental physical constant that plays a critical role in statistical mechanics and thermodynamics.
This constant provides a bridge between macroscopic and microscopic physics, allowing scientists to relate the average kinetic energy of particles to the temperature of the whole system.
Boltzmann constant is valued at \( 1.38 \times 10^{-23} \text{ J/K} \). It defines the scale by which energy is measured in terms of temperature. In formulas like \( v_{rms} = \sqrt{\frac{3kT}{m}} \), it links the thermal energy of single particles to thermodynamic temperature.
  • \( k \) is essential in determining \( v_{rms} \) as it appears in the core formula.
  • It relates to the energy per particle per degree of freedom, grounding the measurement of thermal energy.
  • \( k \) helps explain the rate of particle collisions, which is key in observing phenomena like Brownian motion.
Understanding \( k \) enables a deeper comprehension of how temperature influences particle dynamics and why particles exhibit specific behaviors over various temperature scales.

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

Planetary Atmospheres. (a) Calculate the density of the atmosphere at the surface of Mars (where the pressure is 650 \(\mathrm{Pa}\) and the temperature is typically \(253 \mathrm{K},\) with a \(\mathrm{CO}_{2}\) atmosphere), Venus (with an average temperature of 730 \(\mathrm{K}\) and pressure of 92 atm, with a \(\mathrm{CO}_{2}\) atmosphere), and Saturn's moon Titan (where the pressure is 1.5 atm and the temperature is \(-178^{\circ} \mathrm{C},\) with a \(\mathrm{N}_{2}\) atmosphere). (b) Compare each of these densities with that of the earth's atmosphere, which is 1.20 \(\mathrm{kg} / \mathrm{m}^{3} .\) Consult the periodic chart in Appendix D to determine molar masses.

Modern vacuum pumps make it easy to attain pressures of the order of \(10^{-13}\) atm in the laboratory. Consider a volume of air and treat the air as an ideal gas. (a) At a pressure of \(9.00 \times 10^{-14}\) atm and an ordinary temperature of \(300.0 \mathrm{K},\) how many molecules are present in a volume of 1.00 \(\mathrm{cm}^{3} ?\) (b) How many molecules would be present at the same temperature but at 1.00 atm instead?

(a) How much heat does it take to increase the temperature of 2.50 mol of a diatomic ideal gas by 50.0 \(\mathrm{K}\) near room temperature if the gas is held at constant volume? (b) What is the answer to the question in part (a) if the gas is monatomic rather than diatomic?

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 \(1003^{\circ} \mathrm{C}\) and the pressure is 92 atm?

(a) Oxygen (O) has a molar mass of 32.0 \(\mathrm{g} / \mathrm{mol} .\) What is the average translational kinetic energy of an oxygen molecule at a temperature of 300 \(\mathrm{K}\) ? (b) What is the average value of the square of its speed? (c) What is the root-mean-square speed? (d) What is the momentum of an oxygen molecule traveling at this speed? (e) Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a cubical vessel 0.10 \(\mathrm{m}\) on a side. What is the average force the molecule exerts on one of the walls of the container? (Assume that the molecule's velocity is perpendicular to the two sides that it strikes.) (f) What is the average force per unit area? (g) How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 atm? (h) Compute the number of oxygen molecules that are actually contained in a vessel of this size at 300 \(\mathrm{K}\) and atmospheric pressure. (i) Your answer for part (h) should be three times as large as the answer for part (g). Where does this discrepancy arise?

See all solutions

Recommended explanations on Physics Textbooks

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.