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Chapter 19: Problem 28

At what frequency would the wavelength of sound in air be equal to the mean free path of oxygen molecules at \(1.0 \mathrm{~atm}\) pressure and \(0.00^{\circ} \mathrm{C} ?\) The molecular diameter is \(3.0 \times 10^{-8} \mathrm{~cm}\).

Short Answer

Expert verified
The frequency is approximately \(3.0 \times 10^9 \mathrm{~Hz}\).

Step by step solution

01

Understand the Mean Free Path Formula

The mean free path \( \lambda \) is the average distance travelled by a molecule between collisions. The formula for mean free path in a gas is given by \( \lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P} \), where \( k_B \) is the Boltzmann constant, \( T \) is the temperature in Kelvin, \( d \) is the molecular diameter, and \( P \) is the pressure.
02

Convert Temperature to Kelvin

The temperature provided is \( 0.00^{\circ} \mathrm{C} \). Convert this to Kelvin using the formula \( T(K) = T(^{\circ}C) + 273.15 \). Thus, \( T = 273.15 \) K.
03

Insert Given Values into Mean Free Path Formula

Using the values provided: \( k_B = 1.38 \times 10^{-23} \mathrm{~JK}^{-1} \), \( T = 273.15 \) K, \( d = 3.0 \times 10^{-8} \mathrm{~cm} = 3.0 \times 10^{-10} \mathrm{~m} \), and \( P = 1.0 \mathrm{~atm} = 1.01 \times 10^5 \mathrm{~Pa} \).Calculate \( \lambda = \frac{1.38 \times 10^{-23} \times 273.15}{\sqrt{2} \pi (3.0 \times 10^{-10})^2 \times 1.01 \times 10^5} \).
04

Solve for Mean Free Path

Perform the calculations to find \( \lambda \). The calculation gives \( \lambda \approx 1.1 \times 10^{-7} \mathrm{~m} \).
05

Use the Wave Equation for Sound

The wavelength \( \lambda \) of sound is given by \( \lambda = \frac{v}{f} \), where \( v = 331.5 \) m/s is the speed of sound in air at \( 0^{\circ} \mathrm{C} \) and \( f \) is the frequency we need to find.
06

Solve for Frequency

Equate the wavelength of sound to the mean free path: \( \frac{v}{f} = 1.1 \times 10^{-7} \). Therefore, \( f = \frac{331.5}{1.1 \times 10^{-7}} \).
07

Calculate the Frequency

Calculate \( f \) using the values from the previous step: \( f \approx 3.0 \times 10^9 \mathrm{~Hz} \).

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Key Concepts

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

Molecular Diameter
The notion of molecular diameter is crucial when examining gas molecules. This dimension refers to the effective size of a molecule and is an important factor in understanding molecular interactions. For the oxygen molecules in our exercise, the molecular diameter is given as \(3.0 \times 10^{-8} \text{ cm}\).
This measure helps us grasp how often molecules collide, and thus directly affects the mean free path calculation.
  • The molecular diameter can be expressed in meters for calculations, so here it is \(3.0 \times 10^{-10} \text{ m}\).
  • Given its tiny scale, it underscores how minuscule the gaps between molecules are in gases, impacting how swiftly molecules can travel before a collision occurs.
  • This parameter, combined with pressure and temperature, gives us a clear view of molecular behavior in different conditions.
Understanding molecular diameter helps predict and calculate molecular dynamics in various environments.
Boltzmann Constant
The Boltzmann constant \(k_B\) plays a key role in statistical mechanics and thermodynamics. It connects macroscopic and microscopic physics, providing insights into particle behavior using temperature and energy. In our exercise, the Boltzmann constant is \(1.38 \times 10^{-23} \text{ JK}^{-1}\).
This constant is utilized to calculate the mean free path, indicating how temperature influences molecular collisions.
  • It serves as a bridge between the macroscopic measurements of temperature (such as Kelvin in this context) and the energy of individual particles.
  • In formulas like \(\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}\), \(k_B\) links thermal energy to molecular motion, assisting in calculating mean free path.
  • An understanding of this constant aids in comprehending how temperature fluctuations can alter molecular speeds and collision rates.
The Boltzmann constant is fundamental in understanding molecular movement and statistical dynamics of gases.
Frequency Calculation
Frequency calculation becomes critical when trying to determine the oscillatory nature of phenomena like sound waves. In this exercise, the frequency \(f\) needed to match the mean free path of oxygen molecules and the wavelength of sound
is determined using \(f = \frac{v}{\lambda}\), where \(v\) is the speed of sound.
  • The speed of sound at \(0^{\circ} \text{C}\) is given as \(331.5 \text{ m/s}\).
  • By equating the sound wavelength to the mean free path \(\lambda\), we find \(f\).
  • Using the values, frequency is calculated as \(f \approx 3.0 \times 10^9 \text{ Hz}\).
This simple yet profound calculation highlights how sound properties can be directly derived from molecular behavior.
Sound Wavelength
Sound wavelength refers to the distance between successive crests of a wave and is integral to understanding wave phenomena. In gases, it is closely linked to molecular movement, as explored in our exercise.
  • The wavelength \(\lambda\) is related to frequency and wave speed by \(\lambda = \frac{v}{f}\).
  • At \(0^{\circ} \text{C}\), the speed of sound \(v\) is \(331.5 \text{ m/s}\).
  • This exercise sets the wavelength of sound equal to the mean free path, illustrating a direct relationship between wave behavior and molecular physics.
By considering wavelength, we can appreciate how sound manifests in various environmental contexts like temperature and molecular composition.
Pressure and Temperature Conversion
Pressure and temperature play a pivotal role in the behavior of gases, requiring accurate conversions to ensure correct calculations. In our exercise, we deal with
conversion for temperature and pressure to appropriate units.
  • Temperature in Celsius needs converting to Kelvin using \(T(K) = T(^{\circ}C) + 273.15\), resulting in \(273.15 \text{ K}\).
  • Pressure is converted to Pascals (Pa) for scientific calculations: \(1.0 \text{ atm} = 1.01 \times 10^5 \text{ Pa}\).
  • Such conversions are essential to use uniform units, preventing errors in calculating molecular interactions, like the mean free path.
Grasping these conversions fosters better comprehension of physical properties in gases under different conditions.

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Most popular questions from this chapter

What is the internal energy of \(1.0\) mol of an ideal monatomic gas at \(273 \mathrm{~K}\) ?

The envelope and basket of a hot-air balloon have a combined weight of \(2.45 \mathrm{kN}\), and the envelope has a capacity (volume) of \(2.18 \times 10^{3} \mathrm{~m}^{3}\). When it is fully inflated, what should be the temperature of the enclosed air to give the balloon a lifting capacity (force) of \(2.67 \mathrm{kN}\) (in addition to the balloon's weight)? Assume that the surrounding air, at \(20.0^{\circ} \mathrm{C}\), has a weight per unit volume of \(11.9 \mathrm{~N} / \mathrm{m}^{3}\) and a molecular mass of \(0.028 \mathrm{~kg} / \mathrm{mol}\), and is at a pressure of \(1.0 \mathrm{~atm}\).

In a certain particle accelerator, protons travel around a circular path of diameter \(23.0 \mathrm{~m}\) in an evacuated chamber, whose residual gas is at \(295 \mathrm{~K}\) and \(1.00 \times 10^{-6}\) torr pressure. (a) Calculate the number of gas molecules per cubic centimeter at this pressure. (b) What is the mean free path of the gas molecules if the molecular diameter is \(2.00 \times 10^{-8} \mathrm{~cm} ?\)

When the U. S. submarine Squalus became disabled at a depth of \(80 \mathrm{~m}\), a cylindrical chamber was lowered from a ship to rescue the crew. The chamber had a radius of \(1.00 \mathrm{~m}\) and a height of \(4.00 \mathrm{~m}\), was open at the bottom, and held two rescuers. It slid along a guide cable that a diver had attached to a hatch on the submarine. Once the chamber reached the hatch and clamped to the hull, the crew could escape into the chamber. During the descent, air was released from tanks to prevent water from flooding the chamber. Assume that the interior air pressure matched the water pressure at depth \(h\) as given by \(p_{0}+\rho g h\), where \(p_{0}=1.000 \mathrm{~atm}\) is the surface pressure and \(\rho=\) \(1024 \mathrm{~kg} / \mathrm{m}^{3}\) is the density of seawater. Assume a surface temperature of \(20.0^{\circ} \mathrm{C}\) and a submerged water temperature of \(-30.0^{\circ} \mathrm{C}\). (a) What is the air volume in the chamber at the surface? (b) If air had not been released from the tanks, what would have been the air volume in the chamber at depth \(h=80.0 \mathrm{~m} ?\) (c) How many moles of air were needed to be released to maintain the original air volume in the chamber?

An automobile tire has a volume of \(1.64 \times 10^{-2} \mathrm{~m}^{3}\) and contains air at a gauge pressure (pressure above atmospheric pressure) of \(165 \mathrm{kPa}\) when the temperature is \(0.00^{\circ} \mathrm{C}\). What is the gauge pressure of the air in the tires when its temperature rises to \(27.0^{\circ} \mathrm{C}\) and its volume increases to \(1.67 \times 10^{-2} \mathrm{~m}^{3} ?\) Assume atmospheric pressure is \(1.01 \times 10^{5} \mathrm{~Pa}\).
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