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Chapter 16: Problem 42

A sound wave travels twice as far in neon (Ne) as it does in krypton (Kr) in the same time interval. Both neon and krypton can be treated as monatomic ideal gases. The atomic mass of neon is \(20.2 \mathrm{u}\), and the atomic mass of krypton is \(83.8 \mathrm{u}\). The temperature of the krypton is \(293 \mathrm{K}\). What is the temperature of the neon?

Short Answer

Expert verified
The temperature of the neon is approximately 282 K.

Step by step solution

01

Understanding the Relationship

We are told that a sound wave travels twice as far in neon (Ne) as it does in krypton (Kr) in the same time. This means the speed of sound in neon is twice that in krypton: \(v_{Ne} = 2v_{Kr}\).
02

Formula for Speed of Sound

The speed of sound in a monatomic ideal gas is given by the formula \(v = \sqrt{\frac{\gamma k T}{m}}\), where \(\gamma\) is the adiabatic index (which is \(\frac{5}{3}\) for a monatomic gas), \(k\) is the Boltzmann's constant, \(T\) is the temperature, and \(m\) is the atomic mass.
03

Setting Up the Equation for Neon and Krypton

Substitute the speed of sound expression into our relationship: \(\sqrt{\frac{\gamma k T_{Ne}}{m_{Ne}}} = 2 \sqrt{\frac{\gamma k T_{Kr}}{m_{Kr}}}\).
04

Cancel Out Similar Terms

The terms \(\gamma\) and \(k\) are constant and appear on both sides, so they can be cancelled, giving \(\sqrt{\frac{T_{Ne}}{m_{Ne}}} = 2 \sqrt{\frac{T_{Kr}}{m_{Kr}}}\).
05

Squaring Both Sides

Square both sides of the equation to remove the square root: \(\frac{T_{Ne}}{m_{Ne}} = 4 \frac{T_{Kr}}{m_{Kr}}\).
06

Solve for Neon Temperature

Re-arrange the equation to solve for \(T_{Ne}\): \(T_{Ne} = 4 \frac{m_{Ne}}{m_{Kr}} T_{Kr}\).
07

Substitute Numerical Values

Substitute the given values: \(T_{Ne} = 4 \frac{20.2}{83.8} \times 293\). Compute the value for \(T_{Ne}\).
08

Calculate the Result

Calculate \(T_{Ne} = 4 \times 0.2405 \times 293 \approx 282 \text{ K}\).

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Key Concepts

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

Monatomic Ideal Gases
Understanding monatomic ideal gases is crucial for grasping sound wave propagation in different media. A monatomic gas, as the name suggests, consists of single atoms. Examples include noble gases like neon (Ne) and krypton (Kr). These gases are ideal, meaning they follow the ideal gas law perfectly, without interactions between the atoms except during elastic collisions.
Unlike polyatomic gases, monatomic gases have simpler atomic structures, which are important in thermodynamic calculations. They are characterized by their adiabatic index, denoted by \(\gamma\), which is \(\frac{5}{3}\) for monatomic gases. This adiabatic index is essential when evaluating the speed of sound in these gases.
For students, it's key to remember that the properties of monatomic gases make it easier to predict how sound waves will travel through them due to their straightforward interactions.
Speed of Sound Equation
The speed of sound in any medium is a fundamental concept in physics. For monatomic ideal gases, the speed of sound can be calculated using the formula: \[ v = \sqrt{\frac{\gamma k T}{m}} \]where \(v\) is the speed of sound, \(\gamma\) is the adiabatic index, \(k\) is Boltzmann's constant, \(T\) is the temperature in Kelvin, and \(m\) is the molar mass of the gas.
This equation shows that the speed of sound depends on both the temperature and mass of the molecules:
  • The higher the temperature, the faster the speed of sound. This is because increased temperature leads to higher kinetic energy in gas atoms.
  • The heavier the atomic mass, the slower the speed of sound, since more massive particles move more slowly at the same kinetic energy level.
Hence, in a monatomic ideal gas like neon or krypton, the changes in temperature and molar mass will directly influence how fast sound travels. For those studying sound wave propagation, understanding this relationship is key.
Temperature Calculation
In the given exercise, calculating the temperature of neon (Ne) is necessary to understand how different conditions affect sound wave propagation. We know that a sound wave travels twice as far in neon as it does in krypton under the same conditions, which helps us set up our equation to find the temperature.
Starting from the relationship given by the speeds \(v_{Ne} = 2v_{Kr}\), and using the speed of sound equation, we can syphon out the respective temperatures by involving the molar masses and given conditions. By squaring both sides of this ratio and cancelling constants, the final formula for temperature becomes: \[ T_{Ne} = 4 \frac{m_{Ne}}{m_{Kr}} T_{Kr}\]
We plug in the values
  • Atomic mass of neon (\(m_{Ne}\)) = 20.2
  • Atomic mass of krypton (\(m_{Kr}\)) = 83.8
  • Temperature of krypton (\(T_{Kr}\)) = 293 K
The temperature of neon is calculated to be approximately 282 K, showing how molecular factor differences affect thermal conditions for sound travel.

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Most popular questions from this chapter

For research purposes a sonic buoy is tethered to the ocean floor and emits an infrasonic pulse of sound (speed \(=1522 \mathrm{m} / \mathrm{s}\) ). The period of this sound is 71 ms. Determine the wavelength of the sound.

A jetskier is moving at \(8.4 \mathrm{m} / \mathrm{s}\) in the direction in which the waves on a lake are moving. Each time he passes over a crest, he feels a bump. The bumping frequency is \(1.2 \mathrm{Hz},\) and the crests are separated by \(5.8 \mathrm{m} .\) What is the wave speed?

An observer stands \(25 \mathrm{m}\) behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of the bullets is \(840 \mathrm{m} / \mathrm{s},\) and the air temperature is \(20^{\circ} \mathrm{C} .\) How far does each bullet travel before the observer hears the report of the rifle? Assume that the bullets encounter no obstacles during this interval, and ignore both air resistance and the vertical component of the bullets" motion.

Hearing damage may occur when a person is exposed to intensity level at the center of the field is \(60.0 \mathrm{dB}\). When all the people shout togcther, the intensity level increases to 109 dB. Assuming that each person generates the same sound intensity at the center of the field, how many people are at the game?

The mass of a string is \(5.0 \times 10^{-3} \mathrm{kg}\), and it is stretched so that the tension in it is \(180 \mathrm{N}\). A transverse wave traveling on this string has a frequency of \(260 \mathrm{Hz}\) and a wavelength of \(0.60 \mathrm{m} .\) What is the length of the string?
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