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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 that 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

Understand Sound Wave Speed in Gases

The speed of sound in an ideal monatomic gas is given by \( v = \sqrt{\frac{\gamma k T}{m}} \), where \( \gamma = \frac{5}{3} \) for monatomic gases, \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( m \) is the mass of one molecule of the gas.
02

Expression for Sound Speed Ratio

Since sound travels twice as far in neon as in krypton in the same time interval, the speed of sound in neon (\( v_{Ne} \)) is twice that in krypton (\( v_{Kr} \)). Therefore: \( \frac{v_{Ne}}{v_{Kr}} = 2 \).
03

Substitute Speed Formula

Using the speed formula in both cases: \( \frac{\sqrt{\frac{\gamma k T_{Ne}}{m_{Ne}}}}{\sqrt{\frac{\gamma k T_{Kr}}{m_{Kr}}}} = 2 \). Simplify this to: \( \sqrt{\frac{T_{Ne} m_{Kr}}{T_{Kr} m_{Ne}}} = 2 \).
04

Solve for Temperature of Neon

Square both sides to eliminate the square root: \( \frac{T_{Ne} m_{Kr}}{T_{Kr} m_{Ne}} = 4 \). Then rearrange for \( T_{Ne} \): \( T_{Ne} = 4 \times T_{Kr} \times \frac{m_{Ne}}{m_{Kr}} \).
05

Plug in Given Values

Substitute the given values: \( m_{Ne} = 20.2 \mathrm{u} \), \( m_{Kr} = 83.8 \mathrm{u} \), and \( T_{Kr} = 293 \mathrm{K} \). Calculate \( T_{Ne} = 4 \times 293 \mathrm{K} \times \frac{20.2}{83.8} \).
06

Calculate Result

Perform the calculations: \( T_{Ne} = 4 \times 293 \times \frac{20.2}{83.8} \approx 282 \mathrm{K} \).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation describing the behavior of gases under different conditions. It is expressed as \( PV = nRT \), where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume
  • \( n \) is the number of moles
  • \( R \) is the ideal gas constant
  • \( T \) is the temperature in Kelvin
This law assumes that gas molecules do not interact and occupy no volume. Although these assumptions are idealized, the law provides a good approximation of real gas behavior under most conditions.
If you increase the temperature, you increase the speed and kinetic energy of the gas molecules. This principle is crucial when considering the speed of sound in gases.
Monatomic Gases
Monatomic gases are composed of single atoms. Examples include noble gases like helium, neon, and krypton. These gases have a simple structure with no bonds between atoms.
The simplicity of monatomic gases allows for straightforward calculations. For instance, the speed of sound in monatomic gases can be calculated using the expression \( v = \sqrt{\frac{\gamma k T}{m}} \). Here:
  • \( \gamma = \frac{5}{3} \) is the adiabatic index for monatomic gases
  • \( k \) is the Boltzmann constant
  • \( T \) is the temperature
  • \( m \) is the atomic mass
Their lack of molecular complexity means the calculations often result in more predictable behavior compared to more complex multi-atomic gases.
Temperature and Sound Speed
The speed of sound in a gas depends on its temperature. As the temperature increases, so does the kinetic energy of the gas particles, which in turn increases the speed of sound. This relationship can be seen in the formula \( v = \sqrt{\frac{\gamma k T}{m}} \).
Because the speed of sound in liquid and solid media is generally less sensitive to temperature changes than in gases, temperature is a key factor in gas dynamics. Understanding this allows us to predict how sound will travel through different gases and under different temperature conditions.
In the context of the exercise, knowing the temperature of krypton and the relationship of sound speeds allows us to calculate the temperature at which neon must be to achieve the observed speed ratio.
Atomic Mass
Atomic mass affects the speed of sound in gases significantly. It represents the mass of a single atom and is typically measured in atomic mass units (u).
In the context of sound speed, a lighter gas (lower atomic mass) generally allows sound to travel faster than a heavier gas. This is due to the lower inertia of lighter gas particles. The formula \( v = \sqrt{\frac{\gamma k T}{m}} \) highlights this inverse relationship.
Thus, the atomic mass of a gas is a crucial determinant of how quickly sound waves can propagate through it. In the exercise, the differing atomic masses of neon and krypton directly affect the speed of sound and, consequently, the temperature calculations.

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Most popular questions from this chapter

When one person shouts at a football game, the sound intensity level at the center of the field is \(60.0 \mathrm{~dB}\). When all the people shout together, the intensity level increases to \(109 \mathrm{~dB}\). Assuming that each person generates the same sound intensity at the center of the field, how many people are at the game?

A person lying on an air mattress in the ocean rises and falls through one complete cycle every five seconds. The crests of the wave causing the motion are \(20.0 \mathrm{~m}\) apart. Determine (a) the frequency and (b) the speed of the wave.

A source emits sound uniformly in all directions. A radial line is drawn from this source. On this line, determine the positions of two points, \(1.00 \mathrm{~m}\) apart, such that the intensity level at one point is \(2.00 \mathrm{~dB}\) greater than that at the other.

A wave traveling in the \(+x\) direction has an amplitude of \(0.35 \mathrm{~m},\) a speed of \(5.2 \mathrm{~m} /\) \(\mathrm{s},\) and a frequency of \(14 \mathrm{~Hz}\). Write the equation of the wave in the form given by either Equation 16.3 or 16.4

Concept Questions A uniform rope of mass \(m\) and length \(L\) is hanging straight down from the ceiling. (a) Is the tension in the rope greater near the top or near the bottom of the rope, or is the tension the same everywhere along the rope? Why? (b) A small-amplitude transverse wave is sent up the rope from the bottom end. Is the speed of the wave greater near the bottom or near the top of the rope, or is the speed of the wave the same everywhere along the rope? Explain. (c) Consider a section of the rope between the bottom end and a point that is a distance \(y\) meters above the bottom. What is the weight of this section? Express your answer in terms of \(m, L, y,\) and \(g\) (the acceleration due to gravity). Problem (a) For the rope described in the Concept Questions, derive an expression that gives the speed of the wave on the rope in terms of the distance \(y\) above the bottom end and the acceleration \(g\) due to gravity. (b) Use the expression that you have derived to calculate the speeds at distances of \(0.50 \mathrm{~m}\) and \(2.0 \mathrm{~m}\) above the bottom end of the rope. Be sure that your answers are consistent with your answer to Concept Question (b).

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