/*! 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 44 A monatomic ideal gas \((\gamma=... [FREE SOLUTION] | 91Ó°ÊÓ

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A monatomic ideal gas \((\gamma=1.67)\) is contained within a box whose volume is \(2.5 \mathrm{m}^{3} .\) The pressure of the gas is \(3.5 \times 10^{s} \mathrm{Pa}\). The total mass of the gas is \(2.3 \mathrm{kg}\). Find the speed of sound in the gas.

Short Answer

Expert verified
The speed of sound in the gas is approximately 797 m/s.

Step by step solution

01

Identify the Given Variables

We know the volume \(V = 2.5\, m^3\), pressure \(P = 3.5 \times 10^5\, Pa\), adiabatic index (ratio of specific heats) \(\gamma = 1.67\), and mass \(m = 2.3\, kg\). We aim to find the speed of sound in the gas.
02

Use the Formula for Speed of Sound in an Ideal Gas

The speed of sound \(c\) in an ideal gas is given by the formula:\[ c = \sqrt{\frac{\gamma P}{\rho}} \] where \(\rho\) is the density of the gas.
03

Calculate the Density of the Gas

The density \(\rho\) is calculated using the formula:\[ \rho = \frac{m}{V} \]Substituting the given values:\[ \rho = \frac{2.3}{2.5} = 0.92\, kg/m^3 \]
04

Substitute Values into the Speed of Sound Formula

Substitute the values of pressure \(P\), \(\gamma\), and density \(\rho\) into the speed of sound formula:\[ c = \sqrt{\frac{1.67 \times 3.5 \times 10^5}{0.92}} \]
05

Perform the Calculations

Calculate the value inside the square root first:\[ \frac{1.67 \times 3.5 \times 10^5}{0.92} = 635326.087 \]Now take the square root:\[ c = \sqrt{635326.087} \approx 797\, m/s \]
06

Conclusion

Thus, the speed of sound in the monatomic ideal gas is approximately 797 m/s.

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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 that describes the behavior of an ideal gas. It combines several physical laws into one simple equation: \[ PV = nRT \] Where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume it occupies
  • \( n \) is the number of moles of the gas
  • \( R \) is the ideal gas constant (8.314 J/mol·K)
  • \( T \) is the temperature in Kelvin
    • In this particular problem, we're interested in elements of the ideal gas that contribute to the understanding of sound speed, namely pressure and volume. The Ideal Gas Law helps establish the relationship between these variables when considering constant quantities of gas in a closed system. As pressure or volume changes, the other components will adapt according to this relationship. This equation can be rearranged in various ways depending on the given values and what you are solving for, making it very versatile.
Adiabatic Index
The adiabatic index, denoted as \( \gamma \), is a crucial factor in understanding how gases behave in various thermodynamic processes. It is defined as the ratio of specific heats at constant pressure \( C_p \) and volume \( C_v \):\[ \gamma = \frac{C_p}{C_v} \] In a monatomic gas, \( \gamma \) is often equal to approximately 1.67. This index influences how a gas reacts during adiabatic processes, which involve changes in pressure, volume, and temperature without any heat exchange with the surroundings. For calculating the speed of sound in an ideal gas, the adiabatic index is critical because it helps determine how compressible the gas is when sound waves travel through it. The speed of sound is faster in gases with higher \( \gamma \) because they are generally more resistant to compression. Thus, in our problem, using \( \gamma = 1.67 \) ensures that we accurately reflect the properties of the monatomic ideal gas in our calculations.
Pressure and Volume
Pressure and volume are core concepts in understanding how gases function within a container. Pressure \( P \) is the force exerted by gas molecules against the walls of their container per unit area.Volume \( V \), on the other hand, is the amount of space that a gas occupies. In Boyle's Law, which is a part of the broader Ideal Gas Law, pressure and volume are inversely related given constant temperature and quantity of gas. This relation can be expressed as:\[ PV = k \]For the calculation of the speed of sound in gases, pressure is a vital component. An increase in pressure, with volume and other variables remain constant, typically increases the speed at which sound travels through the gas. This is because higher pressure implies more forceful molecular collisions, and thus greater energy transferability, facilitating faster propagation of sound waves. In this exercise, this relationship is utilized to find the speed of sound given a known pressure and calculated density.
Density Calculation
Density \( \rho \) is the measure of mass per unit volume: \[ \rho = \frac{m}{V} \] Where \( m \) is mass and \( V \) is volume. Density is a crucial factor in determining the speed of sound in a gas. A gas with lower density will generally allow sound to travel faster, as the molecules are spaced further apart, reducing resistance to motion and allowing quicker wave propagation. In this problem, the mass of the gas is provided and, by knowing the volume of the container, we calculate the density using this simple equation. Once the density is known, it can be plugged into the formula for speed of sound in gases: \[ c = \sqrt{\frac{\gamma P}{\rho}} \] Here, it helps us to understand how the physical characteristics of the gas (mass, volume) contribute directly to how fast sound waves can travel through it. This understanding helps predict acoustics, sound speed, and behavior in various substances.

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Most popular questions from this chapter

A source of sound is located at the center of two concentric spheres, parts of which are shown in the drawing. The source emits sound uniformly in all directions. On the spheres are drawn three small patches that may or may not have equal areas. However, the same sound power passes through each patch. The source produces \(2.3 \mathrm{W}\) of sound power, and the radii of the concentric spheres are \(r_{A}=0.60 \mathrm{m}\) and \(r_{\mathrm{B}}=0.80 \mathrm{m}\) (a) Determine the sound intensity at each of the three patches. (b) Thesound power that passes through each of the patches is \(1.8 \times 10^{-3} \mathrm{W}\). Find the area of each patch.

A speedboat, starting from rest, moves along a straight line away from a dock. The boat has a constant acceleration of \(+3.00 \mathrm{m} / \mathrm{s}^{2}\) (see the figure). Attached to the dock is a siren that is producing a \(755-\mathrm{Hz}\) tone. If the air temperature is \(20^{\circ} \mathrm{C},\) what is the frequency of the sound heard by a person on the boat when the boat's displacement from the dock is \(+45.0 \mathrm{m} ?\)

A sound wave travels in air toward the surface of a freshwater lake and enters into the water. The frequency of the sound does not change when the sound enters the water. The wavelength of the sound is \(2.74 \mathrm{m}\) in the air, and the temperature of both the air and the water is \(20^{\circ} \mathrm{C} .\) What is the wavelength in the water?

A spider hangs from a strand of silk whose radius is \(4.0 \times 10^{-6} \mathrm{m}\). The density of the silk is \(1300 \mathrm{kg} / \mathrm{m}^{3} .\) When the spider moves, waves travel along the strand of silk at a speed of \(280 \mathrm{m} / \mathrm{s}\). Ignore the mass of the silk strand, and determine the mass of the spider.

Two wires are parallel, and one is directly above the other. Each has a length of \(50.0 \mathrm{m}\) and a mass per unit length of \(0.020 \mathrm{kg} / \mathrm{m} .\) However, the tension in wire \(\mathrm{A}\) is \(6.00 \times 10^{2} \mathrm{N},\) and the tension in wire \(\mathrm{B}\) is \(3.00 \times 10^{2} \mathrm{N} .\) Transverse wave pulses are generated simultaneously, one at the left end of wire \(\mathrm{A}\) and one at the right end of wire \(\mathrm{B}\). The pulses travel toward each other. How much time does it take until the pulses pass each other?

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