/*! 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 4 Find the speed of sound in a dia... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the speed of sound in a diatomic ideal gas that has a density of \(3.50 \mathrm{~kg} / \mathrm{m}^{3}\) and a pressure of \(215 \mathrm{kPa}\) We know that \(v=\sqrt{\gamma R T / M}\) and can find the temperature from the pressure. Using the gas law \(P V=(m / M) R T\), $$ \frac{R T}{M}=P \frac{V}{m} $$ Moreover, \(\rho=m / V\), and so the expression for the speed becomes $$ v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{(1.40)\left(215 \times 10^{3} \mathrm{~Pa}\right)}{3.50 \mathrm{~kg} / \mathrm{m}^{3}}}=293 \mathrm{~m} / \mathrm{s} $$ We used the fact that \(\gamma \approx 1.40\) for a diatomic ideal gas, as discussed in Chapter 20 .

Short Answer

Expert verified
The speed of sound is approximately 293 m/s.

Step by step solution

01

Identify Given Information

We are given the pressure \( P = 215 \text{ kPa} \) which is equivalent to \( 215,000 \text{ Pa} \). The density of the gas is \( \rho = 3.50 \text{ kg/m}^3 \). We also know that for a diatomic ideal gas, \( \gamma \approx 1.40 \).
02

Formula for Speed of Sound

The formula for the speed of sound \( v \) in an ideal gas is given by \( v = \sqrt{\frac{\gamma P}{\rho}} \). This equation uses the ratio of pressure \( P \) to density \( \rho \).
03

Substitute Values into the Formula

Substitute the known values into the formula:\[v = \sqrt{\frac{1.40 \times 215,000}{3.50}}\]
04

Calculate the Expression

Calculate the expression inside the square root first:\[\frac{1.40 \times 215,000}{3.50} = \frac{301,000}{3.50} = 86,000\]Now take the square root:\[v = \sqrt{86,000}\]
05

Solve for v

Perform the square root calculation to find the speed of sound:\[v \approx 293 \text{ m/s}\]
06

Conclusion: State the Result

The speed of sound in the diatomic ideal gas is approximately 293 meters per second, based on given pressure and density.

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Key Concepts

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

Diatomic Ideal Gas
Diatomic ideal gases are gases that consist of molecules made up of two atoms. Common examples include oxygen (O_2) and nitrogen (N_2). These gases follow the ideal gas law quite closely under typical conditions of temperature and pressure. One essential characteristic of a diatomic gas, especially in the context of thermodynamics, is the heat capacity ratio, denoted by \( \gamma \). This ratio is the relationship between the specific heat at constant pressure (C_p) and the specific heat at constant volume (C_v). For most diatomic gases, \( \gamma \) is approximately equal to 1.40. This value is crucial when calculating the speed of sound in the gas, as it appears in the formula \( v = \sqrt{\frac{\gamma P}{\rho}} \). Understanding the value of \( \gamma \) helps us accurately predict the behavior of the gas under changing conditions.
Density
Density, represented by the symbol \( \rho \), is a measure of mass per unit volume. In the context of gases, it strongly affects the speed of sound, as shown in the equation \( v = \sqrt{\frac{\gamma P}{\rho}} \). A higher density means more mass in a given volume, which typically leads to slower sound speeds. Conversely, a lower density means the gas molecules can move more freely, increasing the speed of sound. Density plays an important role not only in acoustics but also in fields like meteorology and engineering, where the properties of gases must be understood under various conditions. By using the density value provided, alongside the gas pressure, we can accurately compute how quickly sound waves travel through the gas. In our example, the density of the diatomic ideal gas was given as 3.50 ext{ kg/m}^3.
Pressure
Pressure, denoted by \( P \), is the force exerted by the gas molecules per unit area on the walls of its container. In the scenario of calculating the speed of sound in a gas, pressure is a critical factor. It determines how much energy is available for the sound wave to compress and expand the gas as it propagates. For our calculation, we are provided with a pressure value of 215 ext{ kPa}, which equates to 215,000 ext{ Pa}. The formula \( v = \sqrt{\frac{\gamma P}{\rho}} \) illustrates how pressure, alongside density, affects sound speed. The higher the pressure, the faster the sound can travel, due to the energy imparted by more tightly packed molecules. Pressure is a fundamental property in thermodynamics and other disciplines, essential for predicting gas behavior under various conditions.
Ideal Gas Law
The ideal gas law is one of the most important equations in chemistry and physics, because it relates pressure (P), volume (V), and temperature (T) of a gas to the number of moles (n) and the universal gas constant (R). It is expressed as \( PV = nRT \). For a given mass of gas, the law can also be rearranged to \( PV = \frac{m}{M}RT \) where \( m \) is the mass and \( M \) is the molar mass. This relationship helps us understand how changes in one property affect the others. In our example, the ideal gas law allows us to substitute pressure and density into the formula for the speed of sound, \( v = \sqrt{\frac{\gamma P}{\rho}} \). It provides the foundational principles needed to navigate more complex scenarios involving gas dynamics, predicting how the gas will behave under theoretical ideal conditions.

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Most popular questions from this chapter

A sound has an intensity of \(5.0 \times 10^{-7} \mathrm{~W} / \mathrm{m}^{2}\). What is its intensity level?

A tiny sound source emits sound uniformly in all directions. The intensity level at a distance of \(2.0 \mathrm{~m}\) is \(100 \mathrm{~dB}\). How much sound power is the source emitting? The energy emitted by a point source can be considered to flow out through a spherical surface which has the source at its center. Hence, if we find the rate of flow through such a surface, it will equal the flow from the source. Take a concentric sphere of radius \(2.0 \mathrm{~m}\). We know that the sound level on its surface is \(100 \mathrm{~dB}\). You can show that this corresponds to \(I=0.010 \mathrm{~W} / \mathrm{m}^{2}\). Thus, the energy flowing each second through each \(\mathrm{m}^{2}\) of surface is \(0.010 \mathrm{~W}\). The total energy flow through the spherical surface is then \(I\left(4 \pi r^{2}\right)\), where \(I=0.010 \mathrm{~W} / \mathrm{m}^{2}\) and \(r=2.0 \mathrm{~m}\) : $$ \text { Power from source }=\left(0.010 \mathrm{~W} / \mathrm{m}^{2}\right)(4 \pi)(2 \mathrm{~m})^{2}=0.50 \mathrm{~W} $$ Notice how little power issues as sound from even such an intense source.

An increase in pressure of \(100 \mathrm{kPa}\) causes a certain volume of water to decrease by \(5 \times 10^{-3}\) percent of its original volume. ( \(a\) ) What is the bulk modulus of water? (b) What is the speed of sound (compression waves) in water?

A noise-level meter reads the sound level in a room to be \(85.0 \mathrm{~dB}\). What is the sound intensity in the room? Sound level \((\beta)\), in \(\mathrm{dB}\), is given by \(\beta=10 \log _{10}\left(I / I_{0}\right)\) and here it equals \(85.0 \mathrm{~dB}\). Accordingly, and $$ \begin{aligned} \beta &=10 \log _{10}\left(\frac{I}{1.00 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}}\right)=85.0 \mathrm{~dB} \\ \log _{10}\left(\frac{I}{1.00 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}}\right) &=\frac{85.0}{10}=8.50 \\ \frac{I}{1.00 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}} &=\operatorname{antilog}_{10} 8.50=3.16 \times 10^{8} \\ I &=\left(1.00 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}\right)\left(3.16 \times 10^{8}\right)=3.16 \times 10^{-4} \mathrm{~W} / \mathrm{m}^{2} \end{aligned} $$

In an experiment to determine the speed of sound, two observers, \(\mathrm{A}\) and \(\mathrm{B}\), were stationed \(5.00 \mathrm{~km}\) apart. Each was equipped with a gun and a stopwatch. Observer-A heard the report of B's gun \(15.5 \mathrm{~s}\) after seeing its flash. Later, A fired his gun and \(\mathrm{B}\) heard the report \(14.5 \mathrm{~s}\) after seeing the flash. Determine the speed of sound and the component of the speed of the wind along the line joining \(\mathrm{A}\) to \(\mathrm{B}\).

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