GASES

Sound travels through gases, liquids, and solids at considerably different speeds, as Table 16.1 reveals. Near room temperature, the speed of sound in air is 343 m/s (767 mi/h) and is markedly greater in liquids and solids. For example, sound travels more than four times faster in water and more than seventeen times faster in steel than it does in air. In general, sound travels slowest in gases, faster in liquids, and fastest in solids.

Table 16.1   Speed of Sound in Gases, Liquids, and Solids

Substance   Speed (m/s)   Gases      Air (0 °C)    331   Air (20 °C)    343   Carbon dioxide (0 °C)    259   Oxygen (0 °C)    316   Helium (0 °C)    965   Liquids      Chloroform (20 °C)   1004   Ethyl alcohol (20 °C)   1162   Mercury (20 °C)   1450   Fresh water (20 °C)   1482   Seawater (20 °C)   1522   Solids      Copper   5010   Glass (Pyrex)   5640   Lead   1960   Steel   5960

Like the speed of a wave on a guitar string, the speed of sound depends on the properties of the medium. In a gas, it is only when molecules collide that the condensations and rarefactions of a sound wave can move from place to place. It is reasonable, then, to expect the speed of sound in a gas to have the same order of magnitude as the average molecular speed between collisions. For an ideal gas this average speed is the translational rms speed given by Equation 14.6: , where T is the Kelvin temperature, m is the mass of a molecule, and k is Boltzmann’s constant. Although the expression for v_rms overestimates the speed of sound, it does give the correct dependence on Kelvin temperature and particle mass. Careful analysis shows that the speed of sound in an ideal gas is given by

(16.5)

where g=c_P/c_V is the ratio of the specific heat capacity at constant pressure c_P to the specific heat capacity at constant volume c_V.

The factor g is introduced in Section 15.5, where the adiabatic compression and expansion of an ideal gas is discussed. It appears in Equation 16.5 because the condensations and rarefactions of a sound wave are formed by adiabatic compressions and expansions of the gas. The regions that are compressed (the condensations) become slightly warmed, and the regions that are expanded (the rarefactions) become slightly cooled. However, no appreciable heat flows from a condensation to an adjacent rarefaction because the distance between the two (half a wavelength) is relatively large for most audible sound waves and a gas is a poor thermal conductor. Thus, the compression and expansion process is adiabatic. Example 4 illustrates the use of Equation 16.5.

Figure 16.19 shows an ultrasonic ruler that is used to measure the distance between itself and a target, such as a wall. To initiate the measurement, the ruler generates a pulse of ultrasonic sound that travels to the wall and, like an echo, reflects from it. The reflected pulse returns to the ruler, which measures the time it takes for the round-trip. Using a preset value for the speed of sound, the unit determines the distance to the wall and displays it on a digital readout. Suppose the round-trip travel time is 20.0 ms on a day when the air temperature is 23 °C. Assuming that air is an ideal gas for which g=1.40 and that the average molecular mass of air is 28.9 u, find the distance x to the wall. Figure 16.19  An ultrasonic ruler uses sound with a frequency greater than 20 kHz to measure the distance x to the wall. The blue arcs and blue arrow denote the outgoing sound wave, and the red arcs and red arrow denote the wave reflected from the wall. Reasoning  The distance between the ruler and the wall is x=vt, where v is the speed of sound and t is the time for the sound pulse to reach the wall. The time t is one-half the round-trip time, so t=10.0 ms. The speed of sound in air can be obtained directly from Equation 16.5, provided the temperature and mass are expressed in the SI units of kelvins and kilograms, respectively. Problem solving insight When using equation to calculate the speed of sound in an ideal gas, be sure to express the temperature T in kelvins and not in degrees Celsius or Fahrenheit. Solution To convert the air temperature of 23 °C to the Kelvin temperature scale, we add 23 to 273.15 (see Equation 12.1): T=23+273.15=296 K. The mass of a molecule (in kilograms) can be obtained from the conversion relation between atomic mass units and kilograms (see Section 14.1), 1 u=1.6605×10–27 kg: For the speed of sound, we find  (16.5)  The distance to the wall is

Carbon monoxide (CO), hydrogen (H2), and nitrogen (N2) may be treated as ideal gases. Each has the same temperature and nearly the same value for the ratio of the specific heat capacities at constant pressure and constant volume. In which two of the three gases is the speed of sound approximately the same? Background: This question deals with the factors that determine the speed of sound in an ideal gas. Consult the periodic table on the inside of the back cover, as needed. For similar questions (including calculational counterparts), consult Self-Assessment Test 16.2. This test is described at the end of Section 16.9.

Sonar (sound navigation ranging) is a technique for determining water depth and locating underwater objects, such as reefs, submarines, and schools of fish. The core of a sonar unit consists of an ultrasonic transmitter and receiver mounted on the bottom of a ship. The transmitter emits a short pulse of ultrasonic sound, and at a later time the reflected pulse returns and is detected by the receiver. The water depth is determined from the electronically measured round-trip time of the pulse and a knowledge of the speed of sound in water; the depth registers automatically on an appropriate meter. Such a depth measurement is similar to the distance measurement discussed for the ultrasonic ruler in Example 4.

Conceptual Example 5 illustrates how the speed of sound in air can be used to estimate the distance to a thunderstorm, using a handy rule of thumb.

There is a rule of thumb for estimating how far away a thunderstorm is. After you see a flash of lightning, count off the seconds until the thunder is heard. Divide the number of seconds by five. The result gives the approximate distance (in miles) to the thunderstorm. Why does this rule work? Reasoning and Solution Figure 16.20 shows a lightning bolt and a person who is standing one mile (1.6×103 m) away. When the lightning occurs, light and sound (thunder) are produced very nearly at the same instant. Light travels so rapidly (vlight=3.0×108 m/s) that it reaches the observer almost instantaneously. Its travel time is only (1.6×103 m)/(3.0×108 m/s)=5.3×10–6 s. In comparison, sound travels very slowly (vsound=343 m/s). The time for the thunder to reach the person is (1.6×103 m)/(343 m/s)=5 s. Thus, the time interval between seeing the flash and hearing the thunder is about 5 seconds for every mile of travel. This rule of thumb works because the speed of light is so much greater than the speed of sound that the time needed for the light to reach the observer is negligible compared to the time needed for the sound. Figure 16.20  A lightning bolt from a thunderstorm generates a flash of light and sound (thunder). The speed of light is much greater than the speed of sound. Therefore, the light reaches the person first, followed about 5 seconds later by the sound.

LIQUIDS

In a liquid, the speed of sound depends on the density r and the adiabatic bulk modulus B_ad of the liquid:

(16.6)

The bulk modulus is introduced in Section 10.7 in a discussion of the volume deformation of liquids and solids. There it is tacitly assumed that the temperature remains constant while the volume of the material changes; that is, the compression or expansion is isothermal. However, the condensations and rarefactions in a sound wave occur under adiabatic rather than isothermal conditions. Thus, the adiabatic bulk modulus B_ad must be used when calculating the speed of sound in liquids. Values of B_ad will be provided as needed in this text.

Interactive LearningWare 16.1 A rhinoceros is calling to her mate using infrasonic sound whose frequency is 5.0 Hz. Her mate is 480 m away. The air temperature is 35 °C. Assume that air is an ideal gas for which g=1.40 and that the average mass of a molecule in the air is 4.80×10–26 kg. How many cycles of the sound wave are between the two animals? Related Homework: Problem 36

SOLID BARS

When sound travels through a long slender solid bar, the speed of the sound depends on the properties of the medium according to

(16.7)

where Y is Young’s modulus (defined in Section 10.7) and r is the density.

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