In the monsoon, children count the seconds between a flash of lightning and its thunder. The light arrives almost at once. The thunder takes several seconds to cross the same distance. Grandparents say the count tells how far away the storm is. The rule works because sound in air has a definite speed.
The Laplace-corrected formula
Put Laplace's bulk modulus, \(B = \gamma P\), into the wave speed formula:
\[ v = \sqrt{\frac{B}{\rho}} = \sqrt{\frac{\gamma P}{\rho}} \]
Now use the ideal gas law in the form \(P = \rho RT/M\). Here \(\rho\) is the mass density and \(R\) the universal gas constant. \(T\) is the absolute temperature, and \(M\) is the molar mass of the gas. The steps below follow.
\[ v = \sqrt{\dfrac{\gamma P}{\rho}} \]
Why. the speed of sound in a fluid, \(\sqrt{B/\rho}\), with the adiabatic bulk modulus \(B = \gamma P\).
\[ P = \rho\,\dfrac{RT}{M} \]
Why. the ideal gas law \(PV = nRT\), with \(n = m/M\) and \(\rho = m/V\).
\[ \dfrac{P}{\rho} = \dfrac{RT}{M} \]
Why. divide step 2 by \(\rho\).
\[ v = \sqrt{\dfrac{\gamma RT}{M}} \]
Read it. put step 3 into step 1.
In this formula \(T\) is in kelvin, \(M\) is the molar mass, and \(\gamma\) is the adiabatic index. This Laplace-corrected equation gives the accurate speed of sound in gases, checked by experiment.
What affects the speed of sound in air?
This final equation tells us what matters, and what does not.
- Temperature \(T\). The speed is proportional to the square root of the absolute temperature, \(v \propto \sqrt{T}\). On a hot day the molecules move faster, and they hand on the vibrations more quickly.
- Molar mass \(M\). The speed is inversely proportional to the square root of the molar mass, \(v \propto 1/\sqrt{M}\). This is why your voice sounds high after you breathe in helium. Sound travels much faster in the lighter helium gas.
- Pressure \(P\). At a constant temperature, the pressure has no effect. By the ideal gas law, if \(T\) is constant, the ratio \(P/\rho\) is also constant. Raising the pressure only packs the molecules closer together. The density rises by the same factor, and the speed stays the same.
Move the slider across the sim. The curve bends over, because \(v\) grows as the square root of \(T\). At \(0\,°\text{C}\) the readout shows \(331\,\text{m/s}\), and at \(20\,°\text{C}\) it shows \(343\,\text{m/s}\). On a cold winter morning at \(-40\,°\text{C}\), sound crawls at \(306\,\text{m/s}\). The kelvin scale changes much less than the Celsius scale suggests, so the speed changes gently.
Helium does not make your vocal cords vibrate faster. The pitch of their vibration stays almost the same. What changes is the air space of your mouth and throat. Its natural frequencies rise with the speed of sound, nearly three times higher in helium. Those frequencies set the colour of the voice, so the voice sounds squeaky.
At a constant temperature, the pressure of a gas is doubled. The speed of sound in the gas
If the temperature of the air increases, the speed of sound
Helium has \(\gamma = 1.67\) and a molar mass of \(0.004\,\text{kg/mol}\). Air has \(\gamma = 1.40\) and \(M = 0.029\,\text{kg/mol}\). At the same temperature, how many times faster is sound in helium than in air?
Two worked speeds
The two examples below use the formula in its two common ways. The first puts in every constant. The second compares two temperatures by a ratio, so the constants cancel.
Find the speed of sound in dry air at \(20\,°\text{C}\). Take the molar mass \(M = 0.029\,\text{kg/mol}\) and the ratio of specific heats \(\gamma = 1.40\).
Step 1: convert the temperature to kelvin, \(T = 20 + 273 = 293\,\text{K}\).
Step 2: recall the universal gas constant, \(R = 8.314\,\text{J/(mol·K)}\).
Step 3: use the formula \(v = \sqrt{\gamma RT/M}\) and put in the values: \(v = \sqrt{\dfrac{1.40 \times 8.314 \times 293}{0.029}} = \sqrt{\dfrac{3410.4}{0.029}}\).
Then \(v = \sqrt{117\,600} \approx 343\,\text{m/s}\).
Final answer: \(343\,\text{m/s}\) in dry air at \(20\,°\text{C}\). This is the speed used in the first two lessons of this chapter.
How much faster does sound travel in air at \(35\,°\text{C}\) than at \(0\,°\text{C}\)?
Step 1: convert to kelvin. At \(0\,°\text{C}\), \(T_1 = 273\,\text{K}\). At \(35\,°\text{C}\), \(T_2 = 308\,\text{K}\).
Step 2: the formula gives \(v \propto \sqrt{T}\). Thus \(\dfrac{v_2}{v_1} = \sqrt{\dfrac{T_2}{T_1}}\).
Step 3: calculate the ratio, \(\dfrac{v_2}{v_1} = \sqrt{\dfrac{308}{273}} = \sqrt{1.128} \approx 1.062\). So \(v_2\) is about \(6.2\,\%\) faster than \(v_1\).
Step 4: take \(v_1 = 331\,\text{m/s}\), the commonly accepted value at \(0\,°\text{C}\). Then \(v_2 = 1.0622 \times 331 \approx 352\,\text{m/s}\).
Final answer: at \(35\,°\text{C}\) the speed of sound in air is about \(352\,\text{m/s}\), about \(21\,\text{m/s}\) faster than at \(0\,°\text{C}\).
Why does sound travel faster at higher temperatures?
On a monsoon evening at about \(20\,°\text{C}\), the thunder arrives \(6\) seconds after the flash. Estimate the distance of the lightning, in kilometres.
Now return to the count after the lightning. Sound in air travels at about \(343\,\text{m/s}\) at \(20\,°\text{C}\), so three seconds of counting means about a kilometre. On a hot afternoon at \(35\,°\text{C}\), sound is about \(6\,\%\) faster. On a cold morning it is a little slower. The pressure of the air, high or low, makes no difference at all.
- Putting the temperature in °C into \(v = \sqrt{\gamma RT/M}\).\(T\) must be absolute, in kelvin: add 273.
- Thinking that sound travels faster at a higher pressure.At a fixed temperature, \(P\) and \(\rho\) rise together and \(P/\rho = RT/M\) is unchanged.
- Taking the speed to be proportional to \(T\).The speed goes as \(\sqrt{T}\). Doubling the speed needs four times the absolute temperature.
- Using the molar mass in g/mol.With \(R = 8.314\,\text{J/(mol·K)}\), \(M\) must be in kg/mol: \(0.029\) for air, not \(29\).