Pump up a bicycle tyre quickly, and then touch the barrel of the pump. It is warm. Each rapid stroke squeezes the air before the heat has time to leave. Pump very slowly, and the barrel stays cool. The same air, squeezed at two different rates, behaves in two different ways.
The bulk modulus of a gas
To use \(v = \sqrt{B/\rho}\) for a gas, we must find the right value of the bulk modulus \(B\). It must be the value that holds under the physical conditions of a sound wave. The bulk modulus measures how strongly a material resists a uniform compression. For a fluid it is defined as
\[ B = -V\,\frac{dP}{dV} \]
where \(V\) is the volume and \(P\) the pressure. The negative sign makes \(B\) positive. An increase in pressure causes a decrease in volume, so \(dV < 0\) when \(dP > 0\).
For a sound wave in a gas, the link between pressure and volume depends on the thermodynamic process. The process is the way the gas is squeezed and released in the rapid compressions and rarefactions of sound. So we must decide which process it is. It may be an isothermal process, at constant temperature \(T\). Or it may be an adiabatic process, with no exchange of heat.
Why does the definition \(B = -V\,dP/dV\) carry a minus sign?
Newton's isothermal assumption
Newton first assumed that the oscillations of sound in air are slow enough. Each element of gas would then keep a constant temperature. For such an isothermal process, the ideal gas law \(PV = nRT\) holds at constant \(T\). Here \(n\) is the amount of gas in moles and \(R\) the gas constant. Then \(PV\) is a constant, and the bulk modulus follows.
\[ PV = \text{constant} \implies P = \dfrac{\text{const}}{V} \]
Why. the ideal gas law \(PV = nRT\) with \(n\) and \(T\) fixed.
\[ \dfrac{dP}{dV} = -\dfrac{\text{const}}{V^2} \]
Why. differentiate both sides with respect to \(V\). The derivative of \(1/V\) is \(-1/V^2\).
\[ \dfrac{dP}{dV} = -\dfrac{P}{V} \]
Why. the constant is \(PV\), so \(\text{const}/V^2 = PV/V^2 = P/V\).
\[ B_{iso} = -V\left(\dfrac{dP}{dV}\right)_T = -V\left(-\dfrac{P}{V}\right) \]
Why. put step 3 into the definition \(B = -V\,dP/dV\). The subscript \(T\) marks constant temperature.
\[ B_{iso} = P \]
Read it. the two minus signs cancel, and so do the two factors of \(V\).
So under Newton's assumption \(B = P\), and this leads to
\[ v_{\text{iso}} = \sqrt{\frac{P}{\rho}} \]
For air at \(0\,°\text{C}\), \(P = 1.013 \times 10^5\,\text{Pa}\) and \(\rho = 1.29\,\text{kg/m}^3\). This gives \(v_{\text{iso}} \approx 280\,\text{m/s}\). The measured speed is \(331\,\text{m/s}\). Newton's value is about \(15\,\%\) too low.
Air at \(0\,°\text{C}\) has a pressure of \(1.013 \times 10^5\,\text{Pa}\) and a density of \(1.29\,\text{kg/m}^3\). Find the speed of sound that Newton's isothermal formula predicts, in m/s.
Laplace's adiabatic correction
Laplace saw the flaw. The pressure changes in sound are extremely rapid. There is not enough time for the gas to exchange heat with its surroundings. So the process is essentially adiabatic. For an adiabatic process in an ideal gas,
\[ PV^{\gamma} = \text{constant} \]
where \(\gamma = C_P/C_V\) is the ratio of specific heats, at constant pressure and at constant volume.
\[ \dfrac{d}{dV}\left(PV^{\gamma}\right) = 0 \]
Why. \(PV^{\gamma}\) is a constant, so its derivative with respect to \(V\) is zero.
\[ V^{\gamma}\,\dfrac{dP}{dV} + \gamma P V^{\gamma - 1} = 0 \]
Why. the product rule. \(P\) depends on \(V\), and the derivative of \(V^{\gamma}\) is \(\gamma V^{\gamma - 1}\).
\[ \dfrac{dP}{dV} + \gamma\,\dfrac{P}{V} = 0 \]
Why. divide every term by \(V^{\gamma}\). Then \(V^{\gamma - 1}/V^{\gamma} = 1/V\).
\[ \dfrac{dP}{dV} = -\gamma\,\dfrac{P}{V} \]
Why. move the second term to the right side.
\[ B_{adiabatic} = -V\left(\dfrac{dP}{dV}\right)_S = -V\left(-\gamma\,\dfrac{P}{V}\right) \]
Why. put step 4 into the definition. The subscript \(S\) marks an adiabatic change, with no exchange of heat.
\[ B_{adiabatic} = \gamma P \]
Read it. the minus signs cancel, and so do the factors of \(V\).
Thus the correct bulk modulus for sound in a gas is \(B = \gamma P\). For diatomic gases such as air, \(\gamma\) is about \(1.4\). The adiabatic gas is stiffer than the isothermal one, by the factor \(\gamma\). Fig 4.1 shows why.
Read Fig 4.1 at the starting state. Squeeze the gas a little, and the pressure rises along one of the two curves. When no heat leaves, the gas warms as it is squeezed, so its pressure climbs faster. That steeper climb is a larger \(|dP/dV|\), and so a larger bulk modulus. A stiffer gas hands on each push more quickly.
With the numbers of Fig 4.2, Laplace's value is \(\sqrt{1.40 \times 1.013 \times 10^5/1.29} \approx 332\,\text{m/s}\). The measured speed is \(331\,\text{m/s}\). The correction multiplies Newton's value by \(\sqrt{\gamma} = \sqrt{1.40} \approx 1.18\), and that closes the gap.
Why are the compressions and rarefactions of a sound wave in air adiabatic rather than isothermal?
For air, \(\gamma = 1.40\). By what factor does Laplace's correction multiply Newton's speed of sound?
In the Principia of 1687, Isaac Newton gave the first calculation of the speed of sound. He treated air as a spring that kept its temperature. His answer fell well short of the speeds measured with cannon and distant observers. Newton tried to explain the gap away, and the puzzle stayed open for more than a century. In 1816 Pierre-Simon Laplace pointed out that the rapid squeezes of sound heat the air. With the adiabatic stiffness \(\gamma P\), theory and measurement agreed.
Now return to the bicycle pump. Rapid strokes warm the air because the heat has no time to escape. Sound squeezes the air hundreds of times faster still. So each compression is adiabatic, and the air behaves as a stiffer spring. That stiffness, \(\gamma P\) instead of \(P\), raises the speed of sound by a factor \(\sqrt{\gamma}\), about \(18\,\%\).
- Using \(B = P\) for sound in a gas.Sound is adiabatic. Use \(B = \gamma P\), so \(v = \sqrt{\gamma P/\rho}\).
- Forgetting the product rule when differentiating \(PV^{\gamma}\).Both \(P\) and \(V^{\gamma}\) change with \(V\), so there are two terms: \(V^{\gamma}\,dP/dV + \gamma PV^{\gamma - 1}\).
- Taking the minus sign of \(B = -V\,dP/dV\) to make \(B\) negative.\(dP/dV\) is itself negative, so \(B\) comes out positive.