What happens to the speed of sound in air when the pressure increases at a constant temperature

Answer Details

The speed of sound in a gas is governed by how stiff the gas is compared with how heavy it is, expressed as \[v = \sqrt{\frac{\gamma P}{\rho}},\] where \(P\) is the pressure, \(\rho\) the density and \(\gamma\) a constant for the gas. It looks as though raising \(P\) should raise \(v\), and that is exactly the trap in this question.

Pressure and density are not independent. For a fixed mass of gas at constant temperature, Boyle's law gives \(PV = \text{constant}\), and since \(\rho = m/V\) the density rises in exact proportion to the pressure. So the ratio \(P/\rho\) stays the same when the pressure is doubled: the gas becomes stiffer, but it also becomes correspondingly heavier per unit volume, and the two effects cancel. The speed of sound is therefore unchanged when pressure increases at constant temperature.

The same equation shows what does change the speed. Writing \(P/\rho = RT/M\) for an ideal gas gives \(v = \sqrt{\gamma RT/M}\), so the speed depends on the absolute temperature and on the molar mass of the gas, and it is proportional to \(\sqrt{T}\). This is why sound travels faster on a hot day and faster in a light gas such as helium, but is not altered by simply pumping the air to a higher pressure at the same temperature. Exam reminder: whenever a question changes the pressure of a gas at constant temperature, check whether the density changes with it before concluding that a quantity depending on \(P/\rho\) has changed.