Speed of Sound

For deriving a formula to calculate the speed of sounds in gases, we take a look at a tube that is filled with a gas and oriented in $z$-direction. A small pressure difference at one end of the tube leads to a pressure gradient causing an acceleration: $$\frac{\mathrm{d}v}{\mathrm{d}t} = -\frac{1}{\varrho}\frac{\mathrm{d}p}{\mathrm{d}z}$$ Because the mass flow rate on both ends has to be identical, we can assume a constant flux $j = \varrho v = \mathrm{const}$ which leads to $$v\,\mathrm{d}\varrho = -\varrho\,\mathrm{d}v$$ With the help of this correlation, the first equation can be rearranged according to $$\mathrm{d}p = -\varrho\frac{\mathrm{d}z}{\mathrm{d}t}\,\mathrm{d}v = v^2\mathrm{d}\varrho$$ Since $v$ is equal to the speed of sound $c$ this can be written as $$c = \sqrt{\frac{\mathrm{d}p}{\mathrm{d}\varrho}}$$ Using the equation for adiabatic compression $$\frac{p_1}{p_2} = \left(\frac{\varrho_1}{\varrho_2}\right)^\kappa$$ leads to $$p = C\varrho^\kappa$$ where the constant $C$ is defined as $$C = \frac{p}{\varrho^\kappa}$$ Calculating the derivative of $p$ with respect to $\varrho$ leads to $$\frac{\mathrm{d}p}{\mathrm{d}\varrho} = c\kappa^{\kappa - 1} = \kappa\frac{p}{\varrho}$$ This equation results in the possibility to calculate the speed of sound as a function of the pressure and density of the gas: $$c = \sqrt{\frac{\kappa p}{\varrho}}$$ Taking the ideal gas equation $$pV = RT$$ for 1 mol into account, this can be written as $$\boxed{c = \sqrt{\frac{\kappa R T}{M}}}$$ where $M$ is the molar mass of the gas particles. Inserting all constants and a molar mass of $M = 0.028\,\mathrm{kg/mol}$ for air results in a speed of $$c = 345\,\frac{\mathrm{m}}{\mathrm{s}}$$ for a temperature of $20^\circ\mathrm{C}$ which is close the measured one of $343\,\mathrm{m/s}$.