# 11.3.2: Speed of Sound in Ideal and Perfect Gases

The speed of sound can be obtained easily for the equation of state for an ideal gas (also perfect gas as a sub set) because of a simple mathematical expression. The pressure for an ideal gas can be expressed as a simple function of density, $$\rho$$, and a function "molecular structure'' or ratio of specific heats, $$k$$ namely
$P= constant\times \rho^{k} \label{gd:sd:eq:iserhoToP} \tag{11}$
and hence
$\label{gd:sd:eq:idealGas1} c = \sqrt{\dfrac{\partial P}{\partial \rho}} = k \times constant \times \rho^{k-1} =k \times \dfrac{\overbrace{ constant \times \rho^k}^{P} }{ \rho} = k \times \dfrac{P }{\rho } \tag{12}$
Remember that $$P / \rho$$ is defined for an ideal gas as $$RT$$, and equation (12) can be written as

Ideal Gas Speed Sound

$\label{gd:sd:eq:sound} c = \sqrt{ k\, R\, T} \tag{13}$

Example 11.2

Calculate the speed of sound in water vapor at $$20 [bar]$$ and $$350^{\circ}C$$, (a) utilizes the steam table, and (b) assuming ideal gas.

Solution 11.2

The solution can be estimated by using the data from steam table
$c \sim \sqrt{ \dfrac{\Delta P}{ \Delta \rho} }_{s=constant} \tag{14}$
At $$20[bar]$$  and  $$350^{\circ}C$$: s = 6.9563 $$\left[ \dfrac{kJ }{ K\, kg}\right]$$ $$\rho$$ = 6.61376 $$\left[ \dfrac{kg }{ m^3} \right]$$
At $$18[bar]$$ and   $$350^{\circ}C$$: s = 7.0100 $$\left[\dfrac{ kJ }{ K\, kg}\right]$$ $$\rho$$ = 6.46956 $$\left[ \dfrac{ kg }{ m^3}\right]$$
At $$18[bar]$$ and $$300^{\circ}C$$: s = 6.8226 $$\left[\dfrac{ kJ }{ K\, kg}\right]$$ $$\rho$$ = 7.13216 $$\left[ \dfrac{kg }{ m^3} \right]$$
After interpretation of the temperature:
At $$18[bar]$$ and $$335.7^{\circ}C$$: s $$\sim$$ 6.9563 $$\left[\dfrac{ kJ }{ K\, kg} \right]$$ $$\rho \sim$$ 6.94199 $$\left[ \dfrac{kg }{ m^3} \right]$$
and substituting into the equation yields
$c = \sqrt{ 200000 \over 0.32823} = 780.5 \left[ m \over sec \right] \tag{15}$
for ideal gas assumption (data taken from Van Wylen and  Sontag, Classical Thermodynamics, table A 8.)
$c = \sqrt{k\,R\,T} \sim \sqrt{ 1.327 \times 461 \times (350 + 273)} \sim 771.5 \left[ \dfrac{m }{ sec} \right]$
Note that a better approximation can be done with a steam table, and it $$\cdots$$,

### Contributors

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.