Properties of Gases

Particle Velocity of a Gas

The particle velocity of a gas flow at any point can be defined as the average velocity (in meters per second, m/s) of the air molecules passing through a plane cutting orthogonal to the flow. The term ``velocity'' in this book, when referring to air, means ``particle velocity.''

It is common in acoustics to denote particle velocity by lower-case $ u$.


Volume Velocity of a Gas

The volume velocity $ U$ of a gas flow is defined as particle velocity $ u$ times the cross-sectional area $ A$ of the flow, or

$\displaystyle U(t,x) = u(t,x) A(x)
$

where $ x$ denotes position along the flow, and $ t$ denotes time in seconds. Volume velocity is thus in physical units of volume per second (m$ \null^3$/s).

When a flow is confined within an enclosed channel, as it is in an acoustic tube, volume velocity is conserved when the tube changes cross-sectional area, assuming the density $ \rho$ remains constant. This follows directly from conservation of mass in a flow: The total mass passing a given point $ x$ along the flow is given by the mass density $ \rho$ times the integral of the volume volume velocity at that point, or

$\displaystyle M(t_0:t_f,x) = \rho\int_{t_0}^{t_f} U(t,x) dt.
$

As a simple example, consider a constant flow through two cylindrical acoustic tube sections having cross-sectional areas $ A_1$ and $ A_2$, respectively. If the particle velocity in cylinder 1 is $ u_1$, then the particle velocity in cylinder 2 may be found by solving

$\displaystyle u_1 A_1 = u_2 A_2
$

for $ u_2$.

It is common in the field of acoustics to denote volume velocity by an upper-case $ U$. Thus, for the two-cylinder acoustic tube example above, we would define $ U_1\isdeftext u_1A_1$ and $ U_2\isdeftext u_2A_2$, so that

$\displaystyle U_1 = U_2
$

would express the conservation of volume velocity from one tube segment to the next.


Pressure is Confined Kinetic Energy

According the kinetic theory of ideal gases [180], air pressure can be defined as the average momentum transfer per unit area per unit time due to molecular collisions between a confined gas and its boundary. Using Newton's second law, this pressure can be shown to be given by one third of the average kinetic energy of molecules in the gas.

$\displaystyle p = \frac{1}{3}\rho \left<u^2\right>
$

Here, $ \langle u^2\rangle $ denotes the average squared particle velocity in the gas. (The constant $ 1/3$ comes from the fact that we are interested only in the kinetic energy directed along one dimension in 3D space.)


Proof: This is a classical result from the kinetic theory of gases [180]. Let $ M$ be the total mass of a gas confined to a rectangular volume $ V = Aw$, where $ A$ is the area of one side and $ w$ the distance to the opposite side. Let $ \overline{u}_x$ denote the average molecule velocity in the $ x$ direction. Then the total net molecular momentum in the $ x$ direction is given by $ M\vert\overline{u}_x\vert$. Suppose the momentum $ \overline{u}_x$ is directed against a face of area $ A$. A rigid-wall elastic collision by a mass $ M$ traveling into the wall at velocity $ \overline{u}_x$ imparts a momentum of magnitude $ 2M\overline{u}_x$ to the wall (because the momentum of the mass is changed from $ +M\overline{u}_x$ to $ -M\overline{u}_x$, and momentum is conserved). The average momentum-transfer per unit area is therefore $ 2M\overline{u}_x/A$ at any instant in time. To obtain the definition of pressure, we need only multiply by the average collision rate, which is given by $ \overline{u}_x/(2w)$. That is, the average $ x$-velocity divided by the round-trip distance along the $ x$ dimension gives the collision rate at either wall bounding the $ x$ dimension. Thus, we obtain

$\displaystyle p \isdefs \frac{2M\overline{u}_x}{A}\cdot \frac{\overline{u}_x}{2w} \eqsp \rho \overline{u}_x^2
$

where $ \rho=M/V$ is the density of the gas in mass per unit volume. The quantity $ \rho\overline{u}_x^2/2$ is the average kinetic energy density of molecules in the gas along the $ x$ dimension. The total kinetic energy density is $ \rho\overline{u}^2/2$, where $ \overline{u}=\sqrt{\overline{u}_x^2+\overline{u}_y^2+\overline{u}_z^2}$ is the average molecular velocity magnitude of the gas. Since the gas pressure must be the same in all directions, by symmetry, we must have $ \overline{u}_x^2=\overline{u}_y^2=\overline{u}_z^2 = \overline{u}^2/3$, so that

$\displaystyle p = \frac{1}{3}\rho \overline{u}^2.
$


Bernoulli Equation

In an ideal inviscid, incompressible flow, we have, by conservation of energy,

$\displaystyle p + \frac{1}{2}\rho u^2 + \rho g h =$   constant

where

\begin{eqnarray*}
p &=& \mbox{pressure (newtons/m$^2$\ = kg /(m s$^2$))}\\
u &=...
...\
\mbox{\lq\lq Inviscid''} &=& \mbox{\lq\lq Frictionless'', \lq\lq Lossless''}
\end{eqnarray*}

This basic energy conservation law was published in 1738 by Daniel Bernoulli in his classic work Hydrodynamica.

From §B.7.3, we have that the pressure of a gas is proportional to the average kinetic energy of the molecules making up the gas. Therefore, when a gas flows at a constant height $ h$, some of its ``pressure kinetic energy'' must be given to the kinetic energy of the flow as a whole. If the mean height of the flow changes, then kinetic energy trades with potential energy as well.


Bernoulli Effect

The Bernoulli effect provides that, when a gas such as air flows, its pressure drops. This is the basis for how aircraft wings work: The cross-sectional shape of the wing, called an aerofoil (or airfoil), forces air to follow a longer path over the top of the wing, thereby speeding it up and creating a net upward force called lift.

Figure B.8: Illustration of the Bernoulli effect in an acoustic tube.
\includegraphics{eps/bernoulli-effect}

Figure B.8 illustrates the Bernoulli effect for the case of a reservoir at constant pressure $ p_m$ (``mouth pressure'') driving an acoustic tube. Any flow inside the ``mouth'' is neglected. Within the acoustic channel, there is a flow with constant particle velocity $ u$. To conserve energy, the pressure within the acoustic channel must drop down to $ p_m - \rho u^2/2$. That is, the flow kinetic energy subtracts from the pressure kinetic energy within the channel.

For a more detailed derivation of the Bernoulli effect, see, e.g., [179]. Further discussion of its relevance in musical acoustics is given in [144,197].


Air Jets

Referring again to Fig.B.8, the gas flow exiting the acoustic tube is shown as forming a jet. The jet ``carries its own pressure'' until it dissipates in some form, such as any combination of the following:

Pressure recovery refers to the conversion of flow kinetic energy back to pressure kinetic energy. In situations such as the one shown in Fig.B.8, the flow itself is driven by the pressure drop between the confined reservoir (pressure $ p_m$) and the outside air (pressure $ p_m - \rho u^2/2$). Therefore, any pressure recovery would erode the pressure drop and hence the flow velocity $ u$.

For a summary of more advanced aeroacoustics, including consideration of vortices, see [196]. In addition, basic textbooks on fluid mechanics are relevant [171].


Acoustic Intensity

Acoustic intensity may be defined by

$\displaystyle \zbox {\underline{I} \isdefs p \underline{v}}
\quad \left(\frac{\...
...mall Time}}
\eqsp
\frac{\mbox{\small Power Flux}}{\mbox{\small Area}}\right)
$

where

\begin{eqnarray*}
p &=& \mbox{acoustic pressure} \quad \left(\frac{\mbox{\small ...
...ad \left(\frac{\mbox{\small Length}}{\mbox{\small Time}}\right).
\end{eqnarray*}

For a plane traveling wave, we have

$\displaystyle \zbox {p = R v}
$

where

$\displaystyle \zbox {R \isdefs \rho c}
$

is called the wave impedance of air, and

\begin{eqnarray*}
c &=& \mbox{sound speed},\\
\rho &=& \mbox{mass density of ai...
...ume}}\right),\\
v &\isdef & \left\vert\underline{v}\right\vert.
\end{eqnarray*}

Therefore, in a plane wave,

$\displaystyle \zbox {I = p v = Rv^2 = \frac{p^2}{R}.}
$


Acoustic Energy Density

The two forms of energy in a wave are kinetic and potential. Denoting them at a particular time $ t$ and position $ \underline{x}$ by $ w_v(t,\underline{x})$ and $ w_p(t,\underline{x})$, respectively, we can write them in terms of velocity $ v$ and wave impedance $ R=\rho c$ as follows:

\begin{eqnarray*}
w_v &=& \frac{1}{2} \rho v^2 \eqsp \frac{1}{2c} R v^2 \quad\le...
...ad\left(\frac{\mbox{\small Energy}}{\mbox{\small Volume}}\right)
\end{eqnarray*}

More specifically, $ w_v$ and $ w_p$ may be called the acoustic kinetic energy density and the acoustic potential energy density, respectively.

At each point in a plane wave, we have $ p(t,\underline{x})=R\,v(t,\underline{x})$ (pressure equals wave-impedance times velocity), and so

\begin{eqnarray*}
w_v &=& \frac{1}{2c} R v^2 = \frac{1}{2}\cdot \frac{I}{c}\\
w_p &=& \frac{1}{2c} \frac{p^2}{R} = \frac{1}{2} \cdot \frac{I}{c},
\end{eqnarray*}

where $ I(t,\underline{x})\isdef p(t,\underline{x})\,v(t,\underline{x})$ denotes the acoustic intensity (pressure times velocity) at time $ t$ and position $ \underline{x}$. Thus, half of the acoustic intensity $ I$ in a plane wave is kinetic, and the other half is potential:B.30

$\displaystyle \frac{I}{c} = w = w_v+w_p = 2w_v = 2w_p
$

Note that acoustic intensity $ I$ has units of energy per unit area per unit time while the acoustic energy density $ w=I/c$ has units of energy per unit volume.


Energy Decay through Lossy Boundaries

Since the acoustic energy density $ w=w_v+w_p$ is the energy per unit volume in a 3D sound field, it follows that the total energy of the field is given by integrating over the volume:

$\displaystyle E(t) = \iiint\limits_V w(t,\underline{x}) \,dV
$

In reverberant rooms and other acoustic systems, the field energy decays over time due to losses. Assuming the losses occur only at the boundary of the volume, we can equate the rate of total-energy change to the rate at which energy exits through the boundaries. In other words, the energy lost by the volume $ V$ in time interval $ \Delta t$ must equal the acoustic intensity $ \underline{I}(t,\underline{x})$ exiting the volume, times $ \Delta t$ (approximating $ I$ as constant between times $ t$ and $ t+\Delta t$):

$\displaystyle E(t+\Delta t) - E(t) = -\Delta t \iint\limits_A \underline{I}\cdot \underline{n}\, dA
$

The term $ \underline{I}(t,\underline{x})\cdot\underline{n}(\underline{x})$ is the dot-product of the (vector) intensity $ \underline{I}$ with a unit-vector $ \underline{n}$ chosen to be normal to the surface at position $ \underline{x}$ along the surface. Thus, $ \underline{I}\cdot \underline{n}$ is the component of the acoustic intensity $ \underline{I}$ exiting the volume normal to its surface. (The tangential component does not exit.) Dividing through by $ \Delta t$ and taking a limit as $ \Delta t\to 0$ yields the following conservation law, originally published by Kirchoff in 1867:

$\displaystyle \frac{d}{dt}E = \frac{d}{dt}\iiint\limits_V w(t,\underline{x}) \,...
...imits_A \underline{I}(t,\underline{x})\cdot \underline{n}(\underline{x})\, dA.
$

Thus, the rate of change of energy in an ideal acoustic volume $ V$ is equal to the surface integral of the power crossing its boundary. A more detailed derivation appears in [349, p. 37].

Sabine's theory of acoustic energy decay in reverberant room impulse responses can be derived using this conservation relation as a starting point.


Ideal Gas Law

The ideal gas law can be written as

$\displaystyle PV \eqsp nRT \eqsp NkT \protect$ (B.45)

where

\begin{eqnarray*}
P &=& \mbox{total pressure (Pascals)}\\
V &=& \mbox{volume (c...
...ture}\index{absolute temperature\vert textbf} (degrees Kelvin).}
\end{eqnarray*}

The alternate form $ PV=NkT$ comes from the statistical mechanics derivation in which $ N$ is the number of gas molecules in the volume, and $ k$ is Boltzmann's constant. In this formulation (the kinetic theory of ideal gases), the average kinetic energy of the gas molecules is given by $ (3/2) kT$. Thus, temperature is proportional to average kinetic energy of the gas molecules, where the kinetic energy of a molecule $ m$ with translational speed $ v$ is given by $ (1/2)mv^2$.

In an ideal gas, the molecules are like little rubber balls (or rubbery assemblies of rubber balls) in a weightless vacuum, colliding with each other and the walls elastically and losslessly (an ``ideal rubber''). Electromagnetic forces among the molecules are neglected, other than the electron-orbital repulsion producing the elastic collisions; in other words, the molecules are treated as electrically neutral far away. (Gases of ionized molecules are called plasmas.)

The mass $ m$ of the gas in volume $ V$ is given by $ m=nM$, where $ M$ is the molar mass of the gass (about 29 g per mole for air). The air density is thus $ \rho=m/V$ so that we can write

$\displaystyle P \eqsp \frac{R}{M} \rho T.
$

That is, pressure $ P$ is proportional to density $ \rho$ at constant temperature $ T$ (with $ R/M$ being a constant).

We normally do not need to consider the (nonlinear) ideal gas law in audio acoustics because it is usually linearized about some ambient pressure $ P_0$. The physical pressure is then $ P=P_0+p$, where $ p$ is the usual acoustic pressure-wave variable. That is, we are only concerned with small pressure perturbations $ p$ in typical audio acoustics situations, so that, for example, variations in volume $ V$ and density $ \rho$ can be neglected. Notable exceptions include brass instruments which can achieve nonlinear sound-pressure regions, especially near the mouthpiece [198,52]. Additionally, the aeroacoustics of air jets is nonlinear [196,530,531,532,102,101].


Isothermal versus Isentropic

If air compression/expansion were isothermal (constant temperature $ T$), then, according to the ideal gas law $ PV=nRT$, the pressure $ P$ would simply be proportional to density $ \rho$. It turns out, however, that heat diffusion is much slower than audio acoustic vibrations. As a result, air compression/expansion is much closer to isentropic (constant entropy $ S$) in normal acoustic situations. (An isentropic process is also called a reversible adiabatic process.) This means that when air is compressed by shrinking its volume $ V$, for example, not only does the pressure $ P$ increase (§B.7.3), but the temperature $ T$ increases as well (as quantified in the next section). In a constant-entropy compression/expansion, temperature changes are not given time to diffuse away to thermal equilibrium. Instead, they remain largely frozen in place. Compressing air heats it up, and relaxing the compression cools it back down.


Adiabatic Gas Constant

The relative amount of compression/expansion energy that goes into temperature $ T$ versus pressure $ P$ can be characterized by the heat capacity ratio

$\displaystyle \gamma \isdefs \frac{C_P}{C_V}
$

where $ C_P$ is the specific heat (also called heat capacity) at constant pressure, while $ C_V$ is the specific heat at constant volume. The specific heat, in turn, is the amount of heat required to raise the temperature of the gas by one degree. It is derived in statistical thermodynamics [138] that, for an ideal gas, we have $ C_p=C_v+R$, where $ R$ is the ideal gas constant (introduced in Eq.$ \,$(B.45)). Thus, $ \gamma>1$ for any ideal gas. The extra heat absorption that occurs when heating a gas at constant pressure is associated with the workB.2) performed on the volume boundary (fore times distance = pressure times area times distance) as it expands to keep pressure constant. Heating a gas at constant volume involves increasing the kinetic energy of the molecules, while heating a gas at constant pressure involves both that and pushing the boundary of the volume out. The reason not all gases have the same $ \gamma$ is that they have different internal degrees of freedom, such as those associated with spinning and vibrating internally. Each degree of freedom can store energy.

In terms of $ \gamma$, we have

$\displaystyle P_1V_1^\gamma \eqsp P_2V_2^\gamma, \protect$ (B.46)

where $ \gamma\approx 1.4$ for dry air at normal temperatures. Thus, if a volume of ideal gas is changed from $ V_1$ to $ V_2$, the pressure change is given by

$\displaystyle P_2 \eqsp P_1 \left(\frac{V_1}{V_2}\right)^{1/\gamma}
$

and the temperature change is

$\displaystyle T_2 \eqsp T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1}.
$

These equations both follow from Eq.$ \,$(B.46) and the ideal gas law Eq.$ \,$(B.45).

The value $ \gamma=1.4$ is typical for any diatomic gas.B.31 Monatomic inert gases, on the other hand, such as Helium, Neon, and Argon, have $ \gamma\approx 1.6$. Carbon dioxide, which is triatomic, has a heat capacity ratio $ \gamma=1.28$. We see that more complex molecules have lower $ \gamma$ values because they can store heat in more degrees of freedom.


Heat Capacity of Ideal Gases

In statistical thermodynamics [175,138], it is derived that each molecular degree of freedom contributes $ R/2$ to the molar heat capacity of an ideal gas, where again $ R$ is the ideal gas constant.

An ideal monatomic gas molecule (negligible spin) has only three degrees of freedom: its kinetic energy in the three spatial dimensions. Therefore, $ C_v=(3/2)R$. This means we expect

$\displaystyle \gamma\isdefs \frac{C_p}{C_v}\eqsp \frac{C_v+R}{C_v} \eqsp
\frac{3/2+1}{3/2} \eqsp 5/3,
$

a result that agrees well with experimental measurements [138].

For an ideal diatomic gas molecule such as air, which can be pictured as a ``bar bell'' configuration of two rubber balls, two additional degrees of freedom are added, both associated with spinning the molecule about an axis orthogonal to the line connecting the atoms, and piercing its center of mass. There are two such axes. Spinning about the connecting axis is neglected because the moment of inertia is so much smaller in that case. Thus, for diatomic gases such as dry air, we expect

$\displaystyle \gamma\isdefs \frac{C_p}{C_v}\eqsp \frac{C_v+R}{C_v} \eqsp
\frac{5/2+1}{5/2} \eqsp 7/5\eqsp 1.4,
$

as observed to a good degree of approximation at normal temperatures. At high temperatures, new degrees of freedom appear associated with vibrations in the molecular bonds. (For example, the ``bar bell'' can vibrate longitudinally.) However, such vibrations are ``frozen out'' at normal room temperatures, meaning that their (quantized) energy levels are too high and spaced too far apart to be excited by room temperature collisions [138, p. 147].B.32


Speed of Sound in Air

The speed of sound in a gas depends primarily on the temperature, and can be estimated using the following formula from the kinetic theory of gases:B.33

$\displaystyle c = \sqrt{\gamma R_m T},
$

where, as discussed in the previous section, the adiabatic gas constant is $ \gamma=1.4$ for dry air, $ R_m=286$ is the ideal gas constant for air in meters-squared per second-squared per degrees-Kelvin-squared, and $ T$ is absolute temperature in degrees Kelvin (which equals degrees Celsius + 273.15). For example, at zero degrees Celsius (32 degrees Fahrenheit), the speed of sound is calculated to be 1085.1 feet per second. At 20 degrees Celsius, we get 1124.1 feet per second.


Air Absorption

This section provides some further details regarding acoustic air absorption [318]. For a plane wave, the decline of acoustic intensity as a function of propagation distance $ x$ is given by

$\displaystyle I(x) = I_1 e^{-x/\xi},
$

where

\begin{eqnarray*}
I(x) &=& \hbox{intensity $x$\ meters from the source}\\
& & ...
...ox{(depends on frequency, temperature, humidity, and pressure).}
\end{eqnarray*}

Tables B.1 and B.2 (adapted from [314]) give some typical values for air.


Table B.1: Attenuation constant $ m = 1/\xi $ (in inverse meters) at 20 degrees Celsius and standard atmospheric pressure
Relative Frequency in Hz
Humidity 1000 2000 3000 4000
40 0.0013 0.0037 0.0069 0.0242
50 0.0013 0.0027 0.0060 0.0207
60 0.0013 0.0027 0.0055 0.0169
70 0.0013 0.0027 0.0050 0.0145



Table B.2: Attenuation in dB per kilometer at 20 degrees Celsius and standard atmospheric pressure.
Relative Frequency in Hz
Humidity 1000 2000 3000 4000
40 5.6 16 30 105
50 5.6 12 26 90
60 5.6 12 24 73
70 5.6 12 22 63


There is also a (weaker) dependence of air absorption on temperature [183].

Theoretical models of energy loss in a gas are developed in Morse and Ingard [318, pp. 270-285]. Energy loss is caused by viscosity, thermal diffusion, rotational relaxation, vibration relaxation, and boundary losses (losses due to heat conduction and viscosity at a wall or other acoustic boundary). Boundary losses normally dominate by several orders of magnitude, but in resonant modes, which have nodes along the boundaries, interior losses dominate, especially for polyatomic gases such as air.B.34 For air having moderate amounts of water vapor ($ H_2O$) and/or carbon dioxide ($ CO_2$), the loss and dispersion due to $ N_2$ and $ O_2$ vibration relaxation hysteresis becomes the largest factor [318, p. 300]. The vibration here is that of the molecule itself, accumulated over the course of many collisions with other molecules. In this context, a diatomic molecule may be modeled as two masses connected by an ideal spring. Energy stored in molecular vibration typically dominates over that stored in molecular rotation, for polyatomic gas molecules [318, p. 300]. Thus, vibration relaxation hysteresis is a loss mechanism that converts wave energy into heat.

In a resonant mode, the attenuation per wavelength due to vibration relaxation is greatest when the sinusoidal period (of the resonance) is equal to $ 2\pi$ times the time-constant for vibration-relaxation. The relaxation time-constant for oxygen is on the order of one millisecond. The presence of water vapor (or other impurities) decreases the vibration relaxation time, yielding loss maxima at frequencies above 1000 rad/sec. The energy loss approaches zero as the frequency goes to infinity (wavelength to zero).

Under these conditions, the speed of sound is approximately that of dry air below the maximum-loss frequency, and somewhat higher above. Thus, the humidity level changes the dispersion cross-over frequency of the air in a resonant mode.


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Wave Equation in Higher Dimensions
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Wave Equation for the Vibrating String