Question

By applying Newton’s laws to the oscillations of a continuous medium, one can show that the speed of a sound wave is given by

$c_s=\sqrt{\frac{B}{\rho }}$,

where $\rho$is the density of the medium (mass per unit volume) and B is the bulk modulus, a measure of the medium’s stiffness? More precisely, if we imagine applying an increase in pressure $\mathrm{\Delta }P$to a chunk of the material, and this increase results in a (negative) change in volume $\mathrm{\Delta }V$, then B is defined as the change in pressure divided by the magnitude of the fractional change in volume:

$B=\frac{\mathrm{\Delta }P}{–\mathrm{\Delta }V/V}$

This definition is still ambiguous, however, because I haven't said whether the compression is to take place isothermally or adiabatically (or in some other way).

1. Compute the bulk modulus of an ideal gas, in terms of its pressure P, for both isothermal and adiabatic compressions.
2. Argue that for purposes of computing the speed of a sound wave, the adiabatic B is the one we should use.
3. Derive an expression for the speed of sound in an ideal gas, in terms of its temperature and average molecular mass. Compare your result to the formula for the RMS speed of the molecules in the gas. Evaluate the speed of sound numerically for air at room temperature.
4. When Scotland’s Battlefield Band played in Utah, one musician remarked that the high altitude threw their bagpipes out of tune. Would you expect altitude to affect the speed of sound (and hence the frequencies of the standing waves in the pipes)? If so, in which direction? If not, why not?

Short answer

1. The required bulk modulus for the isothermal process is $\boxed{B=P}$and for adiabatic compression is $\boxed{B_{adiabatic}=\gamma P}$.
2. Speed of sound in terms of bulk modulus and density of the medium is $c_s=\sqrt{\frac{B_{adiabatic}}{\rho }}$.
3. the required expression for the speed of sound is $\boxed{c_s=\sqrt{\frac{\gamma kT}{m}}}$, the ratio between root mean square speed and speed of sound in air is$\boxed{\frac{c_s}{v_{rms}}=0.68}$ and speed of sound in air at room temperature is $\boxed{c_s=343{\text{\hspace{0.17em}ms}}^{\text{-1}}}$.
4. The speed of sound decreases with increasing altitude.

Step-by-step solution

Part a. Step 1. Given.

Speed of sound wave is given by

$c_s=\sqrt{\frac{B}{\rho }}$ ……. (1)

Here, is the bulk modulus, is the density of the medium.

Expression for bulk modulus is

$B=\frac{\Delta P}{-\left(\frac{\Delta V}{V}\right)}$ …… (2)

Here,$\mathrm{\Delta }P$ is the change in pressure,$\mathrm{\Delta }V$ is the change in volume, and minus sign in the denominator is there because an increase in pressure causes a decrease in volume.

Part a. Step 2. Formula used.

For an isothermal process, the expression for pressure is

$P=\frac{NkT}{V}$ …… (3)

Here,$k$ is Boltzmann constant,$N$ is the number of molecules,$V\text{\hspace{0.17em}and\hspace{0.17em}}T$ are volume and temperature of the gas respectively.

For adiabatic process,

$P=\text{C}{V}^{-\gamma }$

Where C is constant.

Part a. Step 3. Calculation.

As$T$ is constant for an isothermal process, change in pressure can be written as

$\Delta P=NkT\Delta \left(\frac{1}{V}\right)$

$\Delta P=-NkT\left(\frac{\Delta V}{V^2}\right)$ …… (4)

Substitute$P$ for$\frac{NkT}{V}$ in equation (4)

$\begin{array}{l}\Delta P=-NkT\left(\frac{\Delta V}{V^2}\right)\\ \Delta P=-\left(\frac{NkT}{V}\right)\left(\frac{\Delta V}{V}\right)\\ \Delta P=-P\left(\frac{\Delta V}{V}\right)\end{array}$

$\left(\frac{\Delta V}{V}\right)=-\frac{\Delta P}{P}$ …… (5)

Substitute$\frac{\Delta P}{P}$ for$\left(-\frac{\Delta V}{V}\right)$ in equation (2)

$B=\frac{\cancel{\Delta P}}{\frac{\cancel{\Delta P}}{P}}$

$\boxed{B=P}$

For adiabatic compression,

$P=\text{C}{V}^{-\gamma }$ …… (6)

Change in pressure is calculated as

$\begin{array}{l}\Delta P=C\Delta \left(V^{-\gamma }\right)\\ \Delta P=C\left(-\gamma V^{-\gamma -1}\Delta V\right)\end{array}$

$\Delta P=-\gamma \text{C}\left(\frac{V^{-\gamma }}{V}\Delta V\right)$ …… (7)

Equation (7) can be simplified as

$\Delta P=-\gamma \text{C}{V}^{-\gamma }.\left(\frac{1}{V}\Delta V\right)$ …… (8)

Substitute$P$ for$\text{C}{V}^{-\gamma }$ in equation (8)

$\Delta P=-\gamma P\left(\frac{1}{V}\Delta V\right)$

$\left(\frac{1}{V}\Delta V\right)=-\frac{\Delta P}{\gamma P}$ …… (9)

Substitute $-\frac{\Delta P}{\gamma P}$for$\left(\frac{1}{V}\Delta V\right)$ in equation (2)

$\boxed{B_{adiabatic}=\gamma P}$

Part a. Step 4. Conclusion.

Hence, the required bulk modulus for the isothermal process is $\boxed{B=P}$and for adiabatic compression is $\boxed{B_{adiabatic}=\gamma P}$.

Part b. Step 1. Introduction.

Isothermal compression is so slow that the temperature of the gas doesn’t rise at all and in the adiabatic compression, the process is so fast that no heat escapes from the gas during the process. Most real compression processes will be somewhere between these extremes usually closer to the adiabatic approximation.

Speed of sound wave is given by

$c_s=\sqrt{\frac{B}{\rho }}$ ……. (1)

Here, is the bulk modulus, is the density of the medium.

Part b. Step 2. Explanation.

Since the sound wave is a kind of pressure wave that propagates through the medium at a high speed at, pressure changes occur here and heat exchange proceeds very slowly, so the adiabatic bulk modulus is the perhaps proper one to use in calculating. The speed of sound can be calculated from the bulk modulus and the material's mass density as $c_s=\sqrt{\frac{B_{adiabatic}}{\rho }}$.

Part b. Step 3. Conclusion.

Hence, the speed of sound in terms of bulk modulus and density of the medium is $c_s=\sqrt{\frac{B_{adiabatic}}{\rho }}$.

Part c. Step 1. Given.

Speed of sound wave is given by

$c_s=\sqrt{\frac{B}{\rho }}$ ……. (1)

Here,$B$ is the bulk modulus,$\rho$ is the density of the medium.

Part c. Step 2. Formula used.

Speed of sound in terms of the bulk modulus is

$c_s=\sqrt{\frac{B_{adiabatic}}{\rho }}$ …… (2)

For an ideal gas, the expression for pressure is

$P=\frac{NkT}{V}$ …… (3)

Here,$k$ is Boltzmann constant,$N$ is the number of molecules,$V\text{\hspace{0.17em}and\hspace{0.17em}}T$ are volume and temperature of the gas respectively.

Average molecular mass can be expressed as

$m=\frac{\text{mass\hspace{0.17em}of\hspace{0.17em}one\hspace{0.17em}mole\hspace{0.17em}of\hspace{0.17em}air}}{{\text{number\hspace{0.17em}of\hspace{0.17em}molecules\hspace{0.17em}in\hspace{0.17em}one\hspace{0.17em}mole\hspace{0.17em}N}}_{\text{A}}}$ …… (4)

Part c. Step 3. Calculation.

Density can be expressed in terms of mass and volume is

$\rho =\frac{Nm}{V}$ …… (5)

Here,$N$ is a number of molecules.

Substitute$\frac{Nm}{V}$ for$\rho$ in equation (2)

$c_s=\sqrt{\frac{B_{adiabatic}}{\rho }}$

$c_s=\sqrt{\frac{V{B}_{aiabatic}}{Nm}}$ …… (6)

Substitute$\gamma P$ for$B_{adiabatic}$ in equation (5)

$c_s=\sqrt{\frac{VPV}{Nm}}$ …… (7)

But from equation (3),$PV=NkT$

Substitute $NkT$for$PV$ in equation (7)

$c_s=\sqrt{\frac{VNkT}{Nm}}=\sqrt{\frac{\gamma kT}{m}}$

$\boxed{c_s=\sqrt{\frac{\gamma kT}{m}}}$ …… (8)

The root mean square speed of the molecules in an ideal gas is given by

$v_{rms}=\sqrt{\frac{3kT}{m}}$ …… (9)

Divide equation (8) by equation (9)

$\frac{c_s}{v_{rms}}=\sqrt{\frac{\gamma }{3}}$ …… (10)

But for air,$\gamma =1.4$

Substitute$1.4$ for$\gamma$ in equation (11)

$\begin{array}{l}\frac{c_s}{v_{rms}}=\sqrt{\frac{1.4}{3}}\\ \boxed{\frac{c_s}{v_{rms}}=0.68}\end{array}$

Substitute 0.0289 g for the mass of one mole of air and$6.022\times {10}^{23}$ for$N_A$ in equation (4)

$\begin{array}{c}m=\frac{0.0289}{6.022\times {10}^{23}}\\ m=4.81\times {10}^{-26}\text{\hspace{0.17em}kg}\end{array}$

Substitute$4.81\times {10}^{-26}\text{\hspace{0.17em}kg}$ for $m$,$1.38\times {10}^{-23}{\text{\hspace{0.17em}m}}^{\text{2}}{\text{.kg.s}}^{\text{-2}}{\text{.K}}^{\text{-1}}$ for $k$, 1.4 for $\gamma$,$293\text{\hspace{0.17em}K}$ for$T$ in equation (8)

$\begin{array}{l}{c}_s=\sqrt{\frac{\gamma kT}{m}}\\ =\sqrt{\frac{1.4\times 1.38\times {10}^{-23}\times 293}{4.81\times {10}^{-26}}}\\ \boxed{c_s=343{\text{\hspace{0.17em}ms}}^{\text{-1}}}\end{array}$

Part d. Step 4. Conclusion.

Hence, the required expression for the speed of sound is, ratio between root mean square speed and speed of sound in air is and speed of sound in air in room temperature is.

Part d. Step 1. Introduction.

The speed of sound is dependent on the density of the medium and the temperature of the medium. It also depends on the adiabatic exponent. It directly doesn’t depend on altitude but indirectly it depends.

Speed of sound wave is given by

$c_s=\sqrt{\frac{\gamma kT}{m}}$ ……. (1)

Here,$\gamma$ is the adiabatic exponent,$k$ is Boltzmann constant,$T$ is the temperature in Kelvin, and$m$ its mass of gas.

Part d. Step 2. Explanation.

Since the temperature decreases by about$9.8°\text{C}$ for every 1000 m if it is sunny or dry day and it decreases by$6°\text{C}$ if it’s snowing or raining. Since the speed of sound is proportional to the square root of temperature, so the speed of sound seems to decrease as the temperature decreases, from where it can be concluded that the speed of sound decreases as altitude increases.

Part d. Step 3. Conclusion.

Hence, the speed of sound decreases with increasing altitude.