Question

State whether the following statement is true or false:
Speed of sound can never exceed the average molecular speed in a fluid.
TRUE
FALSE

Answer 1

From Laplace’s correction we know that, the speed of sound of an ideal gas is given by $v = \sqrt {\dfrac{{\gamma P}}{\rho }}$, where P is the pressure and $\rho$ is the density of the gas.
Using the ideal gas equation $PV = nRT$, we can rearrange the formula for velocity as: $v = \sqrt {\dfrac{{\gamma RT}}{M}}$
Also, the average molecular speed of a fluid is given by formula: ${v_{avg}} = \sqrt {\dfrac{{8RT}}{M}}$
Their ratio is: $\dfrac{{{v_{{\text{sound(air)}}}}}}{{{v_{{\text{avg}}}}}} = \dfrac{{\sqrt {\dfrac{{\gamma RT}}{M}} }}{{\sqrt {\dfrac{{8RT}}{M}} }} = \sqrt {\dfrac{\gamma }{8}}$
Since, the maximum possible value for $\gamma = \dfrac{5}{3} = 1.67{\text{ }}\left( {{\text{for monoatomic gas}}} \right)$, which is less than 8.
Hence, speed of sound in a gas is always less than the molecular average speed of the gas.

Complete step by step answer:
Sound is a longitudinal wave, and the speed of sound depends upon how fast the vibrational energy is transferred through a medium.
The speed of sound in a fluid is given by formula:
$v = \sqrt {\dfrac{B}{\rho }}$, where B is the bulk modulus of elasticity and $\rho$ is the density of material.
Bulk modulus is the ratio of pressure applied to the relative change in volume of the body.
$\therefore B = \dfrac{{\Delta P}}{{\dfrac{{\Delta V}}{V}}} = \dfrac{{\Delta P \cdot V}}{{\Delta V}} - - - - - - - - - \left( 1 \right)$
According to Laplace, when the sound waves travel the expansion and compression of gas is an adiabatic process.
Hence, $P{V^\gamma } = {\text{ constant = c}}$
$\Rightarrow \Delta \left( {P{V^\gamma }} \right) = \Delta \left( c \right) \\\ \Rightarrow \Delta P \cdot {V^\gamma } + P\left( {\gamma {V^{\gamma - 1}}} \right).( - \Delta V) = 0{\text{}}\left( {\because {\text{ change in volume is negative}}} \right) \\\ \Rightarrow \Delta P \cdot {V^\gamma } = \gamma P \cdot \Delta V \cdot {V^{\gamma - 1}} \\\ \Rightarrow \dfrac{{\Delta P \cdot {V^\gamma }}}{{\Delta V \cdot {V^{\gamma - 1}}}} = \gamma P \\\ \Rightarrow \dfrac{{\Delta P \cdot V}}{{\Delta V}} = \gamma P - - - - - - - - - - - - \left( 2\right) \\\$
From (1) and (2):
$B = \gamma P$
Hence, velocity of sound in an ideal gas is given by:
$v = \sqrt {\dfrac{{\gamma P}}{\rho }} - - - - - - - - - - - \left( 3 \right)$
Also, for an ideal gas, the ideal gas equation is:
$PV = nRT \\\ \Rightarrow PV = \dfrac{m}{M}RT{\text{ }}\left( {{\text{where n = }}\dfrac{{{\text{mass of material(m)}}}}{{{\text{Molar mass (M)}}}}} \right) \\\ \Rightarrow P = \dfrac{m}{V}\dfrac{{RT}}{M} = \rho \dfrac{{RT}}{M} \\\ \Rightarrow \dfrac{P}{\rho } = \dfrac{{RT}}{M} \\\$
Putting this relation in equation (3), we get:
Speed of sound in an ideal gas is given by: $v = \sqrt {\dfrac{{\gamma RT}}{M}} - - - - - - - - - - - - \left( 4 \right)$
Where, $\gamma$ is the adiabatic constant
R is universal gas constant
T is temperature in kelvin
And M is Molecular mass of the material.
From Maxwell-Boltzmann distribution of molecular speed, we know that:
Average molecular speed of a fluid is:
${v_{avg}} = \sqrt {\dfrac{{8RT}}{M}} - - - - - - - - - - \left( 5 \right)$
From equation (4) and (5) we see that,
The only difference in the formula for speed of sound in an ideal gas and average molecular speed of a fluid is that the former has $\gamma$ and the later has 8.
Since, the maximum possible value for $\gamma = \dfrac{5}{3} = 1.67{\text{ }}\left( {{\text{for monoatomic gas}}} \right)$
And $1.67 < 8$.
Therefore, we can conclude that sound in an ideal gas is lesser than the average molecular speed of a fluid.
Hence, True is the correct answer.

Note: The speed of sound in a solid depends upon the Young’s modulus of elasticity of the medium and the density of the material and is given by:
${v_{solid}} = \sqrt {\dfrac{Y}{\rho }}$