Speed of Sound in Gases

By Jitender Singh on Nov 11, 2019

The speed of sound in a fluid is given by $v=\sqrt{B/\rho}$, where $B$ is bulk modulus of the fluid and $\rho$ is its density. The sound waves travel in a gas by adiabatic compression and rarefaction (expansion). The speed of sound in an ideal gas is given by

where $\gamma$ is the adiabatic index (ratio of specific heats), $R=8.314$ J/mol-K is the universal gas constant, $T$ is the absolute temperature and $M$ is the molecular mass. Note that adiabatic bulk modulus of an ideal gas at pressure $p$ is $B=\gamma p$ and its density is $\rho=pM/RT$.

Newton’s formula for the speed of sound, $v=\sqrt{RT/M}$, is erroneous because it assumes isothermal compression and rarefaction of the gas (isothermal bulk modulus of an ideal is p). It was corrected by Laplace.

The speed of sound in gases is related to the root mean square speed of particles in the gas $v_\mathrm{rms}=\sqrt{3RT/M}$.

For a given gas, $\gamma$, $R$ and $M$ are constants. The speed depends only on temperature. It is independent of the pressure. To compute the speed of sound in a gaseous mixture, use mixture's adiabatic index and mean molecular mass.

Air is a mixture of gases with $\gamma_\mathrm{air}=1.4$ and mean molecular mass $M=28.9\times10^{-3}$ kg/mol (air is mostly diatomic nitrogen and oxygen). The speed of sound in air at 0℃ is 331 m/s. Its value at temperature $T$ ℃ is approximately equal to \begin{align} v=(331+0.6 T)\;\mathrm{m/s}.\nonumber \end{align} The speed of sound in air increases slightly with an increase in humidity. This is due to a decrease in mean molecular mass of the air due to increase in moisture content (molecular mass of air is 29 g/mol whereas it is 18 g/mol for water). The speed of sound in air is independent of the pressure.

The speed of sound is measured in the laboratory by resonance column method.

Solved Problems from IIT JEE

Problem from IIT JEE 2000

Two monatomic ideal gases 1 and 2 of molecular masses $m_1$ and $m_2$ respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to that in the gas 2 is given by,

$\sqrt{m_1/m_2}$
$\sqrt{m_2/m_1}$
$m_1/m_2$
$m_2/m_1$

Solution: The speed of sound in a gas with molecular mass $M$ and kept at temperature $T$ is given by, \begin{align} \label{ipa:eqn:1} v=\sqrt{\gamma RT/M}. \end{align} For the given monatomic gases, $\gamma_1=\gamma_2=5/3$ and temperature $T_1=T_2$. Substitute in above equation to get, \begin{align} {v_1}/{v_2}=\sqrt{m_2/m_1}.\nonumber \end{align}

Problem from IIT JEE 2004

A source of sound of frequency 600 Hz is placed inside water. The speed of sound in water is 1500 m/s and in air it is 300 m/s. The frequency of sound recorded by an observer who is standing in air is

200 Hz
3000 Hz
120 Hz
600 Hz

Solution: The frequency is a characteristic of the source and does not change when sound is transmitted from one medium to another. Since both source and observer are stationary, the frequency of sound recorded by the observer is equal to the source frequency, which is 600 Hz.

Problem from IIT JEE 1999

The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300 K is

$\sqrt{2/7}$
$\sqrt{1/7}$
$\sqrt{3}/5$
$\sqrt{6}/5$

Solution: The speed of sound in a gas is given by \begin{align} v=\sqrt{\gamma R T/M}. \end{align} The nitrogen is a diatomic gas with $\gamma_{\mathrm{N}_2}={7}/{5}$ and $M_{\mathrm{N}_2}=28$. The helium is a monatomic gas with $\gamma_\text{He}={5}/{3}$ and $M_\text{He}=4$. Substitute these values in above equation to get \begin{align} \frac{v_{\mathrm{N}_2}}{v_\text{He}}=\sqrt{\frac{\gamma_{\mathrm{N}_2}}{\gamma_\text{He}}\frac{M_\text{He}}{M_{\mathrm{N}_2}}}=\frac{\sqrt{3}}{5}.\nonumber \end{align}

Problem from IIT JEE 1983

The ratio of the velocity of sound in hydrogen gas $\left(\gamma=7/5 \right)$ to that in helium gas $\left(\gamma=5/3 \right)$ at the same temperature is $\sqrt{21/5}$.

Solution: The speed of sound in a gas with molecular mass $M$, ratio of specific heats $\gamma$, and temperature $T$ is given by \begin{align} %\label{mza:eqn:1} v=\sqrt{\gamma RT/M}. \nonumber \end{align} For hydrogen, $\gamma_\mathrm{H_2}=7/5$ and $M_\mathrm{H_2}=2$ and for helium $\gamma_\text{He}=5/3$ and $M_\text{He}=4$. The ratio of velocities of sound in hydrogen and helium gases at the same temperature is \begin{align} \frac{v_\mathrm{H_2}}{v_\text{He}}&=\sqrt{\frac{\gamma_\mathrm{H_2}}{\gamma_\text{He}}\frac{M_\text{He}}{M_\mathrm{H_2}}}=\sqrt{\frac{42}{25}}.\nonumber \end{align}

Problem from IIT JEE 2014

A student is performing an experiment using a resonance column and a tuning fork of frequency 244 Hz. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is $(0.350\pm0.005) \,\mathrm{m}$, the gas in the tube is (Given: $\sqrt{167RT}=640\,\mathrm{J^{1/2} mol^{-1/2}}$, $\sqrt{140RT}=590\,\mathrm{J^{1/2} mol^{-1/2}}$. The molar masses $M$ in grams are given in the options. Take the value of $\sqrt{10/M}$ for each gas as given there.)

Neon $\big(M\!=\!20, \sqrt{\frac{10}{20}}=\frac{7}{10}\big)$
Nitrogen $\big(M\!=\!28, \sqrt{\frac{10}{28}}=\frac{3}{5}\big)$
Oxygen $\big(M\!=\!32, \sqrt{\frac{10}{32}}=\frac{9}{16}\big)$
Argon $\big(M\!=\!36, \sqrt{\frac{10}{36}}=\frac{17}{32}\big)$

Solution: The speed of sound in a gas with molecular mass $M$, ratio of specific heat $\gamma$, and temperature $T$ is given by \begin{align} %\label{axb:eqn:1} v=\sqrt{\gamma RT/M}. \nonumber \end{align} The minimum height of air column for the resonance to occur is \begin{align} \label{axb:eqn:2} l=\frac{\lambda}{4}=\frac{v}{4\nu}=\frac{1}{4\nu}\sqrt{\frac{\gamma RT}{M}}. \end{align} The ratio of specific heat is $\gamma_m={5}/{3}=1.67$ for monatomic gases and $\gamma_d={7}/{5}=1.4$ for diatomic gases. Substitute these values in above equation to get \begin{align} l_\mathrm{Ne}&=\frac{1}{4(244)}\sqrt{\frac{1.67RT}{20\times{10}^{-3}}}=\frac{\sqrt{167RT}}{4(244)}\sqrt{\frac{10}{20}}\nonumber\\ &=\frac{640}{4(244)}\frac{7}{10}=0.459\,\mathrm{m},\nonumber\\ l_\mathrm{N_2}&=\frac{\sqrt{140RT}}{4(244)}\sqrt{\frac{10}{28}}=0.363 \,\mathrm{m},\nonumber\\ l_\mathrm{O_2}&=\frac{\sqrt{140RT}}{4(244)}\sqrt{\frac{10}{32}}=0.340\,\mathrm{m}, \nonumber\\ l_\mathrm{Ar}&=\frac{\sqrt{167RT}}{4(244)}\sqrt{\frac{10}{36}}=0.348 \,\mathrm{m}.\nonumber \end{align} Thus, only $l_\mathrm{Ar}$ lies in the specified range of $(0.350\pm 0.005)$ m.

Questions on Speed of Sound in Gases

Question 1: If the velocity of sound in air at 0℃ be 330 m/s then the increase in the velocity of sound in air, for 1℃ rise in temperature is?

Question 2: The velocity of sound in air

Question 3: If air molecules are traveling with a root mean square speed of 500 m/s then the speed of sound in air is?

Question 4: What is the speed of sound in a helium gas at a pressure of 150,000 Pa?

References

JEE Physics Solved Problems in Mechanics

Buy JEE Physics PYQ+: Mechanics

Buy Fluids, Waves & Thermodyanmics