Newton's Formula For The Velocity Of Sound In Gas

Isaac Newton's formula for the velocity of sound in a gas is a significant milestone in the understanding of wave propagation through different media. This formula provides insight into how sound waves travel through gases, considering factors like temperature and density. In our daily lives, the speed of sound is more than just an abstract concept—it's something we experience constantly. Whether it's hearing a distant thunderstorm or the echo of a shout across a canyon, the speed at which sound reaches us can reveal much about the environment. Newton’s early attempts to quantify this speed laid the groundwork for advancements in acoustics and engineering, influencing technologies like sonar and even how we design spaces for optimal sound quality. In this article, we will understand that this formula allows us to appreciate the scientific principles behind the sounds that fill our world every day.

Newton's Formula for the Velocity of Sound In Gas

Newton's formula illustrates that the speed of sound in a gas is influenced by the temperature and the properties of the gas itself. As the temperature of the gas increases, the speed of sound also increases, reflecting the fact that warmer gases have molecules moving more rapidly and thus transmit sound waves faster. Additionally, lighter gases (with lower molar masses) allow sound to travel more quickly compared to heavier gases.

Speed of Sound Wave in Gas: Newton's Formula

The main assumption before deriving the equation is when the sound propagates through a gas, temperature variation in compression and rarefaction is negligible. So, Newton assumed that the exchange of heat with the surroundings, the temperature of the layer will remain the same. Hence this process is isothermal. Thus by using the formula that we have studied in the last concept, we can write that

$v=\sqrt{\frac{B_{\text{isothermal}}}{\rho }}\dots \dots (i)$

Where $B_{\text{isothermal}}=$ Isothermal Bulk modulus

Now, in the isothermal process, PV = Constant

Differentiating both sides, we get

$\begin{array}{c}Pdv=V(-dP)\\ B_{\text{isothermal}}=P=\frac{dP}{\frac{dV}{V}}\end{array}$

So from the definition of Bulk modulus, we can say that the $P=B_{\text{isothermal}}$

$(As,B_i=\frac{dP}{\frac{dV}{V}})$

So from equation (i), We can write that

$v=\sqrt{\frac{P}{\rho }}$

This formula is given by Newton, So it is called Newton's formula.

Laplace Correction

Laplace Correction gives correction to the speed of sound in the gas. Newton's formula was formulated taking into consideration that sound travels in isothermal conditions, the result so obtained did not match with the experimental value of the speed of sound.

Thus, Laplace came up with a correction to it that sound travelling through air is a sudden process, it is well known as a Laplace Correction to Newton's Formula.

$$v=\sqrt{\frac{B_{\text{adiabatic}}}{\rho }}$$

Where $B_{\text{adiabatic}}=$ adiabatic bulk modulus

Now, in the adiabatic process, PV^γ = Constant

Differentiating both sides, we get

$\begin{array}{c}{P}_{\gamma }{V}^{\gamma -1}dv=V^{\gamma }(-dP)\\ B_{\text{adiabatic}}=-\frac{dP}{\frac{dV}{V}}=\gamma P\end{array}$

Factors Affecting the Speed of Sound in the Gas

The speed of sound in a gas is influenced by several factors, which include

1. Effect of Pressure

We know that the speed of sound in gas $=\sqrt{\frac{\gamma P}{\rho }}$
Also, for gas,
$$Pv=nRT=\frac{m}{M}RT$$

At constant temperature, we can write $=P\mathrm{\Delta }V=\frac{\mathrm{\Delta }m}{M}RT$
$$\begin{array}{rl}& P=\frac{\mathrm{\Delta }m}{\mathrm{\Delta }V}\frac{RT}{M}\\ & P=\rho \frac{RT}{M}\end{array}$$
$$\frac{P}{\rho }=\text{constant}$$

And as pressure changes, according to this the density changes. Thus we can say that the ratio will remain the same. So pressure does not create any effect on the speed of sound in the gas.

2. Effect of Density

For two gases of densities ${\rho }_1$ and ${\rho }_2$ at the same pressure with ratios of specific heat ${\gamma }_1$ and ${\gamma }_2$

$\frac{v_1}{v_2}=\sqrt{\frac{{\gamma }_1\times {\rho }_2}{{\gamma }_2\times {\rho }_1}}$

3. Effect of temperature

$\begin{array}{rl}& \text{As,}\frac{P}{\rho }=\frac{RT}{M}\\ & \text{so}v=\sqrt{\frac{\gamma RT}{M}}\Rightarrow v\propto \sqrt{T}\end{array}$

So, as the temperature increases the velocity will increase.

4. Effect of humidity

Humidity is the percentage of water vapour present in the air. As the humidity increases, the percentage of water vapor in the air increases and this decreases the density of air resulting in the increased of velocity of sound. So, with an increase in humidity, the density of air will decrease. And as we know that $v\propto \frac{1}{\sqrt{\rho }}$

So, the speed of sound will increase.

5. Effect of frequency

With the change in frequency, the wavelength also changes in the same proportion.

So, a product of both remains constant. From the equation $f\lambda =v$,

So velocity remains constant.

6. Effect of wind

As sound is carried by air, so as the velocity of wind changes then the velocity of the sound will change accordingly. Let the speed of the wind is v_w and it is blowing at an angle of $\theta$ with the direction of the sound. As shown in the figure

The speed of sound gets extra effect from the speed of the wind as - $v_{\text{sound}}+v_w\cos \theta$

$\theta$ may vary from 0 to ${180}^{\circ }$

Solved Example Based on Newton's Formula For The Velocity Of Sound In Gas

Example 1: A granite rod of 60 cm in length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7×10³ kg/m3 and its Young’s modulus is 9.27×10¹⁰ Pa. What will be the fundamental frequency (in Hz) of the longitudinal vibrations?

1) 5

2) 7.5

3) 2.5

4) 10

Solution:

$\begin{array}{rl}{\nu }_o& =\frac{v}{2l}=\frac{1}{2l}\cdot \sqrt{\frac{\gamma }{\rho }}=\frac{1}{2\ast 0.6}\sqrt{\frac{9.27\ast {10}^{10}}{2.7\ast {10}^3}}\\ & =4.9\ast {10}^3\mathrm{HZ}\simeq 5\mathrm{kHZ}\end{array}$

Hence, the answer is the option (1).

Example 2: Calculate the speed (in m/s) of the longitudinal wave in the helium gas of bulk modulus $1.7\times {10}^5\mathrm{Pa}$ and density is 0.18 kg/m³ at $0^{\circ }\mathrm{C}$ and 1 atm pressure.

1)972

2)413

3)314

4)600

Solution:

For longitudinal waves for liquid or gas

$$v=\sqrt{\frac{B}{\rho }}$$
where $\rho =$ Density of the medium

Speed of sound wave

$$v=\sqrt{\frac{B}{\rho }}$$
wherein
$B$ is the bulk modulus that represents the elastic property of the medium
$\rho =$ the density of the medium that represents the inertial property of the medium.
$$V_{He}=\sqrt{\frac{B}{\rho }}=\sqrt{\frac{1.7\times {10}^5}{0.18}}=972\mathrm{m}/\mathrm{s}$$

Hence, the answer is the option (1).

Example 3: Calculate the speed (in m/s) of the longitudinal wave in oxygen at $0^{\circ }\mathrm{C}$ and 1 atm $\left({10}^5\mathrm{Pa}\right)$ having bulk modulus equal to $1.41\times {10}^5\mathrm{Pa}$ and density of 1.43 kg/m³.

1)314

2)612

3)972

4)0

Solution:

Speed of sound wave

$$v=\sqrt{\frac{B}{\rho }}$$
wherein
$B$ is the bulk modulus that represents the elastic property of the medium
$\rho =$ the density of the medium that represents the inertial property of the medium.
$$V_{o_2}=\sqrt{\frac{B}{\rho }}=\sqrt{\frac{1.41\times {10}^5}{1.43}}=314\mathrm{m}/\mathrm{s}$$

Hence, the answer is the option (1).

Example 4: The pulse of a sound wave travels a distance 1 in helium gas in time T at a particular temperature. If at the same temperature, a pulse of the sound wave is propagated in oxygen gas, it will cover the same distance 1 in time -

1) 4.36 T
2) 0.23 T
3) 3 T
4) 0.46 T

Solution:

$\begin{array}{rl}& \frac{{\mathrm{V}}_{{\mathrm{O}}_2}}{{\mathrm{V}}_{\mathrm{HC}}}=\frac{\sqrt{\frac{7/5\mathrm{RT}}{32}}}{\sqrt{\frac{5/3\mathrm{RT}}{4}}}=\sqrt{\frac{7/5\mathrm{RT}}{32}}\times \sqrt{\frac{4}{5/3\mathrm{RT}}}\\ & \frac{{\mathrm{V}}_{{\mathrm{O}}_2}}{{\mathrm{V}}_{\mathrm{HC}}}=0.32\\ & \text{or}{\mathrm{V}}_{02}=0.32{\mathrm{V}}_{\mathrm{Hc}}\\ & \text{Time taken}=\frac{1}{0.32}\times \mathrm{T}=3\mathrm{T}\end{array}$

Hence, the answer is the option (3).

Example 5: In a mixture of gases, the average number of degrees of freedom per molecule is 6. The RMS speed of the molecules of the gas is The velocity of sound in the gas is -

1) $\frac{c}{\sqrt{2}}$
2) $\frac{3\mathrm{C}}{4}$
3) $\frac{2\mathrm{c}}{3}$
4) $\frac{c}{\sqrt{3}}$

Solution:

$\begin{array}{rl}& \text{using,}\gamma =1+\frac{2}{\mathrm{f}}=1+\frac{2}{6}=\frac{4}{3}\\ & \text{As,}c=\sqrt{\frac{3\mathrm{p}}{\rho }}\text{and}\nu =\sqrt{\frac{\mathrm{rp}}{\rho }}\\ & \nu =\mathrm{c}\sqrt{\frac{\gamma }{3}}=\mathrm{c}\sqrt{\frac{4/3}{3}}\\ & =\frac{2\mathrm{c}}{3}\end{array}$

Hence, the answer is the option (3).

Summary
Newton's principles regarding the velocity of sound in a gas assert that the velocity is dependent upon the pressure and the theorem of the density of the gas. Based on the principle, the velocity would increase if conducted at higher pressures and decrease at higher densities. Later, his primitive thinking was modified to include temperature and specific heat capacities of gases, thus coming out with a much more representative equation for their velocity of sound.