Some of the most important Physics Topics include energy, motion, and force.

## What is the Laplace Correction Adiabatic Process?

Scientist Laplace first pointed out that the propagation of sound waves through a gaseous medium takes place under adiabatic condition and not under isothermal condition as assumed by Newton. According to Laplace’s assumption, compressions and rarefactions occur so quickly that the temperature of a gas cannot remain constant. On the other hand, heat exchange does not take place among different portion of the gas during that small time interval. As a result, the temperature of gaseous layers increases and decreases during propagation of sound—the process is essentially adiabatic.

Now, the relation between pressure (p) and volume (V) of a gas under adiabatic condition is given by

pV^{γ} = constant

Here,

γ \(=\frac{\text { molar specific heat of the gas at constant pressure }\left(C_p\right)}{\text { molar specific heat of the gas at constant volume }\left(C_v\right)}\)

Let the initial pressure and volume of a fixed amount of gas be p and V respectively. During propagation of sound pressure increases to p + p_{1} whereas volume decreases to V – u [Fig.]. Here p_{1} is very small with respect to p and u is very small with respect to V.

So we write,

Proof by calculus: The relation between pressure (p) and volume (V) of a gas under adiabatic condition is given by

pV^{γ} = constant

On differentiation we have,

V^{γ}dp + γp^{γ}V^{γ – 1}dV = 0

or, dp + \(\frac{\gamma p d V}{V}\) = 0

or, γp = \(\frac{-d p}{\frac{d V}{V}}\) = bulk modulus of the gas (k)

So, from equation (1) of section 3.8 we get, velocity of sound in a gaseous medium,

c = \(\sqrt{\frac{\gamma p}{\rho}}\)

Using Laplace’s correction, velocity of sound in air at STP,

c = \(\sqrt{\frac{1.4 \times 76 \times 13.6 \times 980}{0.001293}}\) [for air, γ = 1.4]

= 33117 cm ᐧ s^{-1} = 331.2 m ᐧ s^{-1}

This value of c is very close to the experimental value of velocity of sound in air. So we can come to the conclusion that Laplace’s equation for velocity of sound in a gaseous medium is correct.

Effect of Pressure, Temperature, Humidity and Density of a Gas on the Velocity of Sound.

According to Laplace’s equation, if c is the velocity of sound in a gas of density ρ and pressure p,

c = \(\sqrt{\frac{\gamma p}{\rho}}\) ….. (1)

Here γ is the ratio of the two molar specific heats (at constant pressure and constant volume) of the gas. Now if 1 mol of gas is taken,

mass = M, density ρ = \(\frac{M}{V}\)

and pV = RT (T = absolute temperature of the gas, R = universal gas constant)

So from equation (1) we get,

c = \(\sqrt{\frac{\gamma p}{\frac{M}{V}}}\) = \(\sqrt{\frac{\gamma p V}{M}}\) = \(\sqrt{\frac{\gamma R T}{M}}\) …. (2)

(Here, M = mass of 1 mol of gas. i.e., M is the molecular mass of the gas)

Effect of pressure: γ, R and M are constants for a particular gas. So from the equation (2) it is understood that, if temperature of the gas remains constant, the velocity of sound does not change with change in pressure, i.e., pressure of the gas has no effect on the velocity of sound.

It can also be explained in a different way. At constant temperature, if the pressure of a gas is changed, the density of the gas also changes due to a change of volume in such a way that the ratio \(\frac{p}{\rho}\) remains constant. So from equation (1) it can be said that the velocity of sound (c) remains constant with the change in pressure.

Effect of temperature: For a particular gas γ, R and M — these three quantities are constants. So from equation (2) we get, c ∝ \(\sqrt{T}\), i.e., the velocity of sound in the gas is directly proportional to the square root of its absolute temperature.

If c_{1} and c_{2} are the velocities of sound in the gas at absolute temperatures T_{1} and T_{2}, then

\(\frac{c_1}{c_2}\) = \(\sqrt{\frac{T_1}{T_2}}\)

Let, T_{1} = 0°C = 273K, T_{2} = t°C = (273 + t)K

If the velocities of sound at these two temperatures are c_{0} and c, then

The change of the velocity of sound due to any change in temperature of the gas can be calculated with the help of equation (3).

Velocity of sound in air at 0°C, c_{0} = 332 m ᐧ s^{-1}

So, c = 332(1 + 0.00183 t) = (332 + 0.61 t) m ᐧ s^{-1}

Therefore, the velocity of sound in air Increases by 0.61 m ᐧ s^{-1} or 61 cm ᐧ s^{-1} for 1°C rise in temperature (61 cm ᐧ s^{-1} ≈ 2 ft ᐧ s^{-1}).

Effect of humidity: Density of water vapour is less (approximately 0.622 times) than the density of dry air at the same temperature and pressure. So if water vapour is mixed in air, density of air decreases. Hence, the velocity of sound in air increases, i.e. the velocity of sound in moist air is greater than that in dry air.

Let temperature of air = t°C; atmospheric pressure p;

velocity of sound in dry air at t°C = c;

velocity of sound in moist air at t°C = c_{m};

vapour pressure at that temperature = f

It can be shown from theoretical analysis that

c = c_{m}(1 – 0.378\(\frac{f}{p}\))^{1/2} ≈ c_{m}(1 – 0.189\(\frac{f}{p}\)) …(4)

Any experiment for determination of the velocity of sound is made in normal atmosphere which is more or less humid. So, the velocity of sound determined by experiment is the velocity of sound in moist air i.e., c_{m}. Thus the velocity of sound in dry air, i.e., c can be obtained by using equation (4).

Effect of density: Let two different gases be at the same temperature and pressure. Under this condition if ρ_{1} and ρ_{2} are the densities of the two gases and c_{1} and c_{2} are the velocities of sound in the two gases,

c_{1} = \(\sqrt{\frac{\gamma p}{\rho_1}}\) ; c_{2} = \(\sqrt{\frac{\gamma p}{\rho_2}}\)

[It has been assumed that both the gases have the same γ]

So, \(\frac{c_1}{c_2}\) = \(\sqrt{\frac{\rho_2}{\rho_1}}\) i.e., c ∝ \(\frac{1}{\sqrt{\rho}}\)

Therefore, the velocity of sound in a gas is inversely proportional to the square root of its density.

For example, oxygen is 16 times heavier than hydrogen. So velocity of sound in oxygen is \(\frac{1}{\sqrt{16}}\) i.e., \(\frac{1}{4}\) times the velocity of sound in hydrogen.

In the above discussion γ of both the gases has been taken to be equal. If each molecule of the two gases contain the same number of atoms, the values of γ for both are the same. Again if different number of atoms are present in the molecules corresponding values of γ are to be put in the formula.

### Numerical Examples

**Example 1.**

The velocity of sound in a gas at 51°C is 340 m ᐧ s^{-1}. What will be the velocity of sound if pressure is doubled and temperature becomes 127°C?

**Solution:**

Pressure of the gas has no effect on the velocity of sound. In this case velocity of sound will change only due to the change in temperature.

Initial temperature,

T_{1} = 51°C = (273 + 51)K = 324 K

Final temperature,

T_{2} = 127°C = (273 + 127)K = 400 K

We know that the velocity of sound in a gas is directly proportional to the square root of its temperature, i.e., c ∝ \(\sqrt{T}\).

∴ \(\frac{c_1}{c_2}\) = \(\sqrt{\frac{T_1}{T_2}}\)

or, c_{2} = c_{1}\(\sqrt{\frac{T_2}{T_1}}\) = 340 × \(\sqrt{\frac{400}{324}}\)

= 340 × \(\frac{20}{18}\) = 378 ᐧ s^{-1} (approx.).

**Example 2.**

The velocity of sound in a gas at 50°C is 340 m ᐧ s^{-1}. What will be the velocity of sound if pressure is doubled and temperature becomes 125°C? [HS ‘02]

**Solution:**

Velocity of sound does not depend on pressure of the gas.

Initial temperature,

T_{1} = 50°C = (50 + 273) K = 323 K

Final temperature,

T_{2} = 125°C = (125 + 273) K = 398 K

Now, velocity of sound, c ∝ \(\sqrt{T}\),

i.e., \(\frac{c_1}{c_2}\) = \(\sqrt{\frac{T_1}{T_2}}\)

or, c_{2} = c_{1} × \(\sqrt{\frac{T_2}{T_1}}\) = 340 × \(\sqrt{\frac{398}{323}}\) = 377.4 m ᐧ s^{-1}.

**Example 3.**

Determine the velocity of sound In hydrogen at STP. At STP, density of the gas = 0.09 g ᐧ L^{-1}; γ = 1.4.

**Solution:**

Density, ρ = 0.09 g ᐧ L^{-1} = \(\frac{0.09}{1000}\) g ᐧ cm^{-3}

= 0.09 × 10^{-3} g ᐧ cm^{-3}

Standard pressure, p = 76 × 13.6 × 980 dyn ᐧ cm^{-2}

Therefore, velocity of sound,

c = \(\sqrt{\frac{\gamma p}{\rho}}\) = \(\sqrt{\frac{1.4 \times 76 \times 13.6 \times 980}{0.09 \times 10^{-3}}}\)

= 125525 cm ᐧ s^{-1} = 1255.25 m ᐧ s^{-1}.

**Example 4.**

The ratio of γ of oxygen and methane is 21 : 20. The ratio of their densities at the same pressure is 2 : 1. If the velocity of sound in oxygen is 316 m ᐧ s^{-1}, what is the velocity of sound in methane?

**Solution:**

If the velocities of sound in oxygen and methane are c_{1} and c_{2} respectively, then

**Example 5.**

At what temperature the velocity of sound in nitrogen will be equal to its velocity in oxygen at 27°C?

**Solution:**

Since nitrogen and oxygen are both diatomic gases, the values of γ for both of them are equal.

Molecular mass of nitrogen, M_{1} = 28

Molecular mass of oxygen, M_{2} = 32

Velocity of sound, c = \(\sqrt{\frac{\gamma R T}{M}}\)

So, velocity of sound in nitrogen, c_{1} = \(\sqrt{\frac{\gamma R T_1}{M_1}}\)

velocity of sound in oxygen, c_{2} = \(\sqrt{\frac{\gamma R T_2}{M_2}}\)

According to the question, c_{1} = c_{2}

and T_{2} = 27°C = (273 + 27) K = 300 K

So, \(\sqrt{\frac{\gamma R T_1}{28}}\) = \(\sqrt{\frac{\gamma R \times 300}{32}}\)

or, T_{1} = \(\frac{300 \times 28}{32}\) = 262.5 K = (262.5 – 273)°C

= -10.5°C.

**Example 6.**

The velocity of sound in hydrogen at 0°C is 1200 m ᐧ s^{-1}. If a certain amount of oxygen is mixed with hydrogen, the velocity of sound decreases and becomes 500 m ᐧ s^{-1}. What is the ratio of the volumes of hydrogen and oxygen in the mixture? It is given that the density of oxygen is 16 times that of hydrogen.

**Solution:**

We know, velocity of sound, c = \(\sqrt{\frac{\gamma p}{\rho}}\)

In case of hydrogen, 1200 = \(\sqrt{\frac{\gamma p}{\rho_1}}\)

In case of the given mixture, 500 = \(\sqrt{\frac{\gamma p}{\rho_2}}\)

Hence, the ratio of the volumes of hydrogen and oxygen = 1 : (119/256) = 256 : 119.

**Example 7.**

What is the velocity of sound in air saturated with water vapour at 27°C? It ¡s given that velocity of sound in dry air at 0°C is 332 m ᐧ s^{-1} and aqueous tension at 27°C is 112 mmHg.

**Solution:**

Velocity of sound in dry air at 27°C,

c = 332(1 + 0.00183 × 27) = 332 × 1.05 m ᐧ s^{-1}

Aqueous tension at that temperature,

f = 11.2 mm Hg = 1.12 cm Hg

If c_{m} is the velocity of sound in air saturated with water vapour at 27°C,

c = c_{m}(1 – 0.189\(\frac{f}{p}\)) = c_{m}(1 – 0.189 × \(\frac{1.12}{76}\))

= c_{m} × 0.997

or, c_{m} = \(\frac{c}{0.997}\) = \(\frac{332 \times 1.05}{0.997}\) = 350 m ᐧ s^{-1} (approx.).

**Example 8.**

The velocity of sound in helium gas at -173°C is 582 m ᐧ s^{-1}. If the molecular mass of helium is 4, find the value of γ for the gas. [R = 8.31 × 10^{7} erg ᐧ mol^{-1} ᐧ °C^{-1}] [WBJEE 2000]

**Solution:**

T = -173°C = (-173 + 273)K = 100K

Molecular mass of helium, M = 4

Velocity of sound, c = 582 m ᐧ s^{-1} = 58200 cm ᐧ s^{-1}

Now, c = \(\sqrt{\frac{\gamma R T}{M}}\) or, γ = \(\frac{c^2 M}{R T}\) = \(\frac{(58200)^2 \times 4}{8.31 \times 10^7 \times 100}\) = 1.63.

**Example 9.**

At STP, the velocity of sound in oxygen is 317 m ᐧ s^{-1}. What is the velocity of sound in hydrogen at 30°C temperature and 374 mmHg pressure? [HS 2000]

**Solution:**

At 0°C or 273 K velocity of sound in oxygen,

c_{1} = 317 m ᐧ s^{-1}

Let c_{2} be the velocity of sound in oxygen at 30°C or 303 K.

Now, \(\frac{c_1}{c_2}\) = \(\sqrt{\frac{T_1}{T_2}}\) or, c_{2} = c_{1} × \(\sqrt{\frac{T_2}{T_1}}\)

Again at the same temperature, velocity of sound in two gases is inversely proportional to the square root of their densities.

So, if c_{2}’ is the velocity of sound in hydrogen at 30°C,

\(\frac{c_2^{\prime}}{c_2}\) = \(\sqrt{\frac{\rho}{\rho^{\prime}}}\) or, c_{2}‘ = c_{2} × \(\sqrt{\frac{\rho}{\rho^{\prime}}}\)

∴ c_{2}‘ = 317 × \(\sqrt{\frac{303}{273}}\) × \(\sqrt{\frac{16}{1}}\) = 1336 m ᐧ s^{-1} (approx.)

Velocity of sound does not depend on pressure of the gas. So, this velocity will remain the same at 374 mmHg pressure.

**Example 10.**

The masses of 1 L of hydrogen and 1 L of air are 0.0896 g and 1.293 g respectively at the same temperature and pressure. If the velocity of sound in air at that temperature is 330 m ᐧ s^{-1}, what will be the corresponding velocity in hydrogen?

**Solution:**

We know, velocity of sound in a gas, c = \(\sqrt{\frac{\gamma p}{\rho}}\).

γ = 1.41 for both hydrogen and air. If the velocities of sound in hydrogen and air at the same temperature and pressure are c_{H} and c_{air} respectively,

\(\frac{c_{\mathrm{H}}}{c_{\text {air }}}\) = \(\sqrt{\frac{\rho_{\text {air }}}{\rho_{\mathrm{H}}}}\)

or, c_{H} = c_{air} × \(\sqrt{\frac{\rho_{\text {air }}}{\rho_{\mathrm{H}}}}\) = 330 × \(\sqrt{\frac{1.293}{0.0896}}\) = 1235.6 m ᐧ s^{-1}.