Question

Use the formula $v=\sqrt{\frac{\gamma P}{\rho }}$ to explain why the speed of sound in air is independent of pressure, increases with temperature, increases with humidity.

Answer 1

Dependence of speed of sound in the air on the pressure:

1. Take the relation:

The speed of sound in a gas is given by;

$v=\sqrt{\frac{\text{γ}P}{\text{ρ}}}$ .........(1)

Here, γ is the ratio of specific heat at constant pressure to the specific heat at constant volume.

Where,

Density, $\rho =\frac{Mass}{Volume}=\frac{M}{V}$

M =Molecular weight of the gas

V=Volume of the gas

Hence, equation (i) reduces to:

$v=\sqrt{\frac{\gamma PV}{M}}.....\left(ii\right)$

Now from the ideal gas equation for n = 1:

PV = RT

For constant temperature(T),

PV = Constant

Since both M and $\gamma$ are constants,

The speed of sound in a gas is constant.

That is; v = Constant

Hence,

At a constant temperature,

The speed of sound in a gaseous medium is independent of the change in the pressure of the gas.

Dependence of speed of sound in the air on temperature:

1. TakIng the relation:
   $v=\sqrt{\frac{\gamma RT}{V\rho }}=\sqrt{\frac{\gamma RT}{M}}.....\left(iv\right)$
   Where,
   Mass, $M=\rho V$ is a constant
   $\gamma$ and R are also constants
   We conclude from equation (iv) that;

$v\propto \sqrt{T}$
Hence,

The speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium, i.e., the speed of the sound increases with an increase in the temperature of the gaseous medium and vice versa.

Dependence of the speed of sound in the air on humidity:

1. Taking the relation:
   $v=\sqrt{\frac{\gamma P}{\rho }}$

Let $v_m$ and $v_d$ be the speeds of sound in moist air and dry air respectively.

And also ${\rho }_m$ and ${\rho }_d$ be the densities of moist air and dry air respectively

Hence, the speed of sound in moist air is:

$v_m=\sqrt{\frac{\gamma P}{{\rho }_m}}....\left(i\right)$

And the speed of sound in dry air is:

$v_d=\sqrt{\frac{\gamma P}{{\rho }_d}}....\left(ii\right)$

On dividing equations (i) and (ii), we get:

$\frac{v_m}{v_d}=\sqrt{\frac{\gamma P}{{\rho }_m}\times \frac{{\rho }_d}{\gamma P}}=\sqrt{\frac{{\rho }_d}{{\rho }_m}}$

However, the presence of water vapour reduces the density of air, i.e., ${\rho }_d<{\rho }_m$

$\therefore v_m>v_d$

Hence,

The speed of sound in moist air is greater than it is in dry air.

Thus, in a gaseous medium, the speed of sound increases with humidity.