Question
Physics Question on The speed of a travelling wave
Use the formula v=ργp to explain why the speed of sound in air
(a) is independent of pressure,
(b) increases with temperature,
(c) increases with humidity.
Take the relation:
v=ργp ........(i)
Where,
Density, ρ =VolumeMass=VM
M= Molecular weight of that gas
v= Volume of the gas
Hence, equation (i) reduces to:
v=MγPv ........(ii)
Now from the ideal gas equation for n = 1:
PV = RT
For constant T, PV = Constant
Since both M and γ are constants, 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.
Take the relation:
v=ργP ........(i)
For one mole of an ideal gas, the gas equation can be written as:
PV = RT
p=VRT ........(ii)
Substituting equation (ii) in equation (i), we get:
v=VργRT=MγRT ........(iv)
Where,
Mass, M = ρV is a constant
γ and R are also constants
We conclude from equation (iv) that . v∝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.
Let vmandvd be the speeds of sound in moist air and dry air respectively.
Let ρmandρd be the densities of moist air and dry air respectively
Take the relation:
v=ργP
Hence, the speed of sound in moist air is:
vm=ρmγP ........(i)
And the speed of sound in dry air is
vd=ρdγP ........(ii)
On dividing equations (i) and (ii), we get:
vdvd=ρmγP∗yρPd=pmpd
However, the presence of water vapour reduces the density of air, i.e.,
ρd<pm
∴ vm>vd
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.