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Question

Use the formula v=γPρ to explain why the speed of sound in air
(a) is independent of pressure,
(b) increases with temperature,
(c) Increases with humidity .

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Solution

(a) Take the relation:
v=γPρ ....(i)
where,
Density, ρ = Mass/Volume = M/V
M = Molecular weight of the gas
V = Volume of the gas
Hence, equation (i) reduces to:
v=γPVm ....(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.
(b) Take the relation:
v=γPρ ....(i)
For one mole of any ideal gas, the equation can be written as:
PV = RT
P = RT/V ....(ii)
Substituting equation (ii) in equation (i), we get:
v=γRTPρ=γRTm .....(iii)
where,
mass, M = ρV is a constant
γ and R are also constants
We conclude from equation (iii) that VT
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.
(c) Let Vm and Vd be the speed of sound in moist air and dry air respectively.
Let ρm and ρdbe the densities of the moist air and dry air respectively.
Take the relation :
ν=γρρ
Hence , the speed of sound in most air is
νm=γρρm ... (i)
And the speed of sound in dry air is:
νd=γρρd ... (ii)
On dividing equations (i) and (ii), we get:
νmνd=γρρm×ρdγρ=ρdρm
However, the presence of water vapour reduces the density of air, i.e.,
ρd<ρm
Vm>Vd
Hence, the speed of sound in mois air is greater than it is in dry air. Thus, in gaseous medium, the speed of sound increases with humidity.

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