## Use the formula v = √γP/ρ to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-15 Waves

### Q.4:- Use the formula v = √γP/ρ to explain why the speed of sound in air (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.

(a) Take the relation:
v = √γP/ρ          ….(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 = √γPV/      ….(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 = √γP/ρ          ….(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 = √γRT/Vρ  = √γRT/M  …..(iii)
where,
mass, M = ρV is a constant
γ and R are also constants
We conclude from equation (iii) 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.

(c) Let vm and vd be the speed of sound in moist air and dry air respectively.

Let ρm and ρd be the densities of the moist air and dry air respectively.

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.