Use the formula `v=sqrt((gamma P)/(rho))` to explain why the speed of sound in air
(a) is independent of pressure, (b) increases with temperature, (c) increases with humidity.

Explanation

(a) Take the relation
`v=sqrt((gammap)/(rho))" "…(i)`
Where,
Density, `rho=("Mass")/("Volime")=(M)/(V)`
M=Molecular weight of the gas
V= Volume of the gas
Hence, equation (i) reduces to :
`v=sqrt((gammaPV)/(M))" "...(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=sqrt((gammap)/(rho))" "...(i)`
For one mole of an ideal gas, the gas equation can be written as:
PV = RT
`p=(RT)/(V)" "...(ii)`
Substituting equation (ii) in equation (i), we get
`v=sqrt((gammaRT)/(Vrho))=sqrt((gammaRT)/(M))" "...(iv)`
Where,
Mass , M=`rhoV` is a constant
Y and R are also constants
We conclude from equation (iv) that `v prop 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.
(c) Let `v_(m)` and `v_(d)` be the speeds of sound in moist air and dry air respectively
Let `rho_(m)` and `rho_(d)` be the densities of moist air and dry air respectively
Take the relation
`v=sqrt((gammaP)/(rho))" ...(i)`
And the speed of sound in dry air is:
`v_(d)=sqrt((gammaP)/(rho_(d)))" "...(ii)`
On dividing equations (i) and (ii), we get:
`(v_(m))/(v_(d))=sqrt((gammaP)/(rho_(m))xx(rho_(d))/(gammaP))=sqrt((rho_(d))/(rho_(m)))`
However, the presence of water vapour reduces the density of air, i.e.,
`rho_(d) lt rho_(m)`
`therefore v_(m) gt 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.