### 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.

**Answer:-**(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*/*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* = √*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*/*V**ρ* = √*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 *v*_{m} and *v*_{d} be the speed of sound in moist air and dry air respectively.

Let *ρ _{m}* and

*ρ*be the densities of the moist air and dry air respectively.

_{d}*ρ*

_{d < }

*ρ*

_{m}*v*

_{m > }

*v*

_{d}

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.