Question 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

Hense eqation one 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 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  =  √ 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 speeds of sound in moist air and dry air respectively.

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

Take the relation:

v = γP/ρ    

Hence, the speed of sound in moist air is:

vm = γP/ρm    .....(i)

And the speed of sound in dry air is:

vd = γP/ρd    .....(ii)

On dividing equations (i) and (ii), we get:

However, the presence of water vapour reduces the density of air, i.e.,

ρd < ρm

therefore, 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.

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