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# Solution for Use the Formula V = Sqrt((Gamma P)/Rho) to Explain Why the Speed of Sound in Air is Independent of Pressure, Increases with Temperature, Increases with Humidity. - CBSE (Science) Class 11 - Physics

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ConceptThe Speed of a Travelling Wave

#### Question

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.

#### Solution 1

a) Take the relation:

v = sqrt(gamma P)/rho   ....(i)

where

Density, rho = "Mass"/"Volume" = M/V

M = Molecular weight of the gas

V = Volume of the gas

Hence, equation (i) reduces to

v = sqrt((gamma PV)/M) .....(ii)

Now from the ideal gas equation for n = 1:

PV RT

For constant TPV = 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 = ρis a constant

γ and R are also constants

We conclude from equation (iv) that v prop sqrtT.

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((gamma P)/rho)

Hence, the speed of sound in moist air is:

v_m = sqrt((gamma P)/rho_m) ....(i)

And the speed of sound in dry air is:

v_d = sqrt((gamma P)/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 < rho_m

:. v_m > 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.

#### Solution 2

We are given that v = sqrt((gamma p)/rho)

We know PV = nRT            (For n moles of ideal gas)

=> PV = m/M RT

where m is the total mass and M is the molecular mass of the gas

:. P = m/M. "RT"/M= (rhoRT)/M => P/rho =  "RT"/M

a) For a gas at constant temperature P/rho =costant

:. As P increase, rho also increase and vice versa. This implies that v = sqrt((gamma P)/rho) =  constant, i.e velocity is independent of pressure of the gas.

b) Since P/rho = "RT"/M, therefore,  v = sqrt((gamma P)/rho) =  sqrt((gamma RT)/M)

c) Increase in humidity decrease the effective density of air. Therefore the velocity (v prop 1/sqrtrho) increase

Is there an error in this question or solution?

#### APPEARS IN

Solution Use the Formula V = Sqrt((Gamma P)/Rho) to Explain Why the Speed of Sound in Air is Independent of Pressure, Increases with Temperature, Increases with Humidity. Concept: The Speed of a Travelling Wave.
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