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State newton’s formula for velocity of sound in air. Point out the error and hence, discussion Laplace’s correction.

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Newton’s formula for velocity of sound in air:

v = \(\sqrt{\frac{E}{ρ}}\)(1)

Where v = velocity of sound in the medium.

E = coefficient of elasticity of medium.

ρ = density of medium.

Newton’s formula for velocity of sound in gas: v = \(\sqrt{\frac{k_i}{ρ}}\) 

Since, a gas has only one type of elasticity, i.e., bulk modulus (K),

Sound travel through a gas in the form of compression and rarefactions.

Newton assumed that changes in pressure and volume of a gas, when sound waves are propagated through it, are isothermal. Using coefficient of isothermal elasticity, i.e., Ki in (eqn. 1)

v = \(\sqrt{\frac{K_i}{ρ}}\)(2)

Error in Newton’s formula:

Let us consider the velocity of sound in air at N.T.P.

v = \(\sqrt{\frac{P}{ρ}}\)(3)

As, P = hdg

h = 0.76 m of Hg column

d = 13.6 × 103 kgm-3 

∴ P = 0.76 × 13.6 × 10-3 × 9.8 Nm-2

Density of air, P = 1.293 kg/m3 

Here from equ (3) = \(\sqrt{\frac{0.76\times13.6\times 10^3\times9.8}{1.293}}\) = 280 ms-1

The experimental value of the velocity of sound in air at N.T.P. is 332 ms-1 

Difference between the experimental and theoretical value of velocity or sound in air

= (332 – 280) ms-1

= 52 ms-1

Percentage Error = \(\frac{52}{332}\) × 100

= 15.7 %

Or ≈ 16% 

Laplace’s correction: 

According to Laplace, the changes in pressure and volume of a gas, when sound waves are propagated through it, are not isothermal, but adiabatic. This is because.

(i) Velocity of sound in a gas is quite large.

(ii) A gas is a bad conductor of heat.

Using the coefficient of adiabatic elasticity, i.e., Ki instead of Ka :

v = \(\sqrt{\frac{K_i}{ρ}}=\sqrt{\frac{K_a}{ρ}}\) 

Calculation of ‘Ka

Let P be the initial pressure and V be the initial volume of the certain mass of the gas. Under adiabatic condition PVr  = constant …(1)

Where γ = \(\frac{c_p}{c_v}\) = ratio of two principal specific heats of the gas.

Differentiating both sides of eqn. 1. P(γ\(V^{\gamma-1}\)dV) + V′ (dP) = 0

or γ\(PV^{\gamma-1}\)dV = −Vγ(dP)

or γP = − \(\frac{V^\gamma}{V^{\gamma-1}}(\frac{dP}{dV})\) 

= \(\frac{dP}{\frac{dV}{V}}\)= Ka 

(By definition) 

∴ Ka = γP 

Corrected formula: Substituting this value of Ka in

v = \(\sqrt{\frac{K_a}{ρ}}=\sqrt{\frac{\gamma^P}{ρ}}\)

The value of γ depends on nature of the gas.

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