

InterviewSolution
1. |
State newton’s formula for velocity of sound in air. Point out the error and hence, discussion Laplace’s correction. |
Answer» 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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