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286 PHYSICS −2 −2 Liquids and solids generally have higher speed ML T  2 −2 −3 =  T  (14.17) of sound than gases. [Note for solids, the speed ML  being referred to is the speed of longitudinal ρ waves in the solid]. This happens because they Thus, if B and are considered to be the only are much more difficult to compress than gases relevant physical quantities, andsohavemuchhighervaluesofbulkmodulus. B Now, see Eq. (14.19). Solids and liquids have v = C (14.18) higher mass densities ( ) than gases. But the ρ where, as before,C is the undetermined constant corresponding increase in both the modulus (B) from dimensional analysis. The exact derivation of solids and liquids is much higher. This is the reason why the sound waves travel faster in shows that C=1. Thus, the general formula for longitudinal waves in a medium is: solids and liquids. We can estimate the speed of sound in a gas v = B (14.19) in the ideal gas approximation. For an ideal gas, ρ the pressure P, volume V and temperature T are For a linear medium, like a solid bar, the related by (see Chapter 10). lateral expansion of the bar is negligible and we PV = Nk T (14.21) may consider it to be only under longitudinal B where N is the number of molecules in volume strain. In that case, the relevant modulus of elasticity is Young’s modulus, which has the V, k Bs the Boltzmann constant and T the same dimension as the Bulk modulus. temperature of the gas (in Kelvin). Therefore, for an isothermal change it follows from Eq.(14.21) Dimensional analysis for this case is the same that as before and yields a relation like Eq. (14.18), V∆P + P∆V = 0 with an undetermined C, which the exact derivation shows to be unity. Thus, the speed of or − ∆P = P longitudinal waves in a solid bar is given by ∆V/V Hence, substituting in Eq. (14.16), we have v = Y (14.20) ρ B = P Therefore, from Eq. (14.19) the speed of a where Y is the Young’s modulus of the material longitudinal wave in an ideal gas is given by, of the bar. Table 14.1 gives the speed of sound in some media. v = P (14.22) Table 14.1 Speed of Sound in some Media ρ This relation was first given by Newton and is known as Newton’s formula. Example 14.4 Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 ×10 kg. Answer We know that 1 mole of any gas occupies 22.4 litres at STP. Therefore, density of air at STP is: ρo= (mass of one mole of air)/ (volume of one mole of air at STP) 29.0 ×0 −3 kg = − 3 22.4 ×10 m = 1.29 kg m –3 Reprint 2026-27