P ∆ V = or P V/V P ∆ ∆ Hence, substituting in Eq. ( . ), we have B = P Therefore, from Eq. ( .
) the speed of a longitudinal wave in an ideal gas is given by, v = P ( . ) This relation was first given by Newton and is known as Newton’s formula. u Example . Estimate the speed of sound in air at standard temperature and pressure.
The mass of mole of air is . × – kg. Answer We know that mole of any gas occupies . litres at STP.
Therefore, density of air at STP is: ρ o = (mass of one mole of air)/ (volume of one mole of air at STP) . kg . m × × = . kg m – According to Newton’s formula for the speed of sound in a medium, we get for the speed of sound in air at STP, = m s – ( .
) ⊳ The result shown in Eq.( . ) is about % smaller as compared to the experimental value of m s – as given in Table . . Where did we go wrong ?
If we examine the basic assumption made by Newton that the pressure variations in a medium during propagation of sound are isothermal, we find that this is not correct. It was pointed out by Laplace that the pressure variations in the propagation of sound waves are so fast that there is little time for the heat flow to maintain constant temperature. These variations, therefore, are adiabatic and not isothermal. For adiabatic processes the ideal gas satisfies the relation (see Section .
), PV γ = constant i.e. ∆ ( PV γ ) = or P γ V γ – ∆ V + V γ ∆ P = where γ is the ratio of two specific heats, C p /C v . Thus, for an ideal gas the adiabatic bulk modulus is given by, B ad = V/V P ∆ ∆ = γ P The speed of sound is, therefore,