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13 March, 05:25

The velocity of sound in air saturated with water vapour at 30°C

is 340 m/s. If the atmospheric pressure is 65cm of mercury and

saturated vapour pressure of water at 30°C is 31.7 mm of

mercury. Calculate the velocity of sound in dry air at 0°C.

Ans: 319.74m/s

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Answers (1)
  1. 13 March, 05:51
    0
    The velocity of sound depends on the density of the medium. So we need to find the density of air at each set of conditions. The density of air is:

    ρ = (Pd / (Rd T)) + (Pv / (Rv T))

    where Pd and Pv are the partial pressures of dry air and water vapor,

    Rd and Rv are the specific gas constants of dry air and water vapor,

    and T is the absolute temperature.

    At the first condition:

    Pv = 31.7 mmHg = 4226.3 Pa

    Pd = 650 mmHg - 31.7 mmHg = 618.3 mmHg = 82433 Pa

    Rv = 461.52 J/kg/K

    Rd = 287.00 J/kg/K

    T = 30°C = 303.15°C

    ρ = (82433 / 287.00 / 303.15) + (4226.3 / 461.52 / 303.15)

    ρ = 0.94746 + 0.03021

    ρ = 0.97767 kg/m³

    At the second condition:

    Pv = 0 Pa

    Pd = 650 mmHg = 86660 Pa

    Rv = 461.52 J/kg/K

    Rd = 287.00 J/kg/K

    T = 0°C = 273.15°C

    ρ = (86660 / 287.00 / 273.15) + (0 / 461.52 / 273.15)

    ρ = 1.1054 + 0

    ρ = 1.1054 kg/m³

    The square of the velocity of sound is proportional to the ratio between pressure and density:

    v² = k P / ρ

    Since the atmospheric pressure is constant, we can say it's proportional to just the density:

    v² = k / ρ

    Using the first condition to find the coefficient:

    (340) ² = k / 0.97767

    k = 113018.652

    Now finding the velocity of sound at the second condition:

    v² = 113018.652 / 1.1054

    v = 319.75
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