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

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