Question

Example 1 The speed of sound in dry air at NTP is 332 m/s.
Assuming air as composed of 4 parts of nitrogen and one
part of oxygen in volume, calculate the velocity of sound in
oxygen under similar conditions. Given density of oxygen
and nitrogen at NTP are in the ratio 16: 14 respectively.​

Answer 1

Answer:

let the densities of air, nitrogen and oxygen be ρ, ρn and ρo respectively.

If 'V' is the volume of oxygen in air, then in the first case as N to O ratio is 4:1 then volume of nitrogen will be '4V'.

.

so, as mass = density x volume

mass of oxygen = ρoV

mass of nitrogen = ρn4V

mass of air = mass of nitrogen + mass of oxygen = ρn4V + ρoV = ρ5V...........................(1)

now,

when the oxygen to nitrogen densities are of ratio 16:14, we have

ρo/ρn = 16/14

or

ρn = (14/16)ρo = (7/8)ρo

so, equation (1) becomes

(7/8)ρo.4V + ρoV = ρ5V

or density of air

ρ = (7/2)ρo. +ρo = (9/10)ρo........................................(2)

now,

from Laplace's relation of speed of sound

in oxygen

vo2 = (γP/ρo)1/2 .......................................(3)

in air

vair = (γP/ρ)1/2............................................(4)

now, by dividing (3) by (4), we have

(vo2 / vair) = (γP/ρo)1/2 / (γP/ρ)1/2 = (ρ / ρo)1/2

or

(vo2 / vair) = (ρ / ρo)1/2 = [(9/10)ρo / ρo]1/2

so,

(vo2 / vair) = (9/10)1/2 = 0.948

thus,

vo2 = 0.948 x vair = 0.948 x 332 m/s

so, velocity of sound in oxygen will be

vo2 = 314.96 m/s

Explanation:

i hope it helps you......

good morning...........