The molecular weight of oxygen gas (o2) is 32 while that of hydrogen gas (h2) is 2. prove that under same physical conditions, the velocity of sound in hydrogen is four times that in oxygen

use the formula: v^2=(3kT)/m

Where:

v is the velocity of a molecule

k is the Boltzmann constant (1.38064852e-23 J/K)

T is the temperature of the molecule in the air

m is the mass of the molecule

For an H2 molecule at 20.0°C (293 K):

v^2 = 3 × 1.38e-23 J/K × 293 K / (2.00 u × 1.66e-27 kg/u)

v^2 = 3.65e+6 m^2/s^2

v = 1.91e+3 m/s

For an O2 molecule at same temp.:

v^2 = 3 × 1.38e-23 J/K × 293 K / (32.00 u × 1.66e-27 kg/u)

v^2 = 2.28e+5 m^2/s^2

v = 478 m/s

Therefore, the ratio of H2:O2 velocities is:

1.91e+3 / 478 = 4.00

answer: strength i guess

explanation:

k2s for potassium sulphide