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

i dont have a visual in front of me, but i can you reason to try to get the right one. the graph should look like this, 0 to 6 there is a spike, 12 to 14 there should be a very low point. slight increase in the graph at 12 to 14, and flat 14 to 18

explanation: