The speed of sound in dry air at `NTP` is `332m//s`. Assuming air as composed of `4` parts of nitrogen and one parts 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.

Explanation

Oxygen, `rho_(O_(2)) = 16`
Nitrogen, `rho _ (N_(2)) = 14`
`rho_(air) = (rho_(O_(2)) V_(O_(2)) + rho_(N_(2)) V_(N_(2)))/(V_(O_(2)) + V_(N_(2))) = (16 V + 14 xx 4 V)/(V + 4 V) = (72)/(5)`
`v prop sqrt((1)/(rho))`
`(v_(O_(2)))/(v_(air)) = sqrt((rho_(air))/(rho_(O_(2)))) = sqrt((72//5)/(16)) = sqrt((9)/(10))`
`v_(O_(2)) = sqrt(0.9) v_(air) = sqrt(0.9) xx 332`
`= 315 m//s`