A fresh air is composed of ntirogen `N_(2)(78%)` and oxygen `O_(2)(21%)`. Find the rms speed of `N_(2)andO_(2)` at `20^(@)C`.

Explanation

For nitrogen,
Molar mass m = 0.0280 kg/mol
Temperature `T=20^(@)C=20+273=293K`
`R=8.314Jmol^(-1)k^(-1)`
`V_(rms)=sqrt((3RT)/(m))`
`=sqrt((3xx8.31xx293)/(0.0280))=sqrt(2610xx10^(2))`
`=511ms^(-1)`
`V_(rms)" of "N_(2)=511ms^(-1)`
For `O_(2)`
Molar mass `M=0.0320kgmol^(-1)`
`V_(rms)=sqrt((3xx8.31xx293)/(0.0320))=sqrt(2280xx10^(2))`
`V_(rms)" of "O_(2)=478ms^(-1)`