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

`{:("Molar mass"),("of nitrogen"):}}=0.028"kg//mol"`
Temperature T = 20 + 273 = 293 K.
`{:("Universal"),("gas constant"):} }` R = 8.314J/mol/k
R.M.S speed of nitrogen `N_2`
`V_(rms)=sqrt((3RT)/m)`
`=sqrt((3xx8.31xx293)/(0.0280))`
`sqrt(2610xx10^2)`
= 511 m//s
`{:("Molar mass"),("of oxygen"):} } =` 0.032 kg/mol
Temperature = 20+273 = 293K
`V_(rms)=sqrt((3RT)/m)`
`=sqrt((3xx8.31xx293)/(0.032))=sqrt(2280xx10^2)`
`=478m//s`