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

Absolute temperature `T=20^(@)C+273=293K`
Gas constant `R=8.32J mol^(-1)K^(-1)`
For, Nitrogen `(N_(2)`,
Molar mass (M) `=28 g per mol =28xx10^(-3)kg/mol`
`therefore nu_(rms)=sqrt((3RT)/M)=sqrt((3xx8.32xx293)/(28xx10^(-3)))=sqrt((7313.28)/(28xx10^(-3)))`
`(nu_(rms))_(N_2)=511ms^(-1)`
For, Oxygen `(O_(2))`,
Molar mass (M) =32 g per mol `=32xx10^(-3) kg//mol`
`therefore nu_(rms)=sqrt((3RT)/M)=sqrt((3xx8.32xx293)/(32xx10^(-3)))=sqrt((7313.28)/(32xx10^(-3)))`
`(nu_(rms))_(O_2)=478ms^(-1)`