Calculate the average velocity of oxygen molecule at `27^@C`.

Explanation

The average velocity `bar(u)` is given by
`bar(u) =sqrt((8RT)/(pi M))`
In the present case,
`T = 27^@C = 27 + 273 = 300 K, M = 32`
and `R= 8.31 xx 10^7 " ergs " K^(-1) mol^(-1)`
`:. " " bar(u)=sqrt((8xx8.31xx10^7 xx 300)/2)=1.845xx10^5 " cm " s^(-1)`
Hence, the average velocity of oxygen molecule at `27^@C " is " 4.46 xx 10^4 " cm " s^(-1)`