At what temperatue will the rms speed of oxygen molecule will be sufficient for esca-ping from the earth ?[`v_(e)`=11.2 km/s,m=`2.76xx10^(-26)`Kg and k=`1.38xx10^(-23)`J/K]

Explanation

The mean square velocity of gas molecule given by `v_(av)=sqrt((2KT)/(m))`
For the molecule to just escape from the earth `v_(av)=v_(e)`
or `sqrt((2KT)/(m))=v_(e)`
`T=(mv_(e)^(2))/(2k)=(5.34 xxx 10^(-26) xx (11.0 xx 10^(3))^(2))/(2 xx 1.38 xx 10^(-23))`
`=2.34 xx 10^(5)K`