Question

If nitrogen gas molecule goes straight up with its rms speed at `0^@ C` from the surface of the earth and there are no collisions with other molecules, then it will rise to an approximate height of.
A. `8 km`
B. `12 km`
C. `12 m`
D. `8 m`

Answer 1

Correct Answer - B
`v_(r m s) = sqrt((3 R T)/(M)) = sqrt(( 3 xx 8.31 xx 273)/(28 xx 10^-3))`
= `493 m//s`
Now, increase in gravitation `(PE)`
= decrease in kinetic energy
`:. (mg h)/(1 + h/R) = (1)/(2) m v_(r m s)^2 or (2 gh)/(1 + (h//R))=v_(r m s)^2`
or `(2 xx 9.81 xx h)/(1 + (h // 6400 xx 10^3)) = (493)^2`
Solving this equation, we get
`h ~~ 12000 m`
or `12 km`.