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A nitrogen molecules at teh surface of e...

A nitrogen molecules at teh surface of earth happens to have the rms speed for that gas at `0^(@)`. If it were to go straight up without colliding with other molecules, how high would it rise? Mass of nitrogen molecules, `m = 4.65 xx 10^(26)lg, k=1.38 xx 10^(-23)J "molecule"^(-1) K^(-1)`.

Text Solution

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Here, `k=1.38 xx 10^(-23) J "molecule" ^(-1)K^(-1), m=4.65 xx 10^(-26) kg , T = 0 +273 = 273 K`
`upsilon_(rms) = sqrt((3kT)/(m)) = sqrt((3xx1.38 xx 10^(-23)xx273)/(4.65 xx 10^(-26))) = 493.1 m//s`
The molecule goes to a height h till its entire K.E. is converted into P.E.
`:. mgh = 1/2 m upsilon_(rms)^(2)`
`h = (upsilon_(rms)^(2))/(2g) = ((493.1)^(2))/(2 xx 9.8) = 12.4 xx 10^(4) m = 12.4 km`.
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