A gas molecule of mass M at the surface of the earth has kinetic...

A gas molecule of mass

M

at the surface of the earth has kinetic energy equivalent to

0 ° C

. If it were to go up straight without colliding with any other molecules, how high it would rise? Assume that the height attained is much less than the radius of the earth. (

{k}_{B}

is Boltzmann constant)

Options

A $\dfrac{273{k}_{B}}{2Mg}$
B $\dfrac{819{k}_{B}}{2Mg}$
C $0$
D $\dfrac{546{k}_{B}}{3Mg}$

Correct answer

B. $\dfrac{819{k}_{B}}{2Mg}$

Step-by-step solution

Mgh = \dfrac{1}{2}M{v}_{rms}^{2} $$ = \dfrac{1}{2}M\dfrac{3kT}{m} $$ ⇒ {v}_{rms}^{2} = \dfrac{819k}{M} $Now apply$ {v}^{2} = {u}^{2} - 2gh undefined

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