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RMS and KE Expressions

A gas molecul...

Question

A gas molecule of mass $M$ at the surface of the Earth has kinetic energy equivalent to $0^{o} C$ . 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 radius of the earth. ( $k_{B}$ is Boltzmann constant)

A

0

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B

\frac{273 k_{B}}{2 M g}

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C

\frac{546 k_{B}}{3 M g}

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D

\frac{819 k_{B}}{2 M g}

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Solution

The correct option is D $\frac{819 k_{B}}{2 M g}$
Answer is D
Average translational kinetic energy of molecules=
$K E = \frac{3}{2} k_{B} T = \frac{819}{2} k_{B} a s T = 273 K$
Now. KE= PE
Or,
$\frac{819}{2} k_{B} = M g h$
$h = \frac{819 k_{B}}{2 M g}$

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