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. $0$
• B. $\frac{273k_B}{2Mg}$
• C. $\frac{546k_B}{3Mg}$
• D. $\frac{819k_B}{2Mg}$

Answer 1

The correct option is D $\frac{819k_B}{2Mg}$

Answer is D
Average translational kinetic energy of molecules=
$KE=\frac{3}{2}{k}_{B}T=\frac{819}{2}{k}_{B}asT=273K$
Now. KE= PE
Or,
$\frac{819}{2}{k}_B=Mgh$
$h=\frac{819k_B}{2Mg}$