Question

A gas molecule of mass $M$ at the surface of the Earth has kinetic energy equivalent to $0^{\circ} C$. 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. $\frac{273\,k_{B}}{2\,Mg}$
• C. $\frac{546\,k_{B}}{3\,Mg}$
• D. $\frac{819\,k_{B}}{2\,Mg}$

Answer 1

Answer: (D)

$\frac{3}{2} KT =\frac{1}{2} mv ^{2} $
$v ^{2}=\frac{3 KT }{ m }$
$v ^{2}= u ^{2}+2$ as
$O =\frac{3 KT }{ m }-2 gs $
$S =\frac{3 K }{2} \frac{(273)}{ mg } $
$S =\frac{819 K }{2 mg }$