question_answer

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 radius of the earth. (\[{{k}_{B}}\] is Boltzmann constant).                                          [JEE ONLINE 19-04-2014]

A)  0        

B)                   \[\frac{273{{k}_{B}}}{2Mg}\]

C)  \[\frac{546{{k}_{B}}}{3Mg}\]  

D)       \[\frac{819{{k}_{B}}}{2Mg}\]

Correct Answer: D

Solution :

[d] Kinetic energy of each molecule,

In the given problem,

Temperature, T = 0°C = 273 K

Height attained by the gas molecule, h=?

or