Kinetic Theory

JEE Main level, single correct, medium

161 of 168

Question 161 of 168

A gas molecule of mass MM at the surface of the earth has kinetic energy equivalent to 0∘C0^{\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 the radius of the earth. (kB{k}_{B} is Boltzmann constant)

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Answer: (B) 819kB2Mg\dfrac{819{k}_{B}}{2Mg}

Mgh=12Mvrms2Mgh=\dfrac{1}{2}M{v}_{rms}^{2}
=12M3kTm=\dfrac{1}{2}M\dfrac{3kT}{m}
⇒vrms2=819kM\Rightarrow {v}_{rms}^{2}=\dfrac{819k}{M}
Now apply v2=u2−2gh{v}^{2}={u}^{2}-2gh undefined

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