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JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases

A gas molecule of mass \(M\) at the surface of the Earth has kinetic energy equivalent to \(0\,^oC\). 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)

  1. A \(0\)
  2. B \(\frac{{273{k_B}}}{{2Mg}}\)
  3. C \(\frac{{546{k_B}}}{{3Mg}}\)
  4. D \(\frac{{819{k_B}}}{{2Mg}}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{{819{k_B}}}{{2Mg}}\)

Step-by-step Solution

Detailed explanation

Kinetic energy of each molecule, \(\mathrm{K} \mathrm{E}=\frac{3}{2} \mathrm{K}_{\mathrm{B}} \mathrm{T}\) In the given problem, Temperature, \(\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{K}\) Height attained by the gas molecule, \(h=?\)…
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