JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2014
- Medium
- MCQ
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)
Pick an option to check your answer.
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, \(h=?\)
\(\mathrm{KE}=\frac{3}{2} \mathrm{K}_{\mathrm{B}}(273)=\frac{819 \mathrm{K}_{\mathrm{B}}}{2}\)
\(\mathrm{K.E}=\mathrm{PE}\)
\( \Rightarrow \frac{{819{K_B}}}{2} = Mgh\)
\(h=\frac{{819{k_B}}}{{2Mg}}\)
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