Two gases A and B are at absolute temperatures $360 \mathrm{~K}$ and $420 \mathrm{~K}$ respectively. The…

Two gases A and B are at absolute temperatures $360 \mathrm{~K}$ and $420 \mathrm{~K}$ respectively. The ratio of average kinetic energy of the molecules of gas B to that of gas A is
  1. $6: 7$
  2. $\sqrt{7}: \sqrt{6}$
  3. $7: 6$
  4. $49: 36$

Solution

The mean kinetic energy $\frac{1}{2} \mathrm{~m}<\mathrm{v}>^2$ of the gas molecule is proportional to the temperature $\mathrm{T}$. $\frac{\mathrm{K}_{\mathrm{B}}}{\mathrm{K}_{\mathrm{A}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}=\frac{420}{360}=\frac{7}{6}$

Asked in: MHT CET 2022 (05 Aug Shift 2)

Practice more Kinetic Theory of Gases and Radiation questions on Aicharya