The ratio of kinetic energy of a diatomic gas molecule at a high temperature to that of NTP is

The ratio of kinetic energy of a diatomic gas molecule at a high temperature to that of NTP is
  1. $\frac{3}{2}$
  2. $\frac{5}{3}$
  3. $\frac{5}{7}$
  4. $\frac{7}{5}$

Solution

The kinetic energy of a gas molecule is $\mathrm{K}=\mathrm{K} \cdot \mathrm{E} / \text { molecule }=\frac{\mathrm{f}}{2} \mathrm{k}_{\mathrm{B}} \mathrm{~T}$
For diatomic gas, $f_1=7$ (At high temperature) $\mathrm{f}_2=5$ (At low temperature) $\therefore \frac{\mathrm{k}_1}{\mathrm{k}_2}=\frac{\mathrm{f}_1}{\mathrm{f}_2}=\frac{7}{5}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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