Q140 / 160 · AP EAMCET 2024 (22 May Shift 1) · Physics
The ratio of kinetic energy of a diatomic gas molecule at a high temperature to that of NTP is
- A$\frac{3}{2}$
- B$\frac{5}{3}$
- C$\frac{5}{7}$
- Correct answerD$\frac{7}{5}$
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) $\therefore \frac{\mathrm{k}_1}{\mathrm{k}_2}=\frac{\mathrm{f}_1}{\mathrm{f}_2}=\frac{7}{5}$
For diatomic gas, $f_1=7$ (At high temperature) $\mathrm{f}_2=5$ (At low temperature) $\therefore \frac{\mathrm{k}_1}{\mathrm{k}_2}=\frac{\mathrm{f}_1}{\mathrm{f}_2}=\frac{7}{5}$
Hence, the correct answer is option: 'D'.