MHT CET · Physics · Kinetic Theory of Gases
The mean kinetic energy of the molecules of an ideal gas at \(399^{\circ} \mathrm{C}\) is ' E '. The temperature at which the mean kinetic energy of its molecules will be ' \(E / 2\) ', is
- A \(336^{\circ} \mathrm{C}\)
- B \(276^{\circ} \mathrm{C}\)
- C \(123^{\circ} \mathrm{C}\)
- D \(63^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(D) \(63^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\( T_1 = 399 + 273 = 672 \mathrm{K} \) \( K \propto T \implies \frac{E/2}{E} = \frac{T_2}{T_1} \)
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