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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2024
  • Hard
  • MCQ

The temperature of a gas is \(-78^{\circ} \mathrm{C}\) and the average translational kinetic energy of its molecules is \(\mathrm{K}\). The temperature at which the average translational kinetic energy of the molecules of the same gas becomes \(2 \mathrm{~K}\) is _______.

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Answer & Solution

Correct Answer

(B) \(117^{\circ} \mathrm{C}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{K} . \mathrm{E}=\frac{\mathrm{nf}_1 \mathrm{RT}}{2}\)

\(\mathrm{~T}_{\mathrm{i}}=-78^{\circ} \mathrm{C} \rightarrow 273+\left[-78^{\circ} \mathrm{C}\right]=195 \mathrm{~K}\)

\(\mathrm{~K} . \mathrm{E} \alpha \mathrm{T}\)

To double the \(K.E\) energy temp also become double

\(\mathrm{T}_{\mathrm{f}}=390 \mathrm{~K}\)

\(\mathrm{~T}_{\mathrm{f}}=117^{\circ} \mathrm{C}\)

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