JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
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 _______.
- A \(-39^{\circ} \mathrm{C}\)
- B \(117^{\circ} \mathrm{C}\)
- C \(127^{\circ} \mathrm{C}\)
- D \(-78^{\circ} \mathrm{C}\)
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…
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