AP EAMCET · PHYSICS · Kinetic Theory of Gases
The average translational kinetic energy of a molecule in a gas becomes equal to \(0.69 \mathrm{eV}\) at temperature about, [Boltzmann's constant \(=138 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\) ]
- A \(3370^{\circ} \mathrm{C}\)
- B \(3388^{\circ} \mathrm{C}\)
- C \(5333^{\circ} \mathrm{C}\)
- D \(5060^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(D) \(5060^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Given, average translational kinetic energy \(=0.69 \mathrm{eV}=0.69 \times 1.6 \times 10^{-19} \mathrm{~V}\) As we know that, average translational kinetic energy \(=\frac{3}{2} k T\)…
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