COMEDK · Physics · 14. Kinetic Theory of Gases
The average kinetic energy of a molecule in air at room temperature of \(20^{\circ} \mathrm{C}\)
- A \(6 \times 10^{-22} \mathrm{~J}\)
- B \(7.06 \times 10^{-21} \mathrm{~J}\)
- C \(6.07 \times 10^{-21} \mathrm{~J}\)
- D \(6.70 \times 10^{-21} \mathrm{~J}\)
Answer & Solution
Correct Answer
(C) \(6.07 \times 10^{-21} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
The average kinetic energy of a molecule is given by the formula \(K_{avg} = \dfrac{3}{2} k_{B} T\), where \(k_{B}\) is the Boltzmann constant and \(T\) is the absolute temperature in Kelvin. Given temperature \(T = 20^{\circ} \mathrm{C} = 20 + 273.15 = 293.15 \mathrm{~K}\). The…
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