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Question
The average energy for molecules in one degree of freedom is ______.
Options
`3/2kT`
`(kT)/2`
`3/4kT`
kT
MCQ
Fill in the Blanks
Solution
The average energy for molecules in one degree of freedom is `underlinebb((kT)/2).`
Explanation:
Degree of freedom (for 1 molecule)
(f) = 1,
Average energy (E) = `f/2 "NkT" = (kT)/2`
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