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Question

At what temperature will the mean molecular energy of a perfect gas be one-third of its value of 27oC?
  1. 10oC
  2. 102K
  3. 101K
  4. 103J

A
103J
B
101K
C
102K
D
10oC
Solution
Verified by Toppr

Mean KET
therefore at T=27 which in kelvin becomes 300 K
So KE will be 1/3 when temperature becomes 1/3 which is 300/3=100K
option (C)

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