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The average kinetic energy of translation of a molecule of an ideal gas at temperature T is:
  • a)
    (1/2) kT
  • b)
    (5/7) kT
  • c)
    (3/2) kT
  • d)
    (7/2) kT
Correct answer is option 'C'. Can you explain this answer?
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Every degree of freedom has energy equal to 1/2KT
Ideal gas is monoatomic hence it has 3 degree of freedom
Therefore K.E.=3*1/2KT=3/2KT
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The average kinetic energy of translation of a molecule of an ideal gas at temperature T is:a)(1/2) kTb)(5/7) kTc)(3/2) kTd)(7/2) kTCorrect answer is option 'C'. Can you explain this answer?
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