Start learning →

Average kinetic energy per molecule (monatomic gas):

A2kT
B(3/2) kT
C(1/2)kT
DkT
Answer & Solution
Correct answer: B. (3/2) kT
**For 3 translational degrees of freedom:** <KE_trans> = (3/2)kT per molecule. **From equipartition:** each quadratic DOF contributes (1/2)kT.
Practise this question with Quill — JEE Advanced →

Keep practising Kinetic Theory of Gases

Browse all Kinetic Theory of Gases questions for JEE Advanced, or explore more Physics topics.

Start a Kinetic Theory of Gases lesson →
Related questions
Kinetic theory bingoKnock out the true statementsWhich ones belongSpot the false statementProportional or independentWhich one does not belongMonatomic or diatomicOrder the kinetic argument