Temperature and KEMediumQuestion 31005
Question
Average kinetic energy of a molecule of an ideal gas at temperature T is:
- AkT
- B½kT
- C(3/2)kTCorrect
- D(5/2)kT
Correct answer
(3/2)kT
Explanation
For 3 translational degrees of freedom: KE_avg = (3/2)kT where k = Boltzmann constant = 1.38×10⁻²³ J/K.