alleryThe mean translational kinetic energy of one molecule of an ideal gas at temperature $T$ is:
A$\dfrac{1}{2} kT$
B$kT$
C$\dfrac{3}{2} kT$
D$3 kT$
Answer & Solution
Correct answer: C. $\dfrac{3}{2} kT$
Equipartition theorem: each translational degree of freedom contributes $\dfrac{1}{2} kT$ of energy. There are 3 translational degrees ($x, y, z$)
so total mean translational KE per molecule is $\dfrac{3}{2} kT$.
For a mole: $\dfrac{3}{2} RT$.
If the molecule has internal degrees (rotation, vibration), more energy is stored — but those don't add to **translational** KE specifically.
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