alleryAverage translational kinetic energy of a gas molecule at temperature $T$ is:
A$k_B T$
B$\frac{1}{2} k_B T$
C$\frac{3}{2} k_B T$ (= $\frac{3}{2} RT$ per mole)
D$3 k_B T$
Answer & Solution
Correct answer: C. $\frac{3}{2} k_B T$ (= $\frac{3}{2} RT$ per mole)
Average KE per molecule = $\frac{1}{2} m \overline{v^2} = \frac{3}{2} k_B T$.
**Per mole:** $\frac{3}{2} RT$.
Temperature is a direct measure of average kinetic energy.
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