Average translational kinetic energy of a gas molecule at temperature $T$ is:
- $k_B T$
- $\frac{1}{2} k_B T$
- $\frac{3}{2} k_B T$ (= $\frac{3}{2} RT$ per mole)
- $3 k_B T$
Show answer and 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.