Question

The mean translational kinetic energy of a perfect gas molecule at the temperature $TK$ is

• A. $\frac{1}{2}{k}_{B}T$
• B. $k_{B}T$
• C. $\frac{3}{2}{k}_{B}T$
• D. $2k_{B}T$

Answer 1

Answer: (C)

A monoatomic gas has $3$ degrees of freedom, one each along for each axis (translation motion).
The kinetic energy per molecule per degree of freedom is $\frac{1}{2}kT$
i.e. The kinetic energy per molecule for a monoatomic gas is $\frac{3}{2}kT$