Question

The mean translational kinetic energy of a perfect gas molecule at the temperature $T\, K$ is

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

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 $\frac{3}{2} kT$