Q.Find the average translational kinetic energy of a single gas molecule at $400\ \text{K}$. Take $k_B = 1.38 \times 10^{-23}\ \text{J/K}$.

The central result. Comparing the kinetic-theory pressure formula $PV=\tfrac13Nm\overline{v^2}$ with the ideal gas equation $PV=NkT$ gives $kT=\tfrac13m\overline{v^2}$, which rearranges (multiplying by $3/2$) to

This single equation is the bridge between the macroscopic world (temperature, something read off a thermometer) and the microscopic world (the random kinetic energy of individual, invisible molecules): temperature is, quite literally, nothing but a measure of the average translational kinetic energy carried by each molecule of the gas.

Two consequences worth remembering. (i) Average kinetic energy per molecule is directly proportional to absolute temperature -- doubling the absolute temperature exactly doubles the average kinetic energy per molecule. (ii) Average kinetic energy per molecule depends only on temperature, never on the molecule's own mass -- at a given temperature, a heavy oxygen molecule and a light hydrogen molecule carry exactly the same average kinetic energy (though, since KE also depends on speed, the lighter hydrogen molecule must therefore be moving faster on average to compensate).

Internal energy. Multiplying by the total number of molecules $N$ gives the gas's total internal energy, $U=N\overline{KE}=\tfrac32NkT=\tfrac32\mu RT$ for $\mu$ moles -- a result that depends only on the absolute temperature of an ideal gas, completely independent of its pressure or volume. This is precisely why, for an ideal gas, internal energy is treated as a function of temperature alone.