B.1.5—Kelvin temperature and kinetic energy

Absolute temperature and motion

For particles in an ideal gas, Kelvin temperature is proportional to their average random translational kinetic energy:

$$\overline{E_k}=\frac{3}{2}k_BT$$

Here $$k_B$$ is the Boltzmann constant and T must be in kelvin.

What the equation says

If the Kelvin temperature doubles, the average translational kinetic energy doubles. A higher temperature means greater average random kinetic energy, not that every particle has exactly the same kinetic energy.

Scope of the model

The relation describes average random translational motion. It does not include the whole internal energy of a substance, which also contains intermolecular potential energy.

Worked example from local Question Bank row 31728

For helium atoms at $T=320\,\mathrm{K}$ with $m=6.6\times10^{-27}\,\mathrm{kg}$, equate mean translational kinetic energy to $\tfrac12mv^2$:

$$\frac12mv^2=\frac32k_BT\Rightarrow v=\sqrt{\frac{3k_BT}{m}}$$
$$v=\sqrt{\frac{3(1.38\times10^{-23})(320)}{6.6\times10^{-27}}}=1.4\times10^3\,\mathrm{m\,s^{-1}}$$

This is a characteristic speed derived from the average energy, not a claim that every atom has that speed.

Common trap

A Celsius temperature cannot be used in this equation. Convert first; 0 °C corresponds to about 273 K, not zero particle kinetic energy.