# What is the average translational kinetic energy of molecules in an ideal gas at 37°C?

Jun 6, 2017

The average translational energy of a molecule is given by the equipartition theorem as,

$E = \frac{3 k T}{2}$ where $k$ is the Boltzmann constant and $T$ is the absolute temperature.

For $T = \left(37 + 273\right) K$ we are easily in the required high-temperature limit.

Thus, $E = 1.5 \cdot 310 \cdot 1.38 \cdot {10}^{- 23} J$

$\implies E = 641.7 \cdot {10}^{- 23} J$

Or $E = 0.0401$ $e V$