Question

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

Answer 1

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

`E = (3kT)/2` where `k` is the Boltzmann constant and `T` is the absolute temperature.

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

Thus, `E = 1.5 * 310 * 1.38 * 10^(-23) J`

`implies E = 641.7 * 10^(-23) J`

Or `E = 0.0401` `eV`