What is the average translational kinetic energy of molecules in a ideal gas at `37°C`? Given: `T = 37°C = 310 K` Known: `k = 1.38 times 10^(-23)\ J//K` Equation: `overline(KE) = 3/2kT` `overline(KE) = 3/2(1.38 times 10^(-23)\ J//K)(310 K)` Note: A mole of molecules would have a total translational kinetic energy equal to `(6.42 times 10^(-21)\ J)(6.02 times 10^23) = 3900\ J`, which equals the kinetic energy of a `1`-`kg` stone traveling faster than `85\ m//s`.