Answer
$4.8\times10^{-21}J$
Work Step by Step
From equations 14.6 & 14.2, we can write,
$$KE=\frac{3}{2}kT=\frac{3PV}{2N}$$
Let's plug known values into this equation.
$KE=\frac{3PV}{2N}=\frac{3PV}{2nN_{A}}=\frac{3(4.5\times10^{5}Pa)(8.5\times10^{-3}m^{3})}{2(2\space mol)(6.022\times10^{23}mol^{-1})}$
$KE=4.8\times10^{-21}J$ (Translational kinetic energy)
