Answer
$K.E=3.31\times10^{-20}J$
Work Step by Step
The average translational Kinetic energy of Nitrogen can be determined by the formula:
$K.E=\frac{3}{2}(\frac{R}{N_a})(T)$
Substituting the values of $R,N$ and $N_a$ into the formula and solving:
$K.E=\frac{3}{2}(\frac{8.314}{6.023\times10^{23}})(1600)$
$K.E=3.31\times10^{-20}J$
