For a monoatomic gas, the average molecular kinetic energy can be written as
[tex]K= \frac{1}{2}mv^2 = \frac{3}{2} nRT [/tex]
where, in our problem, [tex]m=0.02 kg[/tex] is the mass, [tex]v=1321 m/s[/tex] is the average molecular speed, [tex]n=5 mol[/tex] is the number of moles and [tex]R=8.31 J/(mol K)[/tex]. re-arranging the equation, we can find the temperature T of the gas:
[tex]T= \frac{mv^2}{3nR}= \frac{(0.02 kg)(1321 m/s)^2}{3 (5 mol)(8.31 J/(mol k))}=280 K [/tex]
