Question

The average speed of nitrogen molecule in air is about

Answer 1

Since,

$E=3RT/2N_{A}$; where $N_{A}$ is Avagadros Number; T is temprature; E is AVERAGE kinetic energy of system; R is gas constant(8.314 J/K)

So, $E=3(8.314 J/K)(300K)/2(6.022*10^{23})=6.21*10^{-21} J$

Since,

$E=mv^{2}/2=(28u)(v^{2})/2=6.21*10^{-21}J$

$v=(6.21*10^{-21})/(14*1.672*10^{-27})=515.17 m/s$    [because $1u=1.672*10^_{-27}$Kg]

So, Answer is 515.17 m/s at STP.

Note: I have used SI units so I have to convert everything in SI units.