Question

Calculate the fraction of oxygen molecules moving with a speed between 700 ms^-1 and 710 ms^-1 at 300K using the following equation.

f(u) du= dN/N= 4πu² (M/2π RT)^3/2 e^{-(Mu^2/2RT) du}

Answer 1

Given Data:

Moving with a speed 700 m/s and 710m/s

Temperature(T)=300k

using the following equation

f(u) du= dN/N= 4πu2 (M/2π RT)3/2 e-(Mu^2/2RT) du

We get-

f(u)=N(u)du/N

=4π(m/2πkT)^(3/2)u²exp(-mu²/2kT)du

u=710-700=10m/s

Now,

f=4π{(32×1.67×10^-27)/(2π×1.38×10^-23×300)}^3/2×705²×exp{-(32×1.67×10^-27×705^2)/(2×1.38×10^-23×300)}×10

=7.43×10^-3

Hence the fraction of oxygen molecule is 7.43×10^-3