Answer:
47149435.4884 /s
Explanation:
[tex]k[/tex] = Boltzmann constant = [tex]1.38\times 10^{-23}\ J/K[/tex]
T = Temperature = 20 °C
P = Pressure = [tex]1\times 10^{-2}\ atm[/tex]
Average time between collision is given by
[tex]\Delta t=\dfrac{l_m}{v}[/tex]
Frequency is given by
[tex]f=\dfrac{1}{\Delta t}[/tex]
Mean free path is given by
[tex]l_m=\dfrac{1}{4\pi\sqrt{2}\times r^2\times \dfrac{N}{V}}[/tex]
[tex]\Delta t=\dfrac{\dfrac{1}{4\pi\sqrt{2}\times r^2\times \dfrac{N}{V}}}{v}[/tex]
[tex]f=\dfrac{1}{\dfrac{\dfrac{1}{4\pi\sqrt{2}\times r^2\times \dfrac{N}{V}}}{v}}\\\Rightarrow f=\dfrac{v4\pi\sqrt{2}\times r^2\times N}{V}[/tex]
Hence, proved.
Average velocity is given by
[tex]v=\dfrac{8kT}{\pi m}[/tex]
[tex]PV=NkT[/tex]
[tex]f=\dfrac{4\pi\sqrt{2}\times r^2\times (\dfrac{8kT}{\pi m})^{\dfrac{1}{2}}\times P}{kT}[/tex]
[tex]f=16r^2P(\dfrac{\pi}{kTm})^{\dfrac{1}{2}}\\\Rightarrow f=16(1.5\times 10^{-10})^2\times 1\times 10^{-2}\times 101325(\dfrac{\pi}{1.38\times 10^{-23}\times (20+273.15)\times 28\times 1.66\times 10^{-27}})^{\dfrac{1}{2}}\\\Rightarrow f=47149435.4884\ /s[/tex]
The collision frequency is 47149435.4884 /s
