Answer:
300 nm
Explanation:
R = Gas constant = 8.314 J/molK
r = Atomic radii = [tex]1\times 10^{-10}\ m[/tex]
d = Atomic diameter = [tex]2r=2\times 10^{-10}\ m[/tex]
At STP
T = Temperature = 273.15 K
P = Pressure = 100 kPa
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]
The mean free path is given by
[tex]\lambda=\frac{RT}{\sqrt2d^2N_AP}\\\Rightarrow \lambda=\frac{8.314\times 273.15}{\sqrt2 \pi \times (2\times 10^{-10})^2\times 6.022\times 10^{23}\times 100000}\\\Rightarrow \lambda=2.12165\times 10^{-7}\ m=212.165\times 10^{-9}\ m=212.165\ nm[/tex]
The answer that best represents the mean free path for gas molecules is 300 nm