Answer:
300 nm
Step-by-step explanation:
R = Gas constant = 8.314 J/molK
r = Atomic radii =
![1* 10^(-10)\ m](https://img.qammunity.org/2020/formulas/physics/college/o9ne26qyg67qhrzwzfqmh0qyk1qgf6z1hu.png)
d = Atomic diameter =
![2r=2* 10^(-10)\ m](https://img.qammunity.org/2020/formulas/physics/college/lzj1s7axjux3arr1bfzmlhe337f740lw82.png)
At STP
T = Temperature = 273.15 K
P = Pressure = 100 kPa
= Avogadro's number =
![6.022* 10^(23)](https://img.qammunity.org/2020/formulas/chemistry/high-school/2doyn547wtr4m2pr0z2e1nl42l2rehbl99.png)
The mean free path is given by
![\lambda=(RT)/(\sqrt2d^2N_AP)\\\Rightarrow \lambda=(8.314* 273.15)/(\sqrt2 \pi * (2* 10^(-10))^2* 6.022* 10^(23)* 100000)\\\Rightarrow \lambda=2.12165* 10^(-7)\ m=212.165* 10^(-9)\ m=212.165\ nm](https://img.qammunity.org/2020/formulas/physics/college/cbfyqecs58r8rj6cr6uwuxyz0nwp7yi2lu.png)
The answer that best represents the mean free path for gas molecules is 300 nm