Question

At what pressure will the mean free path in room-temperature (20?c) nitrogen be 1.5 m ?

Answer 1

In physical chemistry, the mean free path is the average distance between the atoms during collision. Its formula is

 $Mean \ free \ path \ = \frac{RT}{ \sqrt{2} \pi d^{2} N_{A}P }$

where d is the diameter of the Nitrogen atom (d = 310 x 10^-7 m), Na is Avogadro's number (Na=6.022 x 10^23), R is the gas constant (8.314 J/mol-K), T is the absolute temperature (T= 20 + 273 = 293 K). Substituting the values,

$1.5 = \frac{(8.314)(293)}{ \sqrt{2} \pi (310 \ x\ 10^{-7} )^{2} (6.022 \ x\ 10^{23} )P }$
P = 6.32 x 10^-13 Pascals