Question

Calculate the mean free path of gas molecule in a chamber of 10-7 mm of
mercury pressure. Assuming the molecular diameter to be 3Ȧ. Two grams of
the molecule of the gas occupy 44.8 L at NTP. Consider the temperature of
the chamber to be 273K.

Answer 1

Pressure $P=133.322\times10^{-7}$ Pa

Diameter $d=3\times10^{-10}m$

Temperature is $T=273K$

Now, mean free path $(l)$ is

$$l=\dfrac{kT}{\sqrt{2}{d^2}P}$$

Therefore,

$$l=\dfrac{1.38×10^{ −23}\times 273}{\sqrt{2}\times 9\times10^{-20}\times 133.332\times 10^{-7}}=2220.31m$$