Answer to Question #169501 in Molecular Physics | Thermodynamics for khushi

Question

The mean speed of the molecules of an ideal gas is 2.0 *10³ ms^-1 .The radius

of a gas molecule is 1.5 *10¹⁰ m.Calculate the (i) collision frequency, and

(ii) mean free path. It is given that n = 4 *10²⁴m^-3.

Answer 1

$\lambda=\frac{1}{\sqrt2 \pi d^2n}=\frac{1}{\sqrt2 \pi (2r)^2n},\\ \lambda=\frac{1}{1.41\cdot 3.14\cdot(2\cdot 1.5\cdot10^{-10})^2\cdot 4\cdot 10^{24}}=0.63~\mu m,$

$\nu=\frac{v}{\lambda},$

$\nu=\frac{2.0\cdot 10^3}{0.63\cdot 10^{-6}}=3.17\cdot 10^9~s^{-1}.$