Question

The mean speed of the molecules of an ideal gas is 2.0×10^3 ms^-1 . The radius of a gas molecule is 1.5×10^-10 m . Calculate the (i) collision frequency, and (ii) mean free path. It is given that 4×10^24 m^-3 .

Answer 1

Answer

Mean free path

$\lambda=\frac{1}{\sqrt{2}\pi d^2 n}$

n=$4×10^{24 }m^{-3}$

d=$3×10^{-10}m$

v=$2.0×10^3 ms^{-1}$

$\lambda=\frac{1}{1.414\times3.14 ( 3×10^{-10}) ^2\times 4×10^{24 }m^{-3} }$

$\lambda=6256\times10^{-10}m$

Now collision frequency

$\nu=\frac{v}{\lambda}\\=\frac{ 2.0×10^3 }{ 6256\times10^{-10} }\\=31.98\times10^8Hz$