Answer to Question #156095 in Molecular Physics | Thermodynamics for AKSHAT

Question

The mean speed of the molecules of an ideal gas is −  3 1 2.0 10 ms .The radius

of a gas molecule is −  10 1.5 10 m.Calculate the (i) collision frequency, and

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

Answer 1

$<v>=312(m/s)$

$n=4\cdot10^{24}(1/m^3)$

$r=1.5\cdot10^{-10}(m)$

(i) $<z>=4\sqrt2\pi n r^2\cdot<v>=$

$=4\cdot\sqrt2\cdot3.14\cdot4\cdot10^{24}\cdot(1.5\cdot10^{-10})^2\cdot312=$

$=500\cdot 10^6(collision/s)$

(ii) $<\lambda>=\frac{<v>}{<z>}=\frac{312}{500\cdot10^6}=624\cdot10^{-9}(m)$