Answer to Question #177244 in Molecular Physics | Thermodynamics for TUHIN SUBHRA DAS

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^-10m. Calculate the (i) collision frequency, and (ii) mean free path. It is given that n=4*10^24m^-3 .

Answer 1

(i) We can find the collision frequency as follows:

$$Z=\sqrt{2}\pi d^2<v>(\dfrac{N}{V})=\sqrt{2}\pi (2r)^2<v>n=4\sqrt{2}\pi r^2<v>n,$$ $$Z=4\sqrt{2}\pi\cdot(1.5\cdot10^{-10}\ m)^2\cdot2\cdot10^3\ \dfrac{m}{s}\cdot4\cdot10^{24}\ m^{-3},$$ $$Z=31.97\cdot10^8\ \dfrac{collision}{s}.$$

(ii) We can find the mean free path as follows:

$$\lambda=\dfrac{1}{\sqrt{2}\pi d^2n}=\dfrac{1}{4\sqrt{2}\pi r^2n},$$ $$\lambda=\dfrac{1}{4\sqrt{2}\pi\cdot(1.5\cdot10^{-10}\ m)^2\cdot4\cdot10^{24}\ m^{-3}}=6.25\cdot10^{-7}\ m.$$