Question #177244

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 .


Expert's answer

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


Z=2πd2<v>(NV)=2π(2r)2<v>n=42πr2<v>n,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=42π⋅(1.5⋅10−10 m)2⋅2⋅103 ms⋅4⋅1024 m−3,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⋅108 collisions.Z=31.97\cdot10^8\ \dfrac{collision}{s}.

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


λ=12πd2n=142πr2n,\lambda=\dfrac{1}{\sqrt{2}\pi d^2n}=\dfrac{1}{4\sqrt{2}\pi r^2n},λ=142π⋅(1.5⋅10−10 m)2⋅4⋅1024 m−3=6.25⋅10−7 m.\lambda=\dfrac{1}{4\sqrt{2}\pi\cdot(1.5\cdot10^{-10}\ m)^2\cdot4\cdot10^{24}\ m^{-3}}=6.25\cdot10^{-7}\ m.
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