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

Question #156095

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.


1
Expert's answer
2021-01-22T11:40:57-0500

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


n=41024(1/m3)n=4\cdot10^{24}(1/m^3)


r=1.51010(m)r=1.5\cdot10^{-10}(m)


(i) <z>=42πnr2<v>=<z>=4\sqrt2\pi n r^2\cdot<v>=


=423.1441024(1.51010)2312==4\cdot\sqrt2\cdot3.14\cdot4\cdot10^{24}\cdot(1.5\cdot10^{-10})^2\cdot312=


=500106(collision/s)=500\cdot 10^6(collision/s)


(ii) <λ>=<v><z>=312500106=624109(m)<\lambda>=\frac{<v>}{<z>}=\frac{312}{500\cdot10^6}=624\cdot10^{-9}(m)











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