Kinetic Theory Of Gases Question 101
Question: For a gas sample with Np number of molecules, function N(V) is given by: $ N( V )=\frac{dN}{dV}=[ \frac{3F _{0}}{V _{0}^{3}} ]V^{2} $ for $ 0\le V\le V _{0} $ and $ N( V )=0 $ for $ V>V _{0} $ Where $ dN $ is number of molecules in speed range V to $ V+dV. $ The rms speed of the gas molecule is
Options:
A) $ \sqrt{\frac{2}{5}}V _{0} $
B) $ \sqrt{\frac{3}{5}}V _{0} $
C) $ \sqrt{2}V _{0} $
D) $ \sqrt{3}V _{0} $
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Answer:
Correct Answer: B
Solution:
[b] $ V _{rms}^{2}=<V^{2}>,=\frac{V _{1}^{2}+V _{2}^{2}+V _{3}^{2}+…….}{N} $
$ =\frac{\int _{{}}^{{}}{V^{2}dN}}{\int _{{}}^{{}}{dN}}\text{ here }\frac{dN}{dV}=N( V ) $
$ V _{rms}^{2}=\frac{1}{N}\int\limits _{0}^{\infty }{N( V )V^{2}dV} $
$ =\frac{1}{N}\int\limits _{0}^{V _{0}}{[ \frac{3N}{V _{0}^{3}}V^{2} ]V^{2}dV=\frac{3}{5}V _{0}^{2}} $
$ \Rightarrow V _{rms}=\sqrt{\frac{3}{5}}V _{0} $
BETA
