Q. If the oxygen $(O_{2})$ has root mean square velocity of $C\ ms^{- 1},$ then root mean square velocity of the hydrogen $(H_{2})$ will be
A) $C\ ms^{- 1}$
B) $\frac{1}{C}ms^{- 1}$
C) $4C\ ms^{- 1}$
D) $\frac{C}{4}ms^{- 1}$
Explanation
(c) $v_{rms} \propto \frac{1}{\sqrt{M}} \Rightarrow \frac{v_{O_{2}}}{v_{H_{2}}} = \sqrt{\frac{M_{H_{2}}}{M_{O_{2}}}} \Rightarrow \frac{C}{v_{H_{2}}} = \sqrt{\frac{2}{32}} = \frac{1}{4}$
$\Rightarrow v_{H_{2}} = 4C\ cm/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.