A) \[S{{O}_{2}}\]
B) \[C{{O}_{2}}\]
C) \[{{O}_{2}}\]
D) \[{{H}_{2}}\]
Correct Answer: D
Solution :
\[{{V}_{rms}}=\sqrt{\frac{3RT}{M}}\therefore {{V}_{rms}}\propto \frac{1}{\sqrt{M}}\]at same T because \[{{H}_{2}}\] has least molecular weight so its r.m.s. velocity should be maximum.
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