A) \[{{H}_{2}}>{{N}_{2}}>{{O}_{2}}>HBr\]
B) \[HBr>{{O}_{2}}>{{N}_{2}}>{{H}_{2}}\]
C) \[HBr>{{H}_{2}}>{{O}_{2}}>{{N}_{2}}\]
D) \[{{N}_{2}}>{{O}_{2}}>{{H}_{2}}>HBr\]
Correct Answer: A
Solution :
Mean square velocity varies inversely with the square root of molar mass. \[{{u}_{rms}}\propto \frac{1}{\sqrt{molar\text{ }mass}}\] Molar mass of \[{{H}_{2}}=2,\text{ }{{N}_{2}}=14,\text{ }{{\text{O}}_{2}}=16,\text{ }HBr=81\]hence, \[{{u}_{rms}}\]velocity order is as . \[{{H}_{2}}>\text{ }{{N}_{2}}>{{O}_{2}}>HBr\]
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