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Three closed vessels \mathrm{A,B} and \mathrm{C} are at the same temperature \mathrm{T} and contains gases which obey maxwellian distribution of velocities. Vessel A contains \mathrm{O}_{2}, vessel \mathrm{B} only \mathrm{N}_{2} and \mathrm{C}  contains mixture of equal quantities of \mathrm{\mathrm{O}_{2}} and \mathrm{\mathrm{N}_{2}}. If the average speed of the \mathrm{\mathrm{O}_{2}} molecules in the vessel \mathrm{A} is \mathrm{V_{1}}, that of \mathrm{\mathrm{N}_{2}}  in vessel \mathrm{B} is \mathrm{V}_{2}. The average speed of the \mathrm{O}_{2} molecules in vessel \mathrm{C} is 

Option: 1

\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right) / 2


Option: 2

\mathrm{V}_{1}


Option: 3

\mathrm{\sqrt{4 v^{2}}}


Option: 4

\mathrm{\sqrt{\frac{3 K T}{m}}}


Answers (1)

best_answer

The average speed \mathrm{V=\sqrt{\frac{2 K T}{m}}} depends only on temperature and molecular weight only. So for oxygen it will be \mathrm{\mathrm{V}_{1}}.

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Rishabh

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