Three closed vessels A, B and C are at the same temperature \[T\] and contain gases which obey the Maxwellian distribution of velocities. Vessel A contains only \[{{O}_{2}}\], vessel B contains only \[{{N}_{2}}\] and vessel C contains a mixture of equal quantities of \[{{O}_{2}}\] and \[{{N}_{2}}\]. If the average speed of \[{{O}_{2}}\] molecules in vessel A is \[{{v}_{1}}\] , average speed of \[{{N}_{2}}\] molecules in vessel B is \[{{v}_{2}}\] then the average speed of \[{{O}_{2}}\] molecules In C will be- (where\[M\]is the mass of oxygen molecule)
(A). \[\dfrac{{{v}_{1}}+{{v}_{2}}}{2}\]
(B). \[{{v}_{1}}\]
(C). \[{{({{v}_{1}}{{v}_{2}})}^{\dfrac{1}{2}}}\]
(D). \[\sqrt{\dfrac{3KT}{M}}\]

Hint: For gases enclosed in a container, the speeds of its molecules vary greatly. It depends on temperature, pressure and mass of the gases. The average speed gives us an idea about the speeds of the majority of the molecules. As \[{{O}_{2}}\] and \[{{N}_{2}}\] molecules are inert their velocities will be independent of each other.

Formula used: \[{{v}_{av}}=\sqrt{\dfrac{3KT}{m}}\]

Complete step by step answer:
Maxwell-Boltzmann distribution gives us the distribution of speeds in molecules of an ideal gas. It is represented by the following graph-
 
 \[{{O}_{2}}\]and\[{{N}_{2}}\] gases are inert gases and cannot react with each other or with other molecules of their kind, therefore their velocities will be independent of each other.
The average of velocity of molecules of gases in a container is given by-
\[{{v}_{av}}=\sqrt{\dfrac{8RT}{\pi m}}\] or \[{{v}_{av}}=\sqrt{\dfrac{3KT}{m}}\]
Here,\[R\] is the gas constant
\[T\] is temperature
\[m\] is mass of the gas
\[K\] is the Boltzmann Constant, its value is \[1.38\times {{10}^{-23}}{{m}^{2}}kg\,{{s}^{-2}}{{K}^{-1}}\]
 As the mass of oxygen is given by \[M\], the average velocity of oxygen molecules in vessel C will be-
\[v=\sqrt{\dfrac{3KT}{M}}m{{s}^{-1}}\]
The velocity of oxygen molecules in a mixture of \[{{O}_{2}}\] and \[{{N}_{2}}\] gases in vessel C is \[\sqrt{\dfrac{3KT}{M}}m{{s}^{-1}}\].

So, the correct answer is “Option D”.

Note: As observed from the Boltzwellian distribution, there are very few molecules with high or low speeds. Most molecules possess moderate speeds. As we increase the temperature, the number of molecules with high speeds increase so the graph moves more towards the right. Similarly, as we decrease the temperature, molecules with low speeds decrease so the graph moves more towards the left.