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 O2 , B only N2 and C a mixture of equal quantities of O2 and N2 . If the average speed of the O2 molecules in genius Kinetic Theory of Gases 22 vessel A is V1 , that of the N2 molecules in vessel B is V2 , the average speed of the O2 molecules in vessel C is (where M is the mass of an oxygen molecule

The correct option is V₁.

According to the Kinetic Theory of Gases, the average speed of gas molecules of mass m at a absolute temperature T is given by the Maxwell-Boltzmann distribution as:
$V_{\text{avg}}=\sqrt{\frac{8k_{B}T}{\pi m}}$
where $k_B$ is the Boltzmann constant.

In vessel A, which contains only oxygen (${\text{O}}_2$) molecules of molecular mass M at temperature T, the average speed of the oxygen molecules is:
$V_1=\sqrt{\frac{8k_{B}T}{\pi M}}$

In vessel C, the system contains a mixture of equal quantities of ${\text{O}}_2$ and ${\text{N}}_2$ at the same temperature T. Because the gases in the mixture are in thermal equilibrium, the average speed of any individual component (like ${\text{O}}_2$) depends solely on the absolute temperature T of the mixture and the mass M of that gas molecule. The presence of nitrogen (${\text{N}}_2$) molecules does not alter the temperature or the mass of the oxygen molecules. Thus, the average speed of the oxygen molecules in vessel C is identical to that in vessel A, which is $V_1$.