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N molecules, each of mass m of gas `A and 2 N` molecules each of mass `2m` of gas `B` are containted in the same vessel which is maintained at temperature T. The mean square velocity of molecules of B type is denoted by `v^(2)` and the mean square velocity of A type is denoted by `(omega)^(2)`. the `omega^(2)//v^(2)` is: |
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Answer» Correct Answer - 2 For gas `A` `V_(A)^(2)=(3KT)/(m)` since `V_(X)^(2)=V_(y)^(2)=V_(z)^(2)` for molecule hence `V_(A)^(2)=V_(X)^(2)+V_(y)^(2)=3V_(X)^(2)` `V_(A)^(2)=3w^(2)=(3KT)/(m)` for gas `B` `V_(B)^(2)=V^(2)=(3KT)/(2m)` divide `=(V_(A)^(2))/(V_(B)^(2))=(w^(2))/(V^(2))=(2)/(3)` |
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