For two gases `A` and `B` with molecular weights `M_(A)` and `M_(B)`, respectively, it is observed that at a certain temperature `T`, the mean velocity of `A` is equal to the `V_(rms)` of `B`. Thus, the mean velocity of `A` can be made equal to the mean velocity of `B`, if

Options

A. `A` is at temperature `T` and `B` at `T' , T gtT'`
B. `A` is lowered to a temperature `T_(2) = (8)/(3pi) T` while `B` is at `T` .
C. Both `A` and `B` are reaised to a higher temperature
D. Both `A` and `B` are placed at lower temperature

Explanation

`u_(AV)-A =sqrt([[8RT)/(piM_(A))]), u_(rms-B) = sqrt([[3RT)/(M_(B))])`
`:' u_(AV) of A =u_(rms) of B`
`(8)/(3pi) = (M_(A))/(M_(B))`
For `AV-A = u_(AV)-A =sqrt([[8RT)/(piM_(A))])`
`u_(AV)-A =sqrt([[8RT)/(piM_(B))]`
`:. (T_(2))/(T) = (M_(A))/(M_(B)) = (8)/(3pi)`
`:. T_(2) = (8)/(3pi).T` .