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 gt T'`
B. `A` is lowered to a temperature `T_(2), T_(2) lt T` while `B` is at `T`
C. Both `A` and `B` are raised to higher temperature
D. Both `A` and `B` are placed at lower temperature

Answer

B

Explanation

`(U_(AV))_(A)=sqrt((8RT)/(piM_(A))) and (U_(rms))_(B)=sqrt((3RT)/(M_(B)))`
`:. (8)/(3pi)=(M_(A))/(M_(B))`
for `A(U_(AV))=sqrt((8RT_(2))/(piM_(A)))`
for `BV_(AV)=sqrt((8RT)/(pi M_(B)))`
`(T_(2))/(T)=(M_(A))/(M_(B))=(8)/(3 pi)`
`:. T_(2)=(8)/(3 pi)T` or `T_(2) lt T`