For two gases A and B with molecular weights `M_(A) and M_(B)`, it is observed that at a certain temperature `T_(1)` the mean velocity of A is equal to the root mean square velocity 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 a higher temperature
D. Both A and B are placed at lower temperature

Explanation

`(U_(AV))_(A)=sqrt((8RT)/(piM_(A))) and (U_(rms))_(B)=sqrt((3RT)/(M_(B)))`
`therefore (8)/(3pi) =(M_(A))/(M_(B))`
For `A (U_(AV))=sqrt((8RT_(2))/(piM_(A)))" or "B V_(AV)=sqrt((8RT)/(piM_(B)))`
`(T_(2))/(T) =(M_(A))/(M_(B))=(8)/(3pi) therefore T_(2) =(8)/(3pi) T" or "T_(2) lt T`
Hence, (B) is the correct answer.