For two gases, A and B with molecular weights `M_(A)` and `M_(B)`. It is observed that at a certain temperature. T, 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:

Explanation

`bar(c_(A))=sqrt((8RT)/(piM_(A)))," " c_(B)=sqrt((3RT)/(M_(B)))`
Given `bar(c_(A))=c_(B)`.Hence,`(8RT)/(piM_(A))=(3RT)/(M_(B))" or "(M_(A))/(M_(B))=(8)/(3pi)`
Now, if B at temperature T and A is at T', then
`bar(c_(A))=bar(c_(B))" "` (Required condition)
`:. " " sqrt((8RT')/(piM_(A)))=sqrt((8RT)/(piM_(B)))" or "(T')/(M_(A))=(T)/(M_(B))" or "(T')/(T)=(M_(A))/(M_(B))=(8)/(3pi)`
or `T'=(8)/(3pi)xxT`. As `(8)/(3pi)lt1,T'ltT`. Hence, temperature of B should be lowered to `(8)/(3pi)` or 0.85 of temperature T.