For two gases A and B with molecular weights `M_A` and `M_B`, it is observed that at 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 : A is lowered to a temperature `T_(2)= (3pi)/(8)T`, A is lowered to a temperature `T_(2)= (8"/"3pi)T`, B is lowered to a temperature `T_(2)= (3piT)/(8)`, B is lowered to a temperature `T_(2)= (8T)/(3pi)`

Options

A. A is lowered to a temperature `T_(2) = (3pi)/(8) T`
B. A is lowered to a temperature `T_(2) = (8//3pi) T`
C. B is lowered to a temperature `T_(2) = (3piT)/(8)`
D. B is lowered to a temperature `T_(2) = (8T)/(3pi)`

Answer

B

Explanation

`sqrt((8RT)/(pi M_(A))) = sqrt((3RT)/(M_(B))) , (8M_(B))/(pi) = 3 M_(A) therefore M_(A) = (8 M_(B))/(3pi)`
Case - II `sqrt((8 RT_(2))/(pi 8 M_(B)//3pi)) = sqrt((8RT)/(pi M_(B))) therefore T_(2) = (8)/(3pi) T`