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