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

Options

A. A is increased to a temperature `T_2 = (3 pi)/8 T`
B. A is lowered to a temperature `T_2 = (8T)/(3pi)`
C. B is increased to a temperature `T_2 = (3pi T)/8`
D. B is lowered to a temperature `T_2 = (8T)/(3pi)`

Answer

B, C

Explanation

`(barC)_A = ( C)_B`
`implies sqrt((8RT)/(pi M_A)) = sqrt((3RT)/(M_B)) implies 8/(pi M_A) = 3/(M_B) `
Case (i) changing T of A
`(barC_A) = (barC)_B implies sqrt((8RT)/(pi M_A)) = sqrt((8RT_2)/(pi M_B))`
`implies (T)/(M_A) = (T_2)/(M_B) implies T_2 = (M_B)/(M_A) T implies T_2 = (3pi)/(8) T > T.`