For two gases `A` and `B` with molecular weights `M_(A)` and `M_(B)`, respectively, it is observed that at a certain temperature `T`, the mean velocity of `A` is equal to the `V_(rms)` of `B`. Thus, the mean velocity of `A` can be made equal to the mean velocity of `B`, if

Options

A. `P` is lowered to a temperature `T_(2)` and `T_(2) lt T and Q` is maintained at temperature `T`
B. `P` is at a temperature `T and Q` at a temperature `T_(2)` where `T gt T_(2)`
C. both `p` and `Q` are raised to higher temperature
D. both `P and Q` are placed at lower temperature

Answer

A

Explanation

Average velocity `= sqrt((8RT)/(pi M))`
Root mean sqyare velocity `= sqrt((3RT)/(pi M))`
`rArr` Average velocity of `P` at temperature `T= u_(av(p))`
`=sqrt((8RT)/(pi M_(P)))`
Root mean square velocity `Q` at temperature `T`
`=sqrt((3RT)/(M_(Q)))= u_(RMS(Q))`
`implies sqrt((8RT)/(piM_(P)))=sqrt((3RT)/(M_(Q)))implies sqrt((8xx7)/(22M_(p)))=sqrt((3)/(M_(Q)))`
`implies (56)/(22M_(p))=(3)/(M_(Q))implies (56)/(66)M_(Q)`
Let the temperature of `Q` be maintained at `T` and that of `P` be changed to `T_(2)`
`implies` Now `RMS` of both `P and Q` should be equal
`implies sqrt((3RT_(2))/(M_(P)))=sqrt((3RT)/(M_(Q)))implies (T_(2))/(M_(P))=(T)/(M_(Q))`
Since `M_(P)=(56)/(66)M_(Q)`
`implies (66T_(2))/(56M_(Q))=(T)/(M_(Q))implies 66T_(2)= 56T`
`T_(2)= (56)/(66) T, T_(2) lt T`