For two gases A and B with molecular masses 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. What should be done to the temperature of A so that its mean velocity becomes equal to the mean velocity of B at temperature T ?

For two gases A and B with molecular masses M_(A) and M_(B), it is obs

1.	For two gases A and B with molecular masses 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. What should be done to the temperature of A so that its mean velocity becomes equal to the mean velocity of B at temperature T ?
Answer» Solution :`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.