For two gases A and B with molecular weights M_(A) and M_(B), it is observed that at a certain temperature T_(1) 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

For two gases A and B with molecular weights M_(A) and M_(B), it is ob

1.	For two gases A and B with molecular weights M_(A) and M_(B), it is observed that at a certain temperature T_(1) 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
Answer» A is at TEMPERATURE T and B at T', `T gt T'` A is lowered to a temperature `T_(2), T_(2) lt T` while B is at T Both A and B are raised to a HIGHER temperature Both A and B are placed at lower temperature Solution :`(U_(AV))_(A)=sqrt((8RT)/(piM_(A))) and (U_(rms))_(B)=sqrt((3RT)/(M_(B)))` `THEREFORE (8)/(3pi) =(M_(A))/(M_(B))` For `A (U_(AV))=sqrt((8RT_(2))/(piM_(A)))" or "B V_(AV)=sqrt((8RT)/(piM_(B)))` `(T_(2))/(T) =(M_(A))/(M_(B))=(8)/(3pi) therefore T_(2) =(8)/(3pi) T" or "T_(2) lt T` Hence, (B) is the correct ANSWER.