A 100 meter hollow tube of uniform thickness has two open ends X and Y. Ammonia gas is sent into the tube from end X and hydrogen chloride from end Y, simultaneously. At what distance from end 'X', the gases first meet to form a smoke ring?

A 100 meter hollow tube of uniform thickness has two open ends X and Y

1.	A 100 meter hollow tube of uniform thickness has two open ends X and Y. Ammonia gas is sent into the tube from end X and hydrogen chloride from end Y, simultaneously. At what distance from end 'X', the gases first meet to form a smoke ring?
Answer» SOLUTION :Two gases diffuse in to the pipe from their respective ends. A, B and MEET at a point .O. ACCORDING to Grahm.s law `(r_(1))/(r_(2)) = sqrt((M_(2))/(M_(1))) = sqrt((36.5)/(17)) = 1.47` The RATIO of rate of diffusion is also equal to the ratio of distance travelled. `(1.47)/(1) = (l)/(100-1),` SOlving l = 59.5 The two gases first meet at a distance from end X = 59.5meter