LPG is made up of propane, C3H8, and butane, C4H10, either stored seperately or together as a mixture. if a sample of C3H8 gas effuses through a tiny hole in 20 seconds, how long will it take a sample of C4H8 gas to effuse through the same hole under the same conditions?
2007-09-17 00:19:04 · 2 answers · asked by hanjew d 1 in Science & Mathematics ➔ Chemistry
LPG is made up of propane, C3H8, and butane, C4H10, either stored seperately or together as a mixture. if a sample of C3H8 gas effuses through a tiny hole in 20 seconds, how long will it take a sample of C4H8 gas to effuse through the same hole under the same conditions?
can anyone solved these with solutions? and how did it come with that answer?
2007-09-17 00:34:49 · update #1
Rate1/Rate2 = (M2/M1)^0.5
C3H8 M(C3H8)=3*12+8*1=44
C4H10 M(C4H10) =4*12+10*1=58
so Rate (C4H10)/Rate(C3H8) = (M(C3H8)/M(C4H10))/^0.5
(C4H10)/Rate(C3H8) =(44/58)^0.5= 0.871
Effusivity of C4H10= Effusivity C3H8 *0.871
and the tile is inversely propotional to effusivity so
time =20/0.871 =22.96s
2007-09-17 00:39:53 · answer #1 · answered by maussy 7 · 0⤊ 0⤋
use graham's law of diffusion
2007-09-17 00:28:02 · answer #2 · answered by ptolemy862000 4 · 0⤊ 0⤋