2 Graham’s Law ra = rate at which substance a travels rb = rate at which substance b travelsma = mass of substance amb = mass of substance b
3 Graham’s Law ta = ma tb mb Since rate is distance over time, this equation can also be rearranged as follows:ta = matb mb
4 Graham’s LawThe practical effect of Graham’s Law is that small molecules travel faster than larger molecules. This is true for both diffusion and effusion.
5 Graham’s Law Diffusion – molecules moving through the air Effusion – molecules moving through very small holes, such as the holes in a balloon
6 Graham’s LawWhat does this mean about air leaking out of a helium balloon?
7 Graham’s LawHelium, with a mass of 4 amu will leave the balloon faster than nitrogen and oxygen gases with masses of 28 and 32 amu.
8 Graham’s LawSolving problems with Graham’s law is mostly just algebra.
9 Graham’s LawFor example: A sample of an unknown gas flows through the wall of a porous cup in 39.9 min. An equal volume of gaseous hydrogen, measured at the same temperature and pressure, flows through in 9.75 min. What is the molar mass of the unknown gas?
10 Graham’s LawWe need the equation using time.ta = matb mb
11 Graham’s Law What do we know? ta = 39.9 min tb = 9.75 min ma = x amu mb = 2 amu
12 Graham’s LawPlug into the equation:39.9 = x2
13 Graham’s Law Cross multiply and solve for x 39.9(2) = 9.75(x)
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