Graham’s Law
ra = mb
rb ma
Graham’s Law
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ra = rate at which substance a travels
rb = rate at which substance b travels
ma = mass of substance a
mb = mass of substance b
Graham’s Law
Since rate is distance over time, this
equation can also be rearranged as
follows:
ta = ma
tb mb
Graham’s Law
The practical effect of Graham’s
Law is that small molecules
travel faster than larger
molecules. This is true for both
diffusion and effusion.
Graham’s Law
Diffusion – molecules moving through the
air
Effusion – molecules moving through very
small holes, such as the holes in a
balloon
Graham’s Law
What does this mean about air
leaking out of a helium balloon?
Graham’s Law
Helium, with a mass of 4 amu will leave
the balloon faster than nitrogen and
oxygen gases with masses of 28 and 32
amu.
Graham’s Law
Solving problems with Graham’s law is
mostly just algebra.
Graham’s Law
For 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?
Graham’s Law
We need the equation using
time.
ta = ma
tb mb
Graham’s Law
What do we know?
ta = 39.9 min
tb = 9.75 min
ma = x amu
mb = 2 amu
Graham’s Law
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Plug into the equation:
39.9 = x
9.75 2

Graham’s Law
Cross multiply and solve for x
39.9(2) = 9.75(x)
56.43 = 9.75(x)
5.78 = (x)
5.782 = x
33.49 = x