I've got a nice example for you:
You have a #V_(i nitial)#-L balloon at room temperature (#22^@C#) and normal pressure (#1 atm#). What will the final volume of the balloon be after you remove 1/3 of the initial number of moles present in the balloon and then double the remaining number of moles in the balloon?
So, let's go about solving this using Avogadro's law; according to this, the volume a gas occupies is proportional to the number of gas molecules present in that respective volume - at constant pressure and temperature.
#V/n = c on st ant -> V_1/n_1 = V_2/n_2#, this describes how volume and number of moles are related for transitions between two stages.
Let's break this problem up into two stages: after the removal of moles (1) and after the addition of moles (2). So, the initial number of moles in the balloon was #n#
#V_(i nitial)/n = V_1/(2/3n) -> V_1 = (2/3n * V_(i nitial))/n = 2/3 V_(i nitial)# - for (1)
The number of moles went from #n# to #(n - 1/3*n) = 2/3n# - notice that the volume decreased to match the drop in the number of moles. Now,
#V_1/(2/3n) = V_2/(2 * 2/3n) -> V_2 = (4/3n * V_1)/(2/3n) = 2V_1# - for (2)
The number of moles doubled - from #2/3n# to #4/3n#, which caused the volume to double. If we express this in terms of #V_(i nitial)# and #n#, we get
#V_(f i nal) = V_2 = 2 * V_1 = 2 * 2/3V_(i nitial) = 4/3 V_(i nitial)# and
#n_(f i nal) = 4/3n#
Again, an overall increase in the number of moles caused the volume to match this increase.
Here's a video of other Avogadro's law examples;
http://socratic.org/chemistry/the-behavior-of-gases/gas-laws/avogadros-law
