Gas law question

Question

Sulfur trioxide, SO3 (g), decomposes into sulfur dioxide gas and oxygen gas according to
the following equation:

2SO3 (g) → 2SO2 (g) + O2 (g)

Assuming the reaction fully completes explain why, when this reaction takes place
at constant pressure and temperature, the volume of the product gases is 50% larger
than the original volume of the SO3.

Answer 1

The volume is proportional to the number of moles of gas present.  Notice that there are 2 moles on the left (reactant side) but 3 moles on the right (product side).  This represents a 50% increase in the number of moles of gas present, thus the volume increases by this amount, if the pressure and temperature remain constant.  Looking at PV = nRT and solving for n (moles of gas), you can see this.

n = PV/RT and if the pressure and temperature remain constant (and R is also a constant), then the volume will increase in proportion to the number of moles (n).