Collection of gas over water When you collect a gas over water, you have to account for the mixed in water vapor by subtracting the room pressure from water vapor pressure. However, why do you not do something similar for volume, where you subtract the total measured volume from the volume of water? 

How can $PV = nRT$ still hold when the pressure you are using is only for the collected gas whereas the volume is for the gas + water vapor volume?
You are right! There is an approximation here in only considering the vapor pressure of the gas. But let's consider the value of vapor pressure of water at $298\ce{K}$:  it's  $13.85\,\ce{mm Hg}$. So, if you neglect the vapor pressure of water, the relative uncertainty is  $$\frac{13.85}{760}\times100=1.83\%$$
As you can see, it's negligible.

P.S. To calculate the vapor pressure of water, I used the first equation in http://en.wikipedia.org/wiki/Vapour_pressure_of_water