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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?

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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

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  • $\begingroup$ Maybe I am misinterpreting your answer, but that is not what I asked - I know that you need to subtract out the vapor pressure of water from the room pressure to correct for the pressure contributed by water vapor, but I want to know why you also do not subtract out the volume contributed by water vapor from the total gas volume. $\endgroup$ – 1110101001 Jan 16 '15 at 23:17
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    $\begingroup$ Any gas occupies the available volume for it. So, for your question water vapor and the gas occupy the same volume. What is different is the partial pressure and the number of moles of each one. $\endgroup$ – Yomen Atassi Jan 17 '15 at 13:59

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