How much water can evaporate from a room with given humidity and temperature?

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Homework Statement


If the humidity in a room of volume 680m^3 at 25C is 80%, what mass of water can still evaporate from an open pan?

Homework Equations



RH = actual VP/Saturated VP
Density = mass/volume

The Attempt at a Solution



I have no idea. I'm assuming there is some external information I'm supposed to use, but I don't know what it is. The answer to this problem is apparently 3.1kg.

RH = actual/saturated, so Actual vapor pressure = saturated*RH = 23.8 torr * 0.8 = 19.04torr.

Now I'm assuming you use the density equation, but do I use the density of water? Water vapor? dry air?

How do I relate density back to the actual pressure?

EDIT: Another thought... using PV=nRT?
If I do a whole lot of converting I get P=0.025atm, V = 680,000L, and T = 298K, where R = 0.0821 L*Atm/Mol*K.
Even at that, I still end up with 12.6kg of water, which is still wrong...

Considering in my review section this is labeled as a "beginner review" problem, I have to believe I'm missing something major here...
 
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to get to maximum saturation I still need an additional 4.76 torr of pressure, but I've run it through using those numbers and I'm getting further away from the right answer.
 
TrpnBils said:
to get to maximum saturation I still need an additional 4.76 torr of pressure, but I've run it through using those numbers and I'm getting further away from the right answer.

I think your 4.76 torr partial pressure is correct. Check your conversion of this into atmospheres. How did you do this?
 
4.76 torr / 760 = 0.006 atm

I'm about 90 minutes into this problem and finally figured out how to do it about 5 minutes ago. Part of it was that I was looking for an answer of 3.1kg when I should have been looking for 3100 on the calculator since the conversion from moles turns it into grams. I ended up with 174 moles of water after a couple of attempts (both in units of atmospheres and liters as well as a shot with units of pascals and cubic meters). when I multiplied that by 18g/mol for water I would end up with 3132 kilograms (or so I thought)...but I still needed to convert that.

I look at this and wonder why the heck I couldn't see this obvious mistake from the get-go...it's really not that hard.