(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

RH = actual VP/Saturated VP

Density = mass/volume

3. 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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# Homework Help: How much water can evaporate?

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