Assume that you have a 1000L container of water open to atmosphere; the water is at an average year round temperature of 175 degrees F; the average ambient air temperature is 60 degrees F; and assume the static concentration of formaldehyde in the water is 100 mg/L.
How would I calculate the potential loss of formaldehyde to the atmosphere from this open heated container of water? My sense is that it would be a function of both water and air temperature plus the vapor pressure of the water and formaldehyde?? Also, would it actually be soluble formaldehyde in the water or would it be an organic acid precursor like acetic or formic acid?
Can you please walk me through the calculations so that I clearly understand?
Please see the attached file.
You need to consider the vapor pressure of solution of formaldehyde as well as pure water in this case.
Since, water is 1000L (assuming density of water = 0.97182 g/ml at 352K)
And vapor pressure of water = 355.1 torr = 0.467 atm
And vapor pressure of pure water is = 0.0313 atm ( I am considering pure VP at 298K)
Similarly, for formaldehyde, pure vapor pressure at 352K is 5.12 atm
Temperature of water = 288.7K
I couldn't get the value of VP for formaldehyde at 353 K ( so, I'll ...
The solution calculates formaldehyde losses to atmosphere from a heated and open container of water.