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Gas solubility

Gas solubility

Gas solubility

(OP)
Hello,

I'm trying to get the concentration (mg/L) of N2, O2 and Ar in air equilibrated water at 1000.9mbar, 20C and 65% humidity  (clear water, salinity~0) using the following literature B and k (Bunsen coef and Henry's coef):

N2
B=15.57 cc.liter-1 atm-1.
k=7.97atm*10-4.

O2
B=29.86 cc.liter-1 atm-1.
k=4.15atm*10-4.

Ar
B=34.03cc.liter-1 atm-1.
k=3.65atm*10-4.

I confuse the unities, once and again, and again, and again...

Which would be the conc (mg/L) if I dissolve 10mL of air in excess (by hydrostatic pressure).

thanks a lot,
merely

RE: Gas solubility

merely:

Your units for Henry's coefficient don't make sense.  If we define the Henry's coefficient (H) as:

H = p/c

where:
p = partial pressure of a gas above water, in atm
c = concentration of a gas dissolved in water, in gram-mols/L (i.e., molarity)

The H for oxygen in water at 25 °C as obtained from three different sources, including Perry's Chemical Engineers Handbook, 6th Edition, is :

H = 984.4 atm/(mol/L)

where:
atm = oxygen partial pressure above water, in atmospheres
mol/L = gram-mols of oxygen dissolved in 1 liter of water = molarity

That H can be converted to these units if you prefer:

H = 43,800 atm/(mol of oxygen/mol of solution)

where we can assume that, for all practical purposes, a mol of solution equals a mol of water since Henry's Law only applies to very dilute solutions.

You must be very careful when looking up Henry's coefficients because it is often defined as:

H' = c/p

As you can see, H and H' are the inverse of each other ... so you must make sure which one you've looked up.

I strongly suggest that you read the discussion of Henry's Law at: http://en.wikipedia.org/wiki/Partial_pressure#Henry.27s_Law_and_the_solubility_of_gases <== click here

As for your question:

Quote:

Which would be the conc (mg/L) if I dissolve 10mL of air in excess (by hydrostatic pressure).
I simply don't understand what you are asking.  Are you looking for the concentration of air dissolved in water (mg of air in one liter of water) when you dissolve 10 milliliters (i.e., ml) of air in water? Is that what you meant? I don't understand what you meant by "in excess". Could you please repeat your question more clearly?

Milton Beychok
(Visit me at www.air-dispersion.com)
.

RE: Gas solubility

(OP)
Dear Milton,

I know they are not the usual units. I took them from:
Benson, B.B. and Krause, D.Jr. 1976. "Empirical laws for dilute aqueous solutions of nonpolar gases." The Journal of Chemical Physics, vol.64, no.2, 689-709.  

What I mean by "extra 10mL of air dissolved in water" is the concentration we would obtain if to a water equilibrated with the atmosphere we would add 10mL of air (atmosphere) and force to be dissolved in the water (i.e. increasing the hydrostatic pressure).
This is a concept known as Excess Air for water infiltrating the subsurface, where small air bubbles can be trapped by the water and dissolved as the water continues its way downwards.

Thanks a lot,
merely

RE: Gas solubility


Mbeychok: there are a variety of units to express Henry's law constants. The units brought by merely are indeed logical if c in H=p/c, represents mol fractions (x).

The only error in merely's quoted H values is in the exponents; eg, for nitrogen at 20oC, the value would be 7.97×104 atm, not as reported.

Thus for air at normal atmospheric pressure, nitrogen's  dissolved mol fraction at 20oC would be:

Nitrogen → x = 0.781÷7.97×104 = 9.8×10-6

Now, 9.8 ×10-6 = (gN2÷28)÷(1000÷18) →

gN2 = (9.8×10-6×28×1000)÷18 = 0.015 g/L = 15 mg/L

The above estimation assumes 78.1% vol nitrogen (MW =28) in air.

In a similar manner for oxygen (20.9%) and argon (0.93%). Do you agree ?

RE: Gas solubility


BTW, the Ostwald absorption coefficients β are expressed in v/V at the T and P conditions at which the gas is dissolved.

For Ar → α =  34.03 mL/(L.atm) = 0.034 L/(L.atm) →
β = 0.034×1 atm×293÷273 = 0.0365 at 20oC and 1 atm.

In all of the above I took the density of water = 1000 g/L.

RE: Gas solubility

25362:

I know that there are literally a dozen or so sets of units which can and are used.

The reason I said his units made no sense are:

(1) The exponents were negative rather than positive, just as you have also noted ... and obviously that makes a big difference.
(2) His values should have been expressed as "atm/mol fraction" or as "atm/(mol of gas/mol of solution)".  This is just another case of failing to express units clearly and correctly.

Regards,

Milton Beychok
(Visit me at www.air-dispersion.com)
.

RE: Gas solubility

(OP)
Hello,

how to translate Henry's K in mol kg-1 atm-1 units to Pascals??

thanks,
merely

RE: Gas solubility


merely: kindly pay attention to point (2) in mbeychok's posting. Henry's law constant has again been erroneously expressed. It should be : (atm)(mol/kg)-1, or (atm)(mol/L)-1.




RE: Gas solubility

(OP)
Hello,

data is from Warner and Weiss, 1985 (Solubilities of chlorofluorocarbons 11 and 12 in water and seawater).
It represents their experimental results for K'.

thanks

RE: Gas solubility


I don't have access to that source. Could it be that it shows H-1 ?

RE: Gas solubility

(OP)
Hello,

well, taking it as if it would be H' I get that the solubility for CFC-11, CFC-12 and CFC-113 are:
CFC-11: 70618140.5Pascal
CFC-12: 286355720.1Pa
CFC-113: 181164133.09Pa

They seem correct, but not sure if they are corect!!

Thanks,

RE: Gas solubility

(OP)
I forgot to tell that this is for 25C and 1 atm data

thanks,
merely

RE: Gas solubility


Henry's constant usually expresses the ratio of vapor pressure to solubility. I'm now aware of the fact that the constant is sometimes expressed in the formula c = H' × p instead of H × c = p.

Whatever H or H' values are selected you must indicate to what concentration units are they related: mol/kg, mass/mass, mol fraction, mg/L, mol/m3, etc., as mbeychok clearly indicated.

Whenever H is expressed in pressure units -as in the case in hand- it may be that the concentration units are given in mol fractions as in the example above, or in mass/mass, or in mol/mol units.

I pressume this time it is mass/mass. Thus, for CFC-12
c = 101,325÷286,355,720 = 3.538 × 10-4 g/g or 0.354 g/L. Where the gas pressure is assumed to be 1 atm = 101,325 Pa.

I hope I'm not wrong. Please check and report. Thanks.

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