Lecture 4 Gases and Gas Exchange
Composition of the atmosphere
Gas solubility
2009
Gas Exchange Fluxes
Effect of wind
Global CO2 fluxes by gas exchange
Emerson and Hedges: Chpts 3 and 10
Sarmiento and Gruber (2002) Sinks for Anthropogenic Carbon
Physics Today August 2002 30-36
Composition of the Atmosphere
More than 95% of all gases except radon reside in the atmosphere.
The atmosphere controls the oceans gas contents for all gases except radon, CO2 and H2O.
Gas
Mole Fraction in Dry Air (fG)
molar volume at STP (l mol-1 )
where fG = moles gas i/total moles
22.414 for an ideal gas (0°C)
N2
O2
Ar
CO2
Ne
He
0.78080
0.20952
9.34 x 10-3
3.3 x 10-4
1.82 x 10-5
5.24 x 10-6
H2O
~0.013
Why is dry air used?
22.391
22.385
22.386
22.296
22.421
22.436
Some comments about units of gases:
In Air
Pressure - Atmospheres
1 Atm = 760 mm Hg
Partial Pressure of Gasi = P (i) /760
In Water
Volume - liters gas at STP / kgsw
STP = standard temperature and pressure
= 1 atm and 0C (= 273ºK)
Volume - liters gas / liters air
(ppmv = ml / l, etc)
Moles - moles gas / kgsw
Conversion: lgas/kgsw / lgas / mole = moles/kgsw
(~22.4 l/mol)
Dalton's Law
Gas concentrations are expressed in terms of pressures.
Total Pressure = SPi = Dalton's Law of Partial Pressures
PT = PN2 + PO2 + PH2O + PAr + .........
Dalton's Law implies ideal behavior -- i.e. all gases behave independently
on one another (same idea as ideal liquid solutions with no electrostatic interactions).
Gases are dilute enough that this is a good assumption.
Variations in partial pressure (Pi) result from:
1) variations in PT (atmospheric pressure highs and lows)
2) variations in water vapor ( PH2O)
We can express the partial pressure (Pi) of a specific gas on a dry air basis as follows:
Pi = [ PT - h/100 Po ] fg
where Pi = partial pressure of gas i
PT = Total atmospheric pressure
h = % relative humidity
Po = vapor pressure of water at ambient T
fg = mole fraction of gas in dry air (see table above)
Example:
Say we have a humidity of 80% today and the temperature is 15C
Vapor pressure of H2O at 15C = Po = 12.75 mm Hg (from reference books)
Then, PH2O = 0.80 x 12.75 = 10.2 mm Hg
If PT = 758.0 mm Hg
PTDry = (758.0 - 10.2) mm Hg
= 747.8 mm Hg
Then: fH2O = PH2O / PT = 10.2 / 758.0 = 0.013
So for these conditions H2O is 1.3% of the total gas in the atmosphere.
That means that water has a higher concentration than Argon (Ar).
This is important because water is the most important greenhouse gas!
Example: Units for CO2
Atmospheric CO2 has increased from 280 (pre-industrial) to 380 (present) ppm.
In the table of atmospheric concentrations (see slide 3)
fG,CO2 = 3.3 x 10-4 moles CO2/total moles
= 330 x 10-6 moles CO2/total moles
= 330 ppm
This can also be expressed in log form as:
= 100.52 x 10-4 = 10-3.48
Example: Units for Oxygen
Conversion from volume to moles
Use O2 = 22,385 L / mol at standard temperature and pressure (STP)
if O2 = 5.0 ml O2/LSW
then
5.0 ml O2 / Lsw x mol O2 / 22,385 ml = 0.000223 mol O2 / Lsw
= 223 mmol O2 / Lsw
Solubility
The exchange or chemical equilibrium of a gas between gaseous and liquid phases
can be written as:
A (g)  A (aq)
At equilibrium we can define the familiar value
K = [A(aq)] / [A(g)]
There are two main ways to express solubility (Henry’s Law and Bunsen Coefficients).
1. Henry's Law:
We can express the gas concentration in terms of partial pressure using the
ideal gas law:
PV = nRT
P = pressure, V = volume, n = # moles
R = gas constant = 8.314 J K-1 mol-1, T = temp
so that the number of moles n divided by the volume is equal to [A(g)]
n/V = [A(g)] = PA / RT where PA is the partial pressure of A
Then
K = [A(aq)] / PA/RT
or
[A(aq)] = (K/RT) PA
[A(aq)] = KH PA
in mol kg-1
units for K are mol kg-1 atm-1;
for PA are atm
Henry's Law states that the concentration of a gas in water is proportional
to its overlying partial pressure.
KH is mainly a function of temperature with a small impact by salinity.
Example (Solubility at 0C):
Partial Pressure = Pi = fG x 1atm total pressure
Gas
Pi
KH (0C , S = 35)
Ci (0C, S = 35; P = 760 mm Hg)
N2
O2
Ar
CO2
0.7808
0.2095
0.0093
0.00033
0.80 x 10-3
1.69 x 10-3
1.83 x 10-3
63 x 10-3
624 x 10-6 mol kg-1
354 x 10-6
17 x 10-6
21 x 10-6
Example
The value of KH for CO2 at 24C is 29 x 10-3 moles kg-1 atm-1 or 2.9 x 10-2 or 10-1.53.
The partial pressure of CO2 in the atmosphere is increasing every day
but if we assume that at some time in the recent past it was 350 ppm
that is equal to 10-3.456 atm.
See Emerson and Hedges: Table 3.6 for 20°C and Table 3A1.1 for regressions fpr all T and S
Example: What is the concentration of CO2 (aq) in equilibrium with the atmosphere?
For PCO2 = 350 ppm = 10-3.456
For CO2 KH = 29 x 10-3 = 2.9 x 10-2 = 10-1.53 moles /kg atm
then
CO2 (aq) = KH x PCO2 = 10-1.53 x 10-3.456 = 10-4.986 mol kg-1
= 10+0.014 10-5 = 1.03 x 10-5 = 10.3 x 10-6 mol/l at 25C
The concentration of CO2(aq) will be dependent only on PCO2 and temperature.
It is independent of pH.
Summary of trends in solubility:
1. Type of gas:
KH goes up as molecular weight
goes up (note that CO2 is anomalous)
2. Temperature:
Solubility goes up as T goes down
3. Salinity:
Solubility goes up as S goes down
Temperature control
on gas concentrations
O2 versus temperature
in surface ocean
solid line equals saturation
for S = 35 at different
temperatures
average supersaturation
≈ 7 mmol/kg (~3%)
Causes of deviations from Equilibrium:
Causes of deviation from saturation can be caused by:
1. nonconservative behavior (e.g. photosynthesis (+) or respiration (-)
or denitrification (+))
2. bubble or air injection (+)
3. subsurface mixing - possible supersaturation due to non linearity of
KH or a vs. T.
4. change in atmospheric pressure - if this happens quickly, surface
waters cannot respond quickly enough to reequilibrate.
Rates of Gas Exchange
Stagnant Boundary Layer Model.
ATM
OCN
Depth (Z)
CSW
well mixed atmosphere
Cg = KH Pgas = equil. with atm
0
Stagnant Boundary
Layer –
transport by
ZFilm
molecular diffusion
well mixed surface SW
Z is positive downward
C/ Z =
F = + (flux into ocean)
Flux of Gas
The rate of transfer across this stagnant film
occurs by molecular diffusion from the region
of high concentration to the region of low concentration.
Transport is described by Fick's First Law
which states simply that flux is proportional to
the concentration gradient..
F = - D d[A] / dZ
where
D = molecular diffusion coefficient in water (= f (gas and T)) (cm2sec-1)
dZ is the thickness of the stagnant film on the ocean side (Zfilm)(cm)
d[A] is the concentration difference across the stagnant film (mol cm-3)
The water at the top of the stagnant film (Cg) is assumed to be
in equilibrium with the atmosphere. We can calculate this value
using the Henry's Law equation for gas solubility.
The bottom of the film has the same concentration as the mixed-layer (CSW).
Thus: Flux = F = - D/Zfilm (Cg - CSW) = - D/Zfilm (KHPg - CSW)
Because D/Zfilm has velocity units, it has been called the Piston Velocity (k)
e.g., D = cm2 sec-1
Z film = cm
Typical values are D = 1 x 10-5 cm2 sec-1 at 15ºC
Zfilm = 10 to 60 mm
Example:
D = 1 x 10-5 cm2 sec-1
Zfilm = 17 mm determined for the average global ocean using 14C data
Thus Zfilm = 1.7 x 10-3 cm
The piston velocity = D/Z = k = 1 x 10-5 cm s-1 /1.7 x 10-3 cm
= 0.59 x 10-2 cm/sec
5m/d
note: 1 day = 8.64 x 104 sec
Each day a 5 m thick layer of water will exchange its gas with the atmosphere.
For a 100m thick mixed layer the exchange will be completed every 20 days.
Gas Exchange and Environmental Forcing: Wind
Wanninkhof, 1992
from 14C
100
Tracer N-2000
Tracer Wat-91
Tracer-Wann
Rn
C-14bomb
C-14 nat
k, 600 W-92
k, 600 L&M
S-86
W-99
k, 600 (cm/hr)
80
~ 5 m d-1
60
40
Liss and Merlivat,1986
from wind tunnel exp.
20
0
0
20 cm hr-1 = 20 x 24 / 102 = 4.8 m d-1
5
10
U
10
(m/s)
15
20
U-Th Series Tracers
222Rn
Example Profile from
North Atlantic
Does Secular Equilibrium Apply?
t1/2 222Rn << t1/2 226Ra
(3.8 d)
222Rn
(1600 yrs)
YES!
A226Ra = A222Rn
226Ra
Why is 222Rn activity
less than 226Ra?
222Rn
is a gas and the 222Rn concentration in the atmosphere
is much less than in the ocean mixed layer ( mixed layer).
Thus there is a net evasion of 222Rn out of the ocean.
The 222Rn balance for the mixed layer, ignoring horizontal
advection and vertical exchange with deeper water, is:
222Rn/dt = sources – sinks
= decay of 226Ra – decay of 222Rn - gas exchange to atmosphere
ml l222Rn [222Rn]/t =  ml l226Ra [226Ra] –  l 222Rn [222RnML]
+ D/Zfilm { [222Rnatm] – [222RnML]}
Knowns: l222Rn, l226Ra, DRn
Measure:  ml, A226Ra, A222Rn, d[222Rn]/dt
Solve for Zfilm
ml l222Rn d[222Rn]/dt =  ml l226Ra [226Ra] – ml l222Rn [222Rn]
+ D/Zfilm { [222Rnatm] – [222RnML]}
ml δA222Rn/ δt = ml (A226Ra – A222Rn) + D/Z (CRn, atm – CRn,ML)
atm Rn = 0
for SS = 0
Then
-D/Z ( – CRn,ml) = ml (A226Ra – A222Rn)
+D/Z (ARn,ml/lRn) = ml (A226Ra – A222Rn)
+D/Z (ARn,ml) = ml lRn (A226Ra – A222Rn)
ZFILM = D (A222Rn,ml) / ml lRn (A226Ra – A222Rn)
ZFILM = (D / ml lRn
)
(
1
)
A226 Ra
1
A222 Rn
Stagnant Boundary Layer Film Thickness
Z = DRn /  l 222Rn
(1/A226Ra/A222Rn) ) - 1
Histogram showing results of film thickness
calculations from many stations.
Organized by Ocean and by Latitude
Average Zfilm = 28 mm
Q. What are limitations of
this approach?
1. unrealistic physical model
2. steady state assumption
One of main goals of JGOFS was
to calculate the CO2 flux across
the air-sea interface
Flux = F = - D/Zfilm (Cg - CSW) = - D/Zfilm (KHPg - CSW)
= - D/Zfilm (KHPo – KHPSW)
= -D/Zfilm KH (Po – PSW)
Expression of Air -Sea CO2 Flux
Magnitude
Mechanism
Apply over larger space time domain
k-transfer velocity
S – Solubility
From Sc # & wind speed
From SST & Salinity
F = k s (pCO2w- pCO2a) = K ∆ pCO2
pCO2w
From measurements
and proxies
pCO2a
From CMDL
CCGG network
Global Map of Piston Velocity (k in m yr-1) times CO2 solubility (mol m-3) = K
from satellite observations (Nightingale and Liss, 2004 from Boutin).
∆pCO2 fields
Overall trends known:
* Outgassing at low latitudes (e.g. equatorial)
* Influx at high latitudes (e.g. circumpolar)
* Spring blooms draw down pCO2 (N. Atl)
* El Niños decrease efflux
JGOFS Gas Exchange Highlight #4 ∆pCO2 fields:Takahashi climatology
Monthly changes in pCO2w
Fluxes: JGOFS- Global monthly fluxes
Combining pCO2 fields with k: F = k s (pCO2w- pCO2a)
On first order flux and ∆pCO2 maps do not look that different
CO2 Fluxes: Status
Do different parameterizations between gas exchange and wind matter?
Global uptakes
Liss and Merlivat-83:
Wanninkhof-92:
Wanninkhof&McGillis-98:
1 Pg C yr-1
1.85 Pg C yr-1
2.33 Pg C yr-1
Zemmelink-03:
2.45 Pg C yr-1
Yes!
Global average k (=21.4 cm/hr):
2.3 Pg C yr-1
We might not know exact parameterization with forcing but forcing is
clearly important
Compare with net flux of 1.3 PgCy-1 (1.9 - 0.6)
in Sarmiento and Gruber (2002), Figure 1
2. Bunson Coefficients
Since oceanographers frequently deal with gas concentrations not only in molar
units but also in ml / l, we can also define
[A(aq)] =
a PA
where a = 22,400 x KH (e.g., one mol of gas occupies 22,400 cm3 at STP)
a is called the Bunsen solubility coefficient. Its units are cm3 mol-1.
Solubilities of Gases in Seawater
from Broecker and Peng, (1982)
Bunson
Coefficient
Solubility increases with
mole weight and
decreasing temperature
Henry’s Law
KH = 29 x 10-3
= 2.9 x 10-2
= 10-1.53
Concentration ratio for
equal volumes of air and
water.
Gas Solubility - CFCs
Is beer carbonated?
Calculate the flux of CO2 (in mol m-2 s-1) out of your favorite, frosty,
carbonated beverage.
The PCO2 in the beverage = 0.125atm
Assume the surface of the beverage is in equilbrium with the PCO2 of the
atmosphere (375 x 10-6 atm.).
Let DCO2= 2 x 10-9 m2 s-1 and let the stagnant boundary layer thickness be Zfilm
= 5 x 10-5 m.
What is the flux? (ans: 0.144 x 10-1 mol m-2 s-1)
Which way does the CO2 flux go? (ans: out of the beer)
Effect of El Nino on ∆pCO2 fields
High resolution pCO2 measurements in the Pacific since Eq. Pac-92
Eq Pac-92 process study
PCO2sw
Always greater
than atmospheric
Cosca et al. in press
-2
0
El Nino Index
1
2
3
4
19 85
19 90
19 95
year
20 00
ENSO INDEX (MEI)
-1