An earlier web page of mine listed absorption
coefficients of interest for atmosphere modeling.
It used all units as given in the original sources, with a few copying mistakes on my part--
sorry about that! Here I step through the math needed to convert everything to a standard unit.
I target the measure I find most congenial, the mass absorption coefficient in m^{2} kg^{-1}.

This source uses cm^{-1} NTP for CO_{2} and O_{3}, but
cm^{2} g^{-1} for H_{2}O.
The H_{2}O conversion is therefore easy:
multiply by 1/10,000 m^{2} cm^{-2} and 1,000 g kg^{-1}, or a factor of 0.1:

μ_{Lo} | μ_{Hi} | k (m^{2} kg^{-1}) |
---|---|---|

50 | inf | 0.18 |

40 | 50 | 0.031 |

20.2 | 29.9 | 1.4 |

18.2 | 20.2 | 3.5 |

9.09 | 10.0 | 0.1 |

6.25 | 9.09 | 0.3 |

5.0 | 6.25 | 1.4 |

3.85 | 5.0 | 20 |

1.25 | 3.85 | 80 |

1.25 | 5.88 | 51 |

NTP, "Normal temperature and pressure," is defined as 293.15 K and 101,325 Pa.
Our first step for CO_{2} and O_{3} would then be to convert from
cm^{-1} to m^{-1} by multiplying by 100.
We now have m^{-1}, and how much CO_{2} is there in one meter at NTP?
The ideal gas law is

P V = n R T

We'll use a volume of 1 m^{3}, reducing this to P = n R T. Solving for n:

n = P / R T = 101325 / (8314.4621 x 293.15) = 0.041571 kmol.
One kmol of CO_{2} has a mass of 44.0098 kg, so we're talking a specific mass of 1.8295 kg m^{-2}.

τ = β s = k SM

k = β s / SM

and since s = 1, k = β / SM, finally giving us, for CO_{2}:

μ_{Lo} | μ_{Hi} | k (m^{2} kg^{-1}) |
---|---|---|

50 | inf | 0.000082 |

40 | 50 | 0.000030 |

29.9 | 40 | 0.037 |

16 | 18.2 | 0.00087 |

15.2 | 16 | 0.00041 |

13.9 | 15.2 | 0.000017 |

12.3 | 13.9 | 0.0028 |

11.4 | 12.3 | 0.052 |

10.9 | 11.4 | 36 |

10.0 | 10.9 | 32 |

9.09 | 10.0 | 0.11 |

6.25 | 9.09 | 0.0026 |

Unfortunately, this gives the lowest figure for the 15-micron absorption band, which should be one of the strongest. Very unlikely, but the error seems to be in the original paper, which itself gives the lowest absorption coefficient for that wavenumber domain.

Lastly, using similar logic, the specific mass for ozone is 1.9953 kg m^{-2} (MW = 47.9982),
giving the ozone absorption coefficient as:

μ_{Lo} | μ_{Hi} | k (m^{2} kg^{-1}) |
---|---|---|

16.0 | 18.2 | 3.8 |

μ_{Lo} | μ_{Hi} | k (m^{2} kg^{-1}) |
---|---|---|

0.175 | 0.225 | 1527 |

0.225 | 0.245 | 9384.1 |

0.245 | 0.28 | 15132 |

0.28 | 0.295 | 2147 |

0.395 | 0.31 | 355.3 |

0.31 | 0.32 | 62.6 |

0.32 | 0.4 | 1.50 |

0.4 | 0.7 | 2.51 |

Using the same reasoning, for STP rather than NTP, and the appropriate molecular weights, Essenhigh's figures become

Gas | μ_{Lo} | μ_{Hi} | k (m^{2} kg^{-1}) |
---|---|---|---|

H_{2}O | 1.8 | 2 | 714 |

2.5 | 3 | 71.4 | |

5 | 8 | 11 | |

12 | 25 | 0.86 | |

CO _{2} | 1.9 | 2.1 | 334 |

2.6 | 2.9 | 71.00 | |

4.1 | 4.5 | 9.356 | |

13 | 17 | 0.754 | |

CH _{4} | 2.2 | 2.5 | 147 |

3.5 | 4.5 | 18.0 | |

6.5 | 7 | 3.2 |

per km per amagat is an odd one. Per km to per meter involves just multiplying by 1/1,000, but what's an amagat? It's a measure of "number density:"

η = n / n_{0}

where η is the number density in amagat (agt) and n_{0} is "Loschmidt's number,"
the number density of an ideal gas at STP (P = 101325 Pa, T = 273.15 K).
The numerical value of n_{0} is 2.6867805 x 10^{25} m^{-3}.

1 m^{3} of pure methane at STP would have a number density of 0.044615 kilomoles.
1 kmol is 1000 N_{A}* or 6.0221413 x 10^{26}, or to put it another way,
the Loschmidt number is 0.044615 kmol--not surprising, since n_{0} is defined at STP.
In other words, we'd have exactly 1 amagat of methane at STP.
The specific mass would be 0.71575 kg m^{-2}. We therefore find:

*N_{A} is Avogadro's number, 6.0221413 x 10^{23}.

μ | k (m^{2} kg^{-1}) |
---|---|

0.83 | 0.000015 |

0.94 | 0.000014 |

1.07 | 0.000007 |

1.28 | 0.000007 |

1.58 | 0.000007 |

2 | 0.000007 |

2.9 | 0.0001 |

This is an easy conversion: multiply by 1/10,000 m^{2} cm^{-2} times 1,000 g kg^{-1}, or a net factor of 0.1.

Gas | μ_{Lo} | μ_{Hi} | k (m^{2} kg^{-1}) |
---|---|---|---|

H_{2}O | 6.33 | 6.50 | 10 |

6.50 | 6.67 | 25 | |

6.67 | 6.85 | 10 | |

10.75 | 11.75 | 0.0000050 | |

21.00 | 25.00 | 1 | |

25.00 | 30.00 | 4 | |

30.00 | 40.00 | 16 | |

40.00 | 125.00 | 40 | |

O _{3} | 8.90 | 9.35 | 0.0063 |

9.35 | 9.9 | 0.16 | |

9.9 | 10.1 | 0.0063 | |

CO _{2} | 14.0 | 14.5 | 0.04 |

14.5 | 15.5 | 0.2 | |

15.5 | 16.0 | 0.04 |

My first absorbers web page incorrectly gave the absorption coefficient units from this source as cm^{-1}, whereas the original paper actually says cm^{-1} atm^{-1}. There are a bunch of different absorbers here, so I summarize the findings below:

Absorber | μ | MW | k (m^{2} kg^{-1}) |
---|---|---|---|

CCl_{2}F_{2} (CFC-12) | 10.8 | 120.9138 | 860 |

CCl_{3}F (CFC-11) | 11.8 | 137.3684 | 1900 |

CCl_{4} (carbon tetrachloride) | 12.6 | 152.823 | 1900 |

C_{2}Cl_{4} (perchloroethylene) | 10.9 | 165.834 | 430 |

CHClF_{2} (CFC-22) | 8.945 | 86.4687 | 2200 |

CH_{3}CCl_{3} (methyl chloroform) | 9.174 | 133.4048 | 270 |

C_{2}H_{2} (acetylene) | 13.9 | 26.0379 | 21000 |

C_{2}H_{4} (ethylene) | 10.5 | 28.0538 | 2500 |

C_{N}H_{2N+2} (paraffinic carbon) | 3.367 | 30.0696* | 310 |

COS (carbonyl sulfide) | 4.866 | 60.0764 | 2800 |

N_{2}O (nitrous oxide) | 3.883 | 44.0128 | 32 |

N_{2}O (nitrous oxide) | 8.688 | 44.0128 | 8.1 |

This requires some logic. Units are cm^{-1} of precipitable water. What the heck does that mean?

Precipitable water is the depth of liquid if the water vapor in the air rained out.
A 1 cm depth of water covering a square meter would have a volume of 0.01 m^{-3}, this would contain 10 kg.
So 1 precipitable cm would be 10 kg to the square meter, or 0.1 m^{2} kg^{-1}.
To find absorption coefficient in m^{2} kg^{-1}, then,
we multiply precipitable cm^{-1} by 0.1:

μ_{Lo} | μ_{Hi} | k (m^{2} kg^{-1}) |
---|---|---|

12 | 13 | 0.025 |

13 | 14 | 0.084 |

14 | 15 | 0.13 |

15 | 16 | 0.165 |

16 | 17 | 0.44 |

17 | 18 | 1.72 |

This is an easy conversion. Multiply by 1,000 g kg^{-1} to find m^{2} kg^{-1}:

Gas | λ (μ) | k (m^{2} kg^{-1}) |
---|---|---|

H_{2}O | 27.97 | 145 ± 8 |

This source gives the absorption cross-section of molecular nitrogen (N_{2}) as σ = 2 x 10^{-21} cm^{2} at wavelengths in the far-ultraviolet of 0.116-0.145 μ. This is an area of 2 x 10^{-25} m^{2}. One molecule of nitrogen has a mass of μ_{N2} = 28.0134 times 1 atomic mass unit, 1.660538921 x 10^{-27} kg, or 4.65173 x 10^{-26} kg. The cross-section for a whole kilogram, then, would be σ divided by this mass:

μ_{Lo} | μ_{Hi} | k (m^{2} kg^{-1}) |
---|---|---|

0.116 | 0.145 | 4.3 |

This is another easy conversion. Change cgs units to mks.

a = 0.42 m k(ν, 296K) = a + b exp(-β ν) k(ν, T) = k(ν, 296K) exp [T where T |

For this source we use the cross-section logic. The mass of one molecule of CFC-12 is 2.00782 x 10^{-25} kg, so the absorption coefficients become:

λ (μ) | k (m^{2} kg^{-1}) |
---|---|

9 | 37,830 |

11 | 28,640 |

A pretty powerful greenhouse gas!

This source uses pure reciprocal cm, which we first change to reciprocal m, for β = 68,000 and 276,000 m^{-1} at 0.076 and 0.0661-0.0796 μ, respectively.

A meter of nitrogen N_{2} at STP holds 1.24982 kg of nitrogen. The absorption coefficients therefore become:

μ_{Lo} | μ_{Hi} | k (m^{2} kg^{-1}) |
---|---|---|

0.076 | - | 54,000 |

0.0661 | 0.0796 | 221,000 |

Normally innocuous nitrogen is dead black in the far ultraviolet!

This team measured CO_{2} absorption coefficients at peak wavelengths in the near infrared.
Their absorption coefficients are in the rare units of m^{2} mol^{-1},
where 1 mole of particles is Avogadro's number, 6.0221413 x 10^{23}.
A mole of CO_{2} has a mass of 0.0440098 kg, so these figures become:

λ_{PEAK} (μ) | k (m^{2} kg^{-1}) |
---|---|

1.437 | > 230 |

1.955 | 5.7 |

2.013 | 9.8 |

2.060 | 9.8 |

This source uses the same units as Kondratiev and Niilisk 1960, for water vapor under the same conditions. We therefore apply the same conversion factor:

μ_{Lo} | μ_{Hi} | k (m^{2} kg^{-1}) |
---|---|---|

12 | 13 | 0.044 |

13 | 14 | 0.052 |

14 | 15 | 0.063 |

15 | 16 | 0.076 |

16 | 17 | 0.094 |

17 | 18 | 0.129 |

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Page created: | 12/15/2014 |

Last modified: | 12/15/2014 |

Author: | BPL |