Determination of Mean Free Path of Quantum/Waves and Total Emissivity of Carbon Dioxide Considering its Molecular Cross Section.
University Professor, Scientist, Scientific Research Director at Biology Cabinet©
Monterrey, N. L., Mexico.
This article was updated on April 8, 2011. This article has been Peer Reviewed by the Faculty of Physics of the University of Nuevo Leon, Mexico.
Nahle, Nasif S. Mean Free Path Length of Photons through the Troposphere and Time of Crossing Path of Photons Leaving the Troposphere Without Colliding with a Molecule of Carbon Dioxide. July 29, 2010. Biology Cabinet Online. Academic Resources. Monterrey, N. L.
* The author is grateful to Dr. Jonathan M. Walsh, PhD in mathematics, for his kind assistance with some calculations, and to TS for his kind assistance with the text; nevertheless, any errors in the text are mine alone.
Through the application of astrophysical formulas, the mean free path length of a Quantum/wave stream leaving the surface of the Earth to the outer space before it has collided with a molecule of carbon dioxide and its total emissivity are calculated. The output of this algorithm indicates a value of about 33 meters. Also calculated is the time taken by a Quantum/wave to exit the atmosphere after it has collided with a molecule of carbon dioxide — which is ~4 milliseconds (ms).
SCarbon dioxide (CO2) is vitally important molecule for life on Earth. Carbon dioxide molecules are taken in from the atmosphere by photosynthetic organisms which employ them to build more complex substances that are used for storing the energy transferred from the quantum/waves to the molecules of chlorophyll.
The current assessment demonstrates that CO2 is so well dispersed throughout the bulk volume of the atmosphere that its efficiency for capturing quantum/waves emitted from the surface is extremely low.
The objectives of this didactic article are to demonstrate that the mean free path length of the Quantum/wave stream does not significantly change the total emissivity of the CO2 and that the time taken by a Quantum/wave to exit the atmosphere to space, without colliding with a molecule of CO2, is extremely low.
I have introduced the molecular cross section of the carbon dioxide σCO2mol, in which case a value for n, calculated from the molar density of the carbon dioxide and the number of molecules per mol of the substance was demanded to make the most accurate solution.
The density of the gas carbon dioxide in the atmosphere is obtained by the following formula:
ρCO2 = (12.187 * Molar mass of CO2 * volumetric fraction of CO2) / (276.69 K) = 756 mg/m^3. (Ref. 7)
Where 12.187 is the molar mass of elemental carbon, 44.01 is the molar mass of carbon dioxide, 390 ppmV is the volumetric fraction of CO2 and 276.69 K is its temperature.
To introduce this value into the formula, which I will describe below, I made use of the following density of CO2 in the atmosphere:
ρCO2 = 756 mg/m^3 = 0.756 g/m^3 = 7.56 x 10^-7 g/cm^3
7.56 x 10^-7 g/cm^3 is the density of the atmospheric carbon dioxide in the atmosphere obtained from the following calculation:
ρCO2 = (12.187 of 12-C * MMCO2 * MFCO2) ÷ Tabs
Where 385 ppmV is MFCO2 or mass fracion of atmospheric CO2, 44.01 are MMCO2 or molar mass of CO2, and 276.69 K is Tabs, or Absolute Temperature. You can make your own calculations at LENNTECH (Ref. 7), calculator online.
Mass of CO2 per cm^3 of air = 7.56 x 10^-7 g/cm^3 * 1 cm^3 = 7.56 x 10^-7 g.
Avogadro's number = 6.02 × 10^23 molecules/mol (Avogadro’s number has not units, however, when it is introduced like a constant it is expressed as 6.02 × 10^23 /mol). 13
CO2 Molar Density = 44.01 g / 22261 cm^3 = 0.001977 g/cm^3.
Molar mass of CO2 = 44.01 g
Number of moles of CO2 = (mass in 1 cm^3 of air) / (molar mass) = 7.56 x 10^-7 (g) / 44.01 (g/mol) = 1.7178 x 10^(-8) moles, or simply, 1.7178 x 10^(-8).
Number of molecules of carbon dioxide per gram = (moles * Avogadro’s number) / molar mass = [(1.7178 x 10^-8 mol) * (6.02 × 10^23 molecules/mol)] / 44.01 g = 2.35 x 10^14 molecules.
Considering the current density of CO2 in the atmosphere, the number of moles of carbon dioxide measured in one cubic centimeter of the atmosphere mass, 1.7178 x 10^(-8) moles are contained in 7.56 x 10^(-7) g of carbon dioxide per each cubic centimeter of the current atmosphere.
CO2 molecule’s cross section (σ) = 5 x 10^-22 cm^2
Determination of the Mean Free Path Length of quantum/waves in the atmosphere before colliding with a molecule of carbon dioxide.
Formula to calculate the mean free path length of quantum/waves before colliding with molecules of CO2:
l = m / (n σ) (References 1 and 2)
Where l is for the mean free path length, m is for the mass of the gas measured in one cubic centimeter of air, n is the effective quantum density, and σ is the cross section of a molecule of CO2 before vibrational dephasing (σ = 5.0 x 10^-22 cm^2).
Notice that n is the total number of states per volume unit, cubic centimeters. For this case, to obtain the effective molar density we must to multiply the number of states per mol by the molar gas density of the carbon dioxide.
CO2 molecule’s cross section = 5 x 10^-22 cm^2
Mass of CO2 (m) corresponding to a mass fraction of 390 ppmV= 7.56 x 10^-7 g
The molar gas density of molecular carbon dioxide is obtained by the following formula:
Molar gas ρCO2 = MMCO2 / MVSTP
Where MMCO2 is molar mass of the carbon dioxide and MVSTP is the molar volume of any gas at Standard Temperature and Pressure (22.4 L).
Molar gas density (ρCO2) = 44.01 g / 22.4 L = 1.9647 g /L = 0.0019647 g /cm^3
The following formula is used to calculate the effective molar density of carbon dioxide per cubic centimeter of air:
nmol CO2 = ρmol * NM
Where nmol CO2 is the effective density of one mole, ρmol is the molar gas density, and NM is the number of molecules per mol.
nmol CO2 =?
ρmol = 0.0019647 (g/(cm^3))
NM = 2.35 x 10^14 molecules in one mole of CO2
nmol CO2 = (0.0019647 (g /cm^3)) * (2.35 x 10^14)= 4.617 x 10^11 g/cm^3
Therefore, 4.617 x 10^11 g/cm^3 is the effective molar density of the carbon dioxide that must be considered for calculating the mean free path length of a quantum/wave crossing the troposphere between collisions with molecules of carbon dioxide.
Determination of the Mean Free Path Length of Quantum/Waves through the Earth’s troposphere.
l = m / (n σ) (References 1 and 2)
l = (7.56 x 10^-7 g) / (4.617 x 10^11 g /cm^3 * 5 x 10^-22 cm^2) = 3274.8 cm
l = 32.75 m
Therefore, the mean free path of the surface quantum/waves stream (l) is 3274 cm, which is the trajectory of a quantum/wave passing through the atmosphere before it collides with a molecule of carbon dioxide.
This means that a quantum/wave shifts about 33 meters to hit onto a molecule of carbon dioxide in the atmosphere.
This distance is well understood if we take into account that as the temperature of the lower troposphere increases during daylight by a highly energized surface quantum/wave stream, the thermal diffusivity of carbon dioxide increases, to be precise, as the temperature increases, the molecules diffuse more quickly into a thermally expanded volume of air.
In consequence, the distance between one molecule of carbon dioxide and another molecule of carbon dioxide is lengthened as the temperature increases. For example, the thermal diffusivity of the carbon dioxide at 255 K (-18 °C or -0.40 F) is 7.83 x 10^-6 (m^2/s), while at 308 K (35 °C, or 95 F) is 1.171 x 10^-5 (m^2/s).14
Determination of the time a quantum/wave takes to leave the atmosphere of the Earth towards space after colliding with molecules of CO2:
Formula to obtain the time taken by a Quantum/wave to leave the Earth’s atmosphere after colliding with molecules of CO2:
t = r^2 / (l * c)(References 1 and 2)
rtrop = 700000 cm
l = 5831 cm
c = 2.99909301 x 10^10 (cm/s)
t = r^2 / (l*c) = 4.9 x 10^11 cm^2 / [3274.8 cm * 2.99909301 x 10^10 (cm/s)] = 0.0049 s
t = 0.0049 s
The result indicates that it takes quantum/waves approximately 5 ms (milliseconds) to cross the troposphere from the surface of the Earth after colliding with or being scattered by molecules of CO2.
In contrast, molecules of Water Vapor (H2O) and solid particles which are also present in the atmosphere, intercept and absorb the quantum/waves emitted from the surface of the Earth long before they could collide with any molecules of carbon dioxide and before the molecules of carbon dioxide interact with other molecules:
Determining the Total Emissivity of the Atmospheric Carbon Dioxide as from the Crossing Time Lapse that a Quantum/Wave takes to leave the Troposphere:
The formula to calculate the total emissivity of fluids, plasma and free electrons is as follows:
ε = (1-e^ (t * (- (1/s))) / √π
Where ε is the total emissivity of emitter particles and t is the crossing time lapse of a quantum/wave.
t = 0.0049 s
ε = (1-e^[(0.0049 s * (-1/s))]) / (√3.1415…) = 0.0027
This result coincides with the result obtained from considering the partial pressure of the carbon dioxide and its instantaneous temperature in the atmosphere; additionally, it coincides with the total emissivity of almost zero obtained from the experiments conducted by H. C. Hottel(8), B. Leckner (9), M. Lapp (10), C. B. Ludwig (11), A. F. Sarofim (12) and their collaborators (11, 12).
The results obtained by experimentation coincide with the results obtained by applying astrophysics formulas. Therefore, both methodologies are reliable to calculate the total emissivity/absorptivity of any gas of any planetary atmosphere.
At an average density, the atmospheric water vapor allows quantum/waves to cross the troposphere to the tropopause in 0.0245 s, i.e. 2.45 cs (centiseconds). By comparing the ability of water vapor to avoid that quantum/waves escape towards the outer space (0.5831 s) with the ability of CO2 (0.0049 s), I can affirm that the role of CO2 on warming the atmosphere or the surface is not possible according to Physics Laws.
The water vapor is five times more efficient on intercepting quantum/waves than the carbon dioxide. Therefore, the carbon dioxide in the atmosphere works like a coolant of the atmospheric water vapor.
By considering also that the carbon dioxide has by far a lower total emissivity than the water vapor I conclude that the carbon dioxide has not an effect on climate changes or warming periods on the Earth.
The low thermal diffusivity of carbon dioxide makes of it to be an inefficient substance to adjust its temperature to the temperature of its surroundings. Consequently, the carbon dioxide can never reach the thermal equilibrium with respect to the remainder molecules of the air.
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