Interesting things about CO2.zip_global warming

<html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta charset="utf-8"> <meta name="generator" content="pdf2htmlEX"> <meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1"> <link rel="stylesheet" href="https://static.pudn.com/base/css/base.min.css"> <link rel="stylesheet" href="https://static.pudn.com/base/css/fancy.min.css"> <link rel="stylesheet" href="https://static.pudn.com/prod/directory_preview_static/6266fc444c65f41259478d36/raw.css"> <script src="https://static.pudn.com/base/js/compatibility.min.js"></script> <script src="https://static.pudn.com/base/js/pdf2htmlEX.min.js"></script> <script> try{ pdf2htmlEX.defaultViewer = new pdf2htmlEX.Viewer({}); }catch(e){} </script> <title></title> </head> <body> <div id="sidebar" style="display: none"> <div id="outline"> </div> </div> <div id="pf1" class="pf w0 h0" data-page-no="1"><div class="pc pc1 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="https://static.pudn.com/prod/directory_preview_static/6266fc444c65f41259478d36/bg1.jpg"><div class="t m0 x1 h2 y1 ff1 fs0 fc0 sc0 ls0 ws0">Determination of Mean Free Path of Quantum/Waves and Total Emissivity of the </div><div class="t m0 x1 h2 y2 ff1 fs0 fc0 sc0 ls0 ws0">Carbon Dioxide Considering the Molecular Cross Section. </div><div class="t m0 x1 h3 y3 ff2 fs1 fc0 sc0 ls0 ws0">By Nasif Nahle </div><div class="t m0 x1 h3 y4 ff2 fs1 fc0 sc0 ls0 ws0">University Professor, Scientific Research Director at Biology Cabinet Division Mexico. </div><div class="t m0 x1 h2 y5 ff1 fs0 fc0 sc0 ls0 ws0">Abstract: </div><div class="t m0 x1 h3 y6 ff2 fs1 fc0 sc0 ls0 ws0">Through the application of astrophysical formulas, the mean free path length of a Quantum/wave stream </div><div class="t m0 x1 h3 y7 ff2 fs1 fc0 sc0 ls0 ws0">leaving the surface of the Earth to the outer space before it has collided with a molecule of carbon dioxide </div><div class="t m0 x1 h3 y8 ff2 fs1 fc0 sc0 ls0 ws0">and its total emissivity are calculated. The output of this algorithm indicates a value of about 33 meters. </div><div class="t m0 x1 h3 y9 ff2 fs1 fc0 sc0 ls0 ws0">Also calculated is the time taken by a Quantum/wave to exit the atmosphere after it has collided with a </div><div class="t m0 x1 h3 ya ff2 fs1 fc0 sc0 ls0 ws0">molecule of carbon dioxide — which is 5 milliseconds (ms). </div><div class="t m0 x1 h2 yb ff1 fs0 fc0 sc0 ls0 ws0">Introduction: </div><div class="t m0 x1 h3 yc ff2 fs1 fc0 sc0 ls0 ws0">Carbon dioxide (CO</div><div class="t m0 x2 h4 yd ff2 fs2 fc0 sc0 ls0 ws0">2</div><div class="t m0 x3 h3 yc ff2 fs1 fc0 sc0 ls0 ws0">) is vitally important molecule for life on Earth. Carbon dioxide molecules are taken in </div><div class="t m0 x1 h3 ye ff2 fs1 fc0 sc0 ls0 ws0">from the atmosphere by photosynthetic organisms which employ them to build more complex substances </div><div class="t m0 x1 h3 yf ff2 fs1 fc0 sc0 ls0 ws0">that are used for storing the energy transferred from the quantum/waves to the molecules of chlorophyll. </div><div class="t m0 x1 h3 y10 ff2 fs1 fc0 sc0 ls0 ws0">The current assessment demonstrates that CO</div><div class="t m0 x4 h4 y11 ff2 fs2 fc0 sc0 ls0 ws0">2</div><div class="t m0 x5 h3 y10 ff2 fs1 fc0 sc0 ls0 ws0"> is so well dispersed throughout the bulk volume of the </div><div class="t m0 x1 h3 y12 ff2 fs1 fc0 sc0 ls0 ws0">atmosphere that its efficiency for capturing quantum/waves emitted from the surface is extremely low. </div><div class="t m0 x1 h3 y13 ff2 fs1 fc0 sc0 ls0 ws0">The objectives of this didactic article are to demonstrate that the mean free path length of the </div><div class="t m0 x1 h3 y14 ff2 fs1 fc0 sc0 ls0 ws0">Quantum/wave stream does not significantly change the total emissivity of the CO</div><div class="t m0 x6 h4 y15 ff2 fs2 fc0 sc0 ls0 ws0">2</div><div class="t m0 x7 h3 y14 ff2 fs1 fc0 sc0 ls0 ws0"> and that the time </div><div class="t m0 x1 h3 y16 ff2 fs1 fc0 sc0 ls0 ws0">taken by a Quantum/wave to exit the atmosphere to space, without colliding with a molecule of CO</div><div class="t m0 x8 h4 y17 ff2 fs2 fc0 sc0 ls0 ws0">2</div><div class="t m0 x9 h3 y16 ff2 fs1 fc0 sc0 ls0 ws0">, is </div><div class="t m0 x1 h3 y18 ff2 fs1 fc0 sc0 ls0 ws0">extremely low. </div><div class="t m0 x1 h2 y19 ff1 fs0 fc0 sc0 ls0 ws0">Methodology </div><div class="t m0 x1 h3 y1a ff2 fs1 fc0 sc0 ls0 ws0">I have introduced the molecular cross section of the carbon dioxide σ</div><div class="t m0 xa h5 y1b ff1 fs2 fc0 sc0 ls0 ws0">CO</div><div class="t m0 xb h6 y1c ff1 fs3 fc0 sc0 ls0 ws0">2</div><div class="t m0 xc h5 y1b ff1 fs2 fc0 sc0 ls0 ws0">mol</div><div class="t m0 xd h7 y1a ff2 fs1 fc0 sc0 ls0 ws0">, in which case a value for n, </div><div class="t m0 x1 h3 y1d ff2 fs1 fc0 sc0 ls0 ws0">calculated from the molar density of the carbon dioxide and the number of molecules per mol of the </div><div class="t m0 x1 h3 y1e ff2 fs1 fc0 sc0 ls0 ws0">substance was demanded to make the most accurate solution. </div><div class="t m0 x1 h8 y1f ff4 fs4 fc0 sc0 ls0 ws0">α = (ε/εb) = ε </div><div class="t m0 x1 h3 y20 ff2 fs1 fc0 sc0 ls0 ws0"> </div><div class="t m0 x1 h2 y21 ff1 fs0 fc0 sc0 ls0 ws0">Analysis </div><div class="t m0 x1 h3 y22 ff2 fs1 fc0 sc0 ls0 ws0">Preliminary Data: </div><div class="t m0 x1 h3 y23 ff2 fs1 fc0 sc0 ls0 ws0">The density of the gas carbon dioxide in the atmosphere is obtained by the following formula: </div><div class="t m0 x1 h9 y24 ff3 fs1 fc0 sc0 ls0 ws0">ρ</div><div class="t m0 xe h5 y25 ff1 fs2 fc0 sc0 ls0 ws0">CO</div><div class="t m0 xf h6 y26 ff1 fs3 fc0 sc0 ls0 ws0">2</div><div class="t m0 x10 h7 y24 ff2 fs1 fc0 sc0 ls0 ws0"> = (12.187 * Molar mass of CO</div><div class="t m0 x11 h4 y25 ff2 fs2 fc0 sc0 ls0 ws0">2</div><div class="t m0 x12 h7 y24 ff2 fs1 fc0 sc0 ls0 ws0"> * volumetric fraction of CO</div><div class="t m0 x13 h5 y25 ff1 fs2 fc0 sc0 ls0 ws0">2</div><div class="t m0 x14 h7 y24 ff2 fs1 fc0 sc0 ls0 ws0">) ÷ (276.69 K) = 756 mg/m^3. </div><div class="t m0 x15 h4 y27 ff2 fs2 fc0 sc0 ls0 ws0">7 </div><div class="t m0 x1 h3 y28 ff2 fs1 fc0 sc0 ls0 ws0">Where 12.187 is the molar mass of elemental carbon, molar mass of carbon dioxide is 44.01, and the </div><div class="t m0 x1 h3 y29 ff2 fs1 fc0 sc0 ls0 ws0">volumetric fraction of CO2 is 390 ppmV and 276.69 K is for temperature. </div><div class="t m0 x1 h3 y2a ff2 fs1 fc0 sc0 ls0 ws0">To introduce this value into the formula that I will describe below, I made use of the following important </div><div class="t m0 x1 h3 y2b ff2 fs1 fc0 sc0 ls0 ws0">magnitudes: </div></div><div class="pi" data-data='{"ctm":[1.611639,0.000000,0.000000,1.611639,0.000000,0.000000]}'></div></div> </body> </html>

<div id="pf2" class="pf w0 h0" data-page-no="2"><div class="pc pc2 w0 h0"><img class="bi x0 y0 w1 h1" alt="" src="https://static.pudn.com/prod/directory_preview_static/6266fc444c65f41259478d36/bg2.jpg"><div class="t m0 x1 h8 y2c ff4 fs4 fc0 sc0 ls0 ws0">ρ</div><div class="t m0 xe ha y2d ff4 fs5 fc0 sc0 ls0 ws0">CO2</div><div class="t m0 x16 h8 y2c ff5 fs4 fc0 sc0 ls0 ws0"> = 756 mg/m^3 = 0.756 g/m^3 = 7.56 x 10^-7 g/cm^3 </div><div class="t m0 x1 h8 y2e ff5 fs4 fc0 sc0 ls0 ws0">7.56 x 10^-7 g/cm^3 is the density of the atmospheric carbon dioxide in the atmosphere obtained from </div><div class="t m0 x1 h3 y2f ff2 fs1 fc0 sc0 ls0 ws0">the following calculation: </div><div class="t m0 x1 h8 y30 ff4 fs4 fc0 sc0 ls0 ws0">ρ</div><div class="t m0 xe ha y4 ff4 fs5 fc0 sc0 ls0 ws0">CO2 </div><div class="t m0 x17 hb y30 ff5 fs4 fc0 sc0 ls0 ws0">= (12.187 of </div><div class="t m0 x18 hc y31 ff5 fs5 fc0 sc0 ls0 ws0">12</div><div class="t m0 x19 h8 y30 ff5 fs4 fc0 sc0 ls0 ws0">C * MM</div><div class="t m0 x1a ha y4 ff4 fs5 fc0 sc0 ls0 ws0">CO2</div><div class="t m0 x1b h8 y30 ff5 fs4 fc0 sc0 ls0 ws0"> * MF</div><div class="t m0 x1c ha y4 ff4 fs5 fc0 sc0 ls0 ws0">CO2</div><div class="t m0 x1d h8 y30 ff5 fs4 fc0 sc0 ls0 ws0">) ÷ T</div><div class="t m0 x1e ha y4 ff4 fs5 fc0 sc0 ls0 ws0">abs</div><div class="t m0 x1f hb y30 ff5 fs4 fc0 sc0 ls0 ws0"> </div><div class="t m0 x1 h7 y32 ff2 fs1 fc0 sc0 ls0 ws0">Where 385 ppmV is MF</div><div class="t m0 x20 h5 y33 ff1 fs2 fc0 sc0 ls0 ws0">CO2</div><div class="t m0 x21 h7 y32 ff2 fs1 fc0 sc0 ls0 ws0">, 44.01 are for MM</div><div class="t m0 x22 h5 y33 ff1 fs2 fc0 sc0 ls0 ws0">CO2</div><div class="t m0 x23 h7 y32 ff2 fs1 fc0 sc0 ls0 ws0"> and 276.69 K is T</div><div class="t m0 x24 h5 y33 ff1 fs2 fc0 sc0 ls0 ws0">abs</div><div class="t m0 x25 h3 y32 ff2 fs1 fc0 sc0 ls0 ws0">. You can make your own </div><div class="t m0 x1 h7 y34 ff2 fs1 fc0 sc0 ls0 ws0">calculations at LENNTECH</div><div class="t m0 x26 h4 y35 ff2 fs2 fc0 sc0 ls0 ws0"> (7)</div><div class="t m0 x21 h3 y34 ff2 fs1 fc0 sc0 ls0 ws0">, calculator online. </div><div class="t m0 x1 h7 y36 ff1 fs1 fc0 sc0 ls0 ws0">Mass of CO</div><div class="t m0 x27 h5 y37 ff1 fs2 fc0 sc0 ls0 ws0">2</div><div class="t m0 x28 h7 y36 ff1 fs1 fc0 sc0 ls0 ws0"> per cm^3 of air = 7.56 x 10^-7 g/cm^3 * 1 cm^3 = 7.56 x 10^-7 g. </div><div class="t m0 x1 h8 y38 ff4 fs4 fc0 sc0 ls0 ws0">Avogadro's number = 6.02 × 10^23 molecules/mol (Avogadro’s number has not units, however, when it </div><div class="t m0 x1 h8 y39 ff2 fs1 fc0 sc0 ls0 ws0">is introduced like a constant it is expressed as 6.02 × 10^23 /mol). </div><div class="t m0 x29 h4 y3a ff2 fs2 fc0 sc0 ls0 ws0">13</div><div class="t m0 x2a h3 y39 ff2 fs1 fc0 sc0 ls0 ws0"> </div><div class="t m0 x1 h8 y3b ff4 fs4 fc0 sc0 ls0 ws0">CO</div><div class="t m0 x2b ha y3c ff4 fs5 fc0 sc0 ls0 ws0">2</div><div class="t m0 x2c h8 y3b ff5 fs4 fc0 sc0 ls0 ws0"> Molar Density = 44.01 g / 22261 cm^3 = 0.001977 g/cm^3. </div><div class="t m0 x1 h8 y3d ff4 fs4 fc0 sc0 ls0 ws0">Molar mass of CO</div><div class="t m0 x2d ha y3e ff4 fs5 fc0 sc0 ls0 ws0">2</div><div class="t m0 x2e h8 y3d ff5 fs4 fc0 sc0 ls0 ws0"> = 44.01 g </div><div class="t m0 x1 h8 y3f ff4 fs4 fc0 sc0 ls0 ws0">Number of moles of CO</div><div class="t m0 x2f ha y40 ff4 fs5 fc0 sc0 ls0 ws0">2</div><div class="t m0 x30 h8 y3f ff5 fs4 fc0 sc0 ls0 ws0"> = (mass in 1 cm^3 of air) / (molar mass) = 7.56 x 10^-7 (g) / 44.01 (g/mol) = </div><div class="t m0 x1 h8 y41 ff5 fs4 fc0 sc0 ls0 ws0">1.7178 x 10^-8 moles </div><div class="t m0 x1 h8 y42 ff4 fs4 fc0 sc0 ls0 ws0">Number of molecules of carbon dioxide per gram = moles * Avogadro’s number / molar mass = </div><div class="t m0 x1 h8 y43 ff5 fs4 fc0 sc0 ls0 ws0">(1.7178 x 10^-8 moles) * (6.02 × 10^23 molecules/mol) / 44.01 g = 2.35 x 10^14 molecules/mol. </div><div class="t m0 x1 h3 y44 ff2 fs1 fc0 sc0 ls0 ws0">Considering the density of CO</div><div class="t m0 x31 h4 y45 ff2 fs2 fc0 sc0 ls0 ws0">2</div><div class="t m0 x32 h3 y44 ff2 fs1 fc0 sc0 ls0 ws0"> in the atmosphere, the number of moles of carbon dioxide measured in </div><div class="t m0 x1 h7 y46 ff2 fs1 fc0 sc0 ls0 ws0">one cubic centimeter in the atmosphere mass, 1.7178 x 10^-8 moles are contained in 7.56 x 10^-7 g of </div><div class="t m0 x1 h3 y47 ff2 fs1 fc0 sc0 ls0 ws0">carbon dioxide per each cubic centimeter of the current atmosphere. </div><div class="t m0 x1 h8 y48 ff4 fs4 fc0 sc0 ls0 ws0">CO</div><div class="t m0 x2b ha y49 ff4 fs5 fc0 sc0 ls0 ws0">2</div><div class="t m0 x2c h8 y48 ff4 fs4 fc0 sc0 ls0 ws0"> molecule’s cross section (σ) = 5 x 10^-22 cm^2 </div><div class="t m0 x1 h7 y4a ff1 fs1 fc0 sc0 ls0 ws0">Determination of the Mean Free Path Length of quantum/waves in the atmosphere before colliding with a </div><div class="t m0 x1 h7 y4b ff1 fs1 fc0 sc0 ls0 ws0">molecule of carbon dioxide. </div><div class="t m0 x1 h3 y4c ff2 fs1 fc0 sc0 ls0 ws0">Formula to calculate the mean free path length of quantum/waves before colliding with molecules of CO</div><div class="t m0 x33 h4 y4d ff2 fs2 fc0 sc0 ls0 ws0">2</div><div class="t m0 x34 h3 y4c ff2 fs1 fc0 sc0 ls0 ws0">: </div><div class="t m0 x1 h8 y4e ff4 fs4 fc0 sc0 ls0 ws0">l = m / (n σ) (References 1 and 2). </div><div class="t m0 x1 h8 y4f ff2 fs1 fc0 sc0 ls0 ws0">Where l is for the mean free path length, m is for the mass of the gas measured in one cubic centimeter </div><div class="t m0 x1 h8 y50 ff2 fs1 fc0 sc0 ls0 ws0">of air, n is the effective quantum density, and σ is the cross section of a molecule of CO</div><div class="t m0 x35 h4 y51 ff2 fs2 fc0 sc0 ls0 ws0">2</div><div class="t m0 x36 h3 y50 ff2 fs1 fc0 sc0 ls0 ws0"> before </div><div class="t m0 x1 h8 y52 ff2 fs1 fc0 sc0 ls0 ws0">vibrational dephasing (σ = 5.0 x 10^-22 cm^2). </div><div class="t m0 x1 h7 y53 ff2 fs1 fc0 sc0 ls0 ws0">[Notice that n is the total number of states per volume unit, cubic centimeters in this case: according to </div><div class="t m0 x1 h7 y54 ff1 fs1 fc0 sc0 ls0 ws0">quantum mechanics, to obtain the effective density we must to multiply the number of molecules per mol </div><div class="t m0 x1 h7 y55 ff1 fs1 fc0 sc0 ls0 ws0">by the molar gas density of the carbon dioxide and consider other parameters, n = N</div><div class="t m0 x37 h5 y56 ff1 fs2 fc0 sc0 ls0 ws0">C </div><div class="t m0 x38 h7 y55 ff1 fs1 fc0 sc0 ls0 ws0">* (e^-((E</div><div class="t m0 x39 h5 y56 ff1 fs2 fc0 sc0 ls0 ws0">C</div><div class="t m0 x3a h7 y55 ff1 fs1 fc0 sc0 ls0 ws0"> – E</div><div class="t m0 x3b h5 y56 ff1 fs2 fc0 sc0 ls0 ws0">F</div><div class="t m0 x3c h7 y55 ff1 fs1 fc0 sc0 ls0 ws0">)/kT). </div><div class="t m0 x1 h7 y57 ff1 fs1 fc0 sc0 ls0 ws0">If we introduce an incorrect effective density, the results will be infinite numbers that have nothing to do </div><div class="t m0 x1 h7 y58 ff1 fs1 fc0 sc0 ls0 ws0">with reality. In this instance, I have applied the short procedure for simplification and more clarity of the </div><div class="t m0 x1 h7 y59 ff1 fs1 fc0 sc0 ls0 ws0">text]. </div><div class="t m0 x1 h7 y5a ff1 fs1 fc0 sc0 ls0 ws0"> </div></div><div class="pi" data-data='{"ctm":[1.611639,0.000000,0.000000,1.611639,0.000000,0.000000]}'></div></div>