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Atmospheric Chemistry 1

Atmospheric Chemistry 1. Objectives of Atmospheric Chemistry: Can we understand how does the atmosphere maintain its clarity in spite of the huge influx of anthropogenic and natural emissions?. Atmospheric Chemistry 1. Literature

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Atmospheric Chemistry 1

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  1. Atmospheric Chemistry 1 Objectives of Atmospheric Chemistry: Can we understand how does the atmosphere maintain its clarity in spite of the huge influx of anthropogenic and natural emissions?

  2. Atmospheric Chemistry 1 • Literature • B. J. Finlayson-Pitts and J. N. Pitts, ‘Chemistry of the upper and lower atmosphere’, Academic Press, 2000 • J. H. Seinfeld and S. N. Pandis, ‘Atmospheric Chemistry and Physics’, Wiley Interscience, 1998.

  3. Atmospheric Chemistry 1

  4. Atmospheric Chemistry 1The physical structure of the atmosphere

  5. http://www.hytti.uku.fi/ ~tornroos/images/Asteroid.jpg http://www.student.oulu.fi/~jkorteni/space/boundary/dinosaur.jpg

  6. http://www.mywallpapers.org/ animals/dinosaurs/Dinosaur-004.jpg

  7. http://www.ancientexplorations.com/ Tannery.JPG

  8. London 1859 http://www.ph.ucla.edu/epi/snow/ 1859map/londonindex_a.html

  9. London 1952 www.npr.org/.../2002/ dec/londonfog/bobby.html

  10. The Great Smog, December 1952

  11. www.mazel.com/ _transportation.htm

  12. Taipei, Taiwan medg.lcs.mit.edu/people/ chris/asiaphotos.html

  13. Vietnam www.gingerb.com/ vietnam_saigon_road_trip.htm

  14. Los Angeles www.mybytes.de/usa/ pages/Los%20AngelesSmog.htm

  15. Dec 27, 2002

  16. At risk? Does grey haze over China reduce rainfall? Beijing

  17. http://news.bbc.co.uk/olmedia/ 1430000/images/_1430268_girl150ap.jpg

  18. http://www.curesarcoma.org/ RideforReid/images/NewGallery/ 13windy%20desert.JPG

  19. Atmospheric Chemistry 1 The Gas Law: Determines the relationship between volume pressure and the number of molecules PV=nRT=w/MRT=NkT

  20. Atmospheric Chemistry 1 The Kinetic Theory of the Gases: Molecular velocity v=(8RT/pM)0.5 At 1atm v=4.7x105 cm/sec Mean free path l=1/(20.5pNd2) At 1atm l=8x10-5 cm Number of collisions Z11=1/2 20.5p N2d2 v Number of collisions between 2 kinds of molecules Z12=p N1N2d1,2(8kT/pm)0.5 2nd order rate constant: k2=AT0.5exp{-Ea/RT}

  21. Atmospheric Chemistry 1 P-dP P A The Barometric Equation: dP=wg/A w=rV dP=rVg/A dP= rAdxg/A= rgdx dx

  22. Atmospheric Chemistry 1 The Barometric Equation: PV=nRT=wRT/M r= w/V=(PM/RT) dP=(PM/RT) g dx dP/P=-Mgh/RT dx P=P0e-Mgh/RT

  23. Atmospheric Chemistry 1 The Barometric Equation: Units M=molecular weight (80%N2+20%O2) g=cm/sec2 h=cm r=erg T=oK Conditions for Jerusalem:

  24. Atmospheric Chemistry 1 P-dP P A The Barometric Equation: The barometric Pressure in Jerusalem (h=750m) P=760 exp-{28.8x981x75,000/8.31x107x298} P=760*0.92=698torr dx

  25. Atmospheric Chemistry 1 Number of molecules in the Earth’s Atmosphere N=N0e-Mgh/RT In a column that its base is 1 cm2 and its heightis  Q’ =Ndh =N0e-Mgh/RTdh= -N0RT/(Mg) [0-1] = N0RT/(Mg)

  26. Atmospheric Chemistry 1 Q’=N0RT/(Mg) Number of molecules in the entire atmosphere: Q=Q’x4pr2 =(6.02x1023x8.31x107x280x4px(6.37x108)2/(22400x28.5x981) Q=1.14x1044 molecules Question: What is the height of the layer that contains 90% of the earth molecules?

  27. Atmospheric Chemistry 1 Q’=N0RT/(Mg) Number of molecules in the entire atmosphere: Q=Q’x4pr2=N0RT/(Mg)x4pr2 =(6.02x1023x8.31x107x280x4px(6.37x108)2/(22400x28.5x981)

  28. Atmospheric Chemistry 1 =(6.02x1023x8.31x107x280x4px(6.37x108)2/(22400x28.5x981) Q=1.14x1044 molecules Question: How many molecules are in the troposphere? (h=15 km)

  29. Atmospheric Chemistry 1 Number of molecules in the Earth’s Atmosphere In a column that its base is 1 cm2 and its heightis  Q’ =Ndh =N0e-Mgh/kTdh=-N0kT/(Mg) [0-1] Q=Q’x4pr2=(6.02x1023x8.31x107x280x4px(6.37x108)2/(22400x28.5x981) Q=1.14x1044 molecules Question: What is the height of the layer that contains 90% of the earth molecules?

  30. Atmospheric Chemistry 1 Number of water molecules in the Jordan Valley: L=300kmx50kmx0.8km=1200km3=12x1012m3 =12x1018gr=6.67x1017moles x6.02x1023=4x1041

  31. Atmospheric Chemistry 1 Number of CO2 molecules in the Atmosphere P{CO2}=380ppm N{CO2}=380x10-6x1.41x1044 N=4.33x1040

  32. Atmospheric Chemistry 1 Number of CFC molecules (F-11) in the Atmosphere: P{F-11}=15 ppt N{F-11}=15x10-12x1.41x1044 N=1.7x1033 molecules m=N/No=1.7x1033/6.02x1023=2.84x109 moles =(x101)=2.9x1011 gr= 2.9x105 tones 0.3 tg (CFC-11)

  33. Atmospheric Chemistry 1 • First Order Chemical Reaction AP R{P}=d[A]/dt=-k1[A]  d[A]/[A]=- k1dt [A]=[A]0exp{-k1t}

  34. Atmospheric Chemistry 1 • First Order Chemical Reaction AP • Life time: Lets define t=t When [A]=[A]0/e Then: [A]0/e=[A]0exp{-k1t} 1=k1t ….  t=1/k1

  35. Atmospheric Chemistry 1 • Second Order Chemical Reaction A+BP d[A]/dt=-d[P]/dt=-k2[A][B] Assume: [A]=[A]0-X, B=[B]0-X d[A]/dt=-k2([A]0-X)([B]0-X) Solution: k2t ={1/([A]0-[B]0)} ln{([A]0-X)[B]0/([B]0-X)[A]0}

  36. Atmospheric Chemistry 1 Pseudo First Order Chemical Reaction: If [B]0-X~[B]0 then: d[A]/dt=-k2[A][B]0=-’k’[A] d[A]/[A] =-’k’t  [A]=[A]0e-’k’t Or: [A]=[A]0exp{-k2[B]0t}

  37. Atmospheric Chemistry 1 Atmospheric half lifetime of CO CO + HO. CO2 +H. k2=4x10-14cm3 molec-1 sec-1 HO.=107 molec/cm3 t= 1/’k’=1/k2[HO.]=1/(4 x10-14 x107)=2,500,000 sec Ort=30 days

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