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

3. Atmospheric Chemistry 1

4. Atmospheric Chemistry 1 The physical structure of the atmosphere

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

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

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

21. Atmospheric Chemistry 1 The Kinetic Theory of the Gases: Molecular velocity v=(8RT/  M) 0.5 At 1atm v=4.7x10 5 cm/sec Mean free path =1/(2 0.5  Nd 2 ) At 1atm =8x10 -5 cm Number of collisionsZ 11 =1/2 2 0.5    d  v Number of collisions between 2 kinds of molecules Z 12 =      d 1,2  (8kT/  ) 0.5 2 nd order rate constant: k 2 =AT 0.5 exp{-E a /RT}

22. Atmospheric Chemistry 1 The Barometric Equation: dP=wg/A w  V dP=  Vg/A dP=  Adxg/A=  gdx P-dP P A dx

23. Atmospheric Chemistry 1 The Barometric Equation: PV=nRT=wRT/M  w/V=(PM/RT) dP=(PM/RT) g dx dP/P=-Mgh/RT dx P=P 0 e -Mgh/RT

24. Atmospheric Chemistry 1 The Barometric Equation: Units M=molecular weight (80%N2+20%O2) g=cm/sec 2 h=cm r=erg T= o K Conditions for Jerusalem:

25. Atmospheric Chemistry 1 The Barometric Equation: The barometric Pressure in Jerusalem (h=750m) P=760 exp-{28.8x981x75,000/8.31x10 7 x298} P=760*0.92=698torr P-dP P A dx

26. Atmospheric Chemistry 1 Number of molecules in the Earth’s Atmosphere N=N 0 e -Mgh/RT In a column that its base is 1 cm 2 and its height is  Q’ =  Ndh =  N 0 e -Mgh/RT dh= -N 0 RT/(Mg) [0-1] = N 0 RT/(Mg)

27. Atmospheric Chemistry 1 Q’=N 0 RT/(Mg) Number of molecules in the entire atmosphere: Q=Q’x4  r 2 =(6.02x10 23 x8.31x10 7 x280x4  x(6.37x10 8 ) 2 /(22400x28.5x981) Q=1.14x10 44 molecules Question: What is the height of the layer that contains 90% of the earth molecules?

28. Atmospheric Chemistry 1 Q’=N 0 RT/(Mg) Number of molecules in the entire atmosphere: Q=Q’x4  r 2 =N 0 RT/(Mg)x4  r 2 =(6.02x10 23 x8.31x10 7 x280x4  x(6.37x10 8 ) 2 /(22400x28.5x981)

29. Atmospheric Chemistry 1 =(6.02x10 23 x8.31x10 7 x280x4  x(6.37x10 8 ) 2 /(22400x 28.5x981) Q=1.14x10 44 molecules Question: How many molecules are in the troposphere? (h=15 km)

30. Atmospheric Chemistry 1 Number of molecules in the Earth’s Atmosphere In a column that its base is 1 cm 2 and its height is  Q’ =  Ndh =  N 0 e -Mgh/kT dh=-N 0 kT/(Mg) [0-1] Q=Q’x4  r 2 =(6.02x10 23 x8.31x10 7 x280x4  x(6.37x10 8 ) 2 /(22400x28.5x981) Q=1.14x10 44 molecules Question: What is the height of the layer that contains 90% of the earth molecules?

31. Atmospheric Chemistry 1 Number of water molecules in the Jordan Valley: L=300kmx50kmx0.8km=1200km 3 =12x10 12 m 3 =12x10 18 gr=6.67x10 17 moles x6.02x10 23 =4x10 41

32. Atmospheric Chemistry 1 Number of CO 2 molecules in the Atmosphere P{CO 2 }=380ppm N{CO 2 }=380x10 -6 x1.41x10 44 N=4.33x10 40

33. Atmospheric Chemistry 1 Number of CFC molecules (F-11) in the Atmosphere: P{F-11}=15 ppt N{F-11}=15x10 -12 x1.41x10 44 N=1.7x10 33 molecules m=N/N o =1.7x10 33 /6.02x10 23 =2.84x10 9 moles =(x101)=2.9x10 11 gr= 2.9x10 5 tones 0.3 tg (CFC-11)

34. Atmospheric Chemistry 1 First Order Chemical Reaction A  P R{P}=d[A]/dt=-k 1 [A]  d[A]/[A]=- k 1 dt [A]=[A] 0 exp{-k 1 t}

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

36. Atmospheric Chemistry 1 Second Order Chemical Reaction A+B  P d[A]/dt=-d[P]/dt=-k 2 [A][B] Assume: [A]=[A] 0 -X, B=[B] 0 -X d[A]/dt=-k 2 ([A] 0 -X)([B] 0 -X) Solution: k 2 t ={1/([A] 0 -[B] 0 )} ln{([A] 0 -X)[B] 0 /([B] 0 -X)[A] 0 }

37. Atmospheric Chemistry 1 Pseudo First Order Chemical Reaction: If [B] 0 -X~[B] 0 then: d[A]/dt=-k 2 [A][B] 0 =-’k’[A] d[A]/[A] =-’k’t  [A]=[A] 0 e -’k’t Or: [A]=[A] 0 exp{-k 2 [B] 0 t}

38. Atmospheric Chemistry 1 Atmospheric half lifetime of CO CO + HO.  CO 2 +H. k 2 =4x10 -14 cm 3 molec -1 sec -1 HO. =10 7 molec/cm 3  = 1/’k’=1/k 2 [HO. ]=1/(4 x10 -14 x10 7 )=2,500,000 sec Or t=30 days