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

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Presentation on theme: "Atmospheric Chemistry 1 Objectives of Atmospheric Chemistry: Can we understand how does the atmosphere maintain its clarity in spite of the huge influx."— Presentation transcript:

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, 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 ~tornroos/images/Asteroid.jpg

6 animals/dinosaurs/Dinosaur-004.jpg

7 Tannery.JPG

8 London map/londonindex_a.html

9 London dec/londonfog/bobby.html

10 The Great Smog, December 1952

11 _transportation.htm

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

13 Vietnam vietnam_saigon_road_trip.htm

14 Los Angeles pages/Los%20AngelesSmog.htm

15 Dec 27, 2002

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

17 /images/_ _girl150ap.jpg

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19 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/    d  v Number of collisions between 2 kinds of molecules Z 12 =      d 1,2  (8kT/  ) 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}=15x 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 =4x cm 3 molec -1 sec -1 HO. =10 7 molec/cm 3  = 1/’k’=1/k 2 [HO. ]=1/(4 x x10 7 )=2,500,000 sec Or t=30 days


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