SOLUTIONS TO PROBLEMS INTRODUCTION TO …

1 SOLUTIONS TO PROBLEMSINTRODUCTION TOATMOSPHERIC CHEMISTRYby Daniel J. JacobHarvard UniversityPrinceton University Press20001 SOLUTIONS TO PROBLEMS , CHAPTER 11. 1 Fog formation1. The saturation vapor pressure of water at 293 K isPH2O,SAT= 23 hPa. At sunset the air is at 50%relative humidity, thereforePH2O= hPa. The dew point corresponding to this water vapor pressureis 282 K. The air must cool to 282 K in order for fog to From the phase diagram we find that the stable phase of water is liquid. The fact that the atmospherecontains water vapor simply means that it is not in equilibrium. Under these conditions water willcondense to produce a liquid phase, andPH2 Owil decrease until the gas-liquid equilibrium line isreached. Under isothermal conditions (20oC) equilibrium will be reached forPH2O, SAT= 23 hPa, so that77% of the water vapor initially present will have 2 Phase of water in a cloudThe saturation vapor pressure at 273 K isPH2O,SAT= 6 hPa.

2 The corresponding mass concentration H2 OisConsidering that cloud liquid water contents are in the range g m-3, we conclude that most of thewater in a cloud is present as 3 The ozone column1. At the peak of the ozone layer,nO3 = 5x1012molecules cm-3. The density of air at that altitude isand the corresponding O3 mixing ratio isCO3 =nO3/na = ppmv = 4200 ppbv. This is fify times the O3air quality standard for surface air!2. In surface air,nO3 = 1x1012molecules cm-3. The density of air at that altitude isand the corresponding O3 mixing ratio isCO3 =nO3/na = 42 ppbv, which is in compliance with the O3 airquality standard. The relative decrease of CO3 from 25 to 0 km is much larger than the relative decreaseofnO3 because of the change in atmospheric pressure:.3. The O3 columnCis the total number of O3 molecules per unit area of Earth s surface: H2 OMH2 OPH2 ORT------------------------------------1 8x103 6x102 273 ---------------------------------------- -4.

3 8 x 1 03 kg m3 g m3 == = = 220 m3 cm3 ==== 300 m3 cm3 ====CO3z2()CO3z1()---------------------- -nO3z2()nO3z1()----------------------naz 1()naz2()----------------- nO3z2()nO3z1()----------------------Pz1( )Tz2()Pz2()Tz1()------------------------ ------ ()nO3z1()----------------------between 0 and 25 km===2 For the triangular function proposed here as an approximation to the O3profile, the integral is simply thearea of the triangle (1/2 x base x height). We carry out the numerical calculation in SI units:4. Consider the above O3column brought to sea level as a layer of unit horizontal areaAand verticalthicknessh(volumeV = Ah)atP= ,T= 273 K. The column contains ,corresponding toN = = moles. Apply the ideal gas law to obtainh:CnO3zd0 =12---x30km()x 5x1012molecules cm3 () 30x1035x1018 m2 ==PVNRTh 273 mm=====3 SOLUTIONS TO PROBLEMS , CHAPTER 22.

4 1 Scale height of the Martian atmospherewhich is larger than the scale height of the Earth s atmosphere. The atmosphere of Mars extends deeperthan that of the Earth because of the .smaller size of Mars and hence its weaker gravitational pull on 2 Scale height and atmospheric mass1. The massdmof the species in an elementary slab of atmosphere of unit area and vertical thicknessdzisdm = (z)dz. We integrate over the depth of the atmosphere and over the areaA of the Earth: (1) From the ideal gas law:whereMa= kg/mol;T= 288 K; andP= We use SI units for all quantities to avoid unitconversion mistakes. In this manner we We substitute in equation (1)A=4 R2= , whereR= 6400 km is the radius of the Earth; (0)= kg m-3; and the atmospheric scale height ish= The resulting mass of the atmosphere ism= , somewhat lower than the value of derived in the text.

5 The reason for thedifference lies in the assumption Pmade when applying the barometric law to air density. In fact P/T(ideal gas law); aszincreases andTdecreases, (z)deviates upward from the value (0)exp(-z/h)predicted from the barometric law and hence equation (1) We apply equation (1) to sea salt, taking into consideration that sea salt is emitted over only 70% ofthe 220 44x103 km===mA z()zd0 A 0()ezh--- zd0 A 0()hezh--- 0 A 0()h==== aMaPRT------------= a 0() 10Pa() K ------------------ 288K() ss0() () 10m2()10x109 kgm3------- 500m() 10kg===4 SOLUTIONS TO PROBLEMS , CHAPTER 33. 1 Ventilation of pollution from the United States1. LetWandhrepresent the width and height of the box respectively. The volume of the box unit time a volumeUWh of air flows out of the box. The residence time of air in the box is therefore:2.

6 Letmrepresent the mass of the pollutant in the United States box. The pollutant is removed bychemical loss (time constant chem, loss ratem/ chem) and export out of the box (time constant out,lossratem/ out). The fractionfof the pollutant removed by export isThe efficiency with which a pollutant emitted from the United States is exported to the global atmosphereL = 5000kmFinFoutU = 10m/sU = 10m/sWh outLWhUWh-------------LU----5106 10------------------51 05 days==== =fexport loss ratetotal loss rate------------------------------------ m out-----------m chem----------------m out-----------+------------------------- ---------=11 out chem----------------+------------------- -------== out/ chem0 12 345depends on its rate of chemical loss relative to the rate of ventilation. For the typical wind speedconsidered here, pollutants with lifetimes longer than 6 days against chemical loss are efficientlyexported out of the United States (f> ) while pollutants with shorter lifetimes are mostly removedwithin the United States and have relatively little global 2 Stratosphere-troposphere exchange1.

7 The90Sr inventories in the stratosphere and troposphere are affected by loss from radioactive decay(L), transfer from the stratosphere to the troposphere (FST), reverse transfer from the troposphere to thestratosphere (FTS), and deposition (D). Deposition applies only in the troposphere. There were no90 Sremissions during the post-1962 period. The mass balance equations are: The transfer ratesFST andFTSare related to the transfer rate constantskST andkTS:The loss rates from radioactive decay areHerekdis the rate constant for radioactive decay;kd= ln2/t1/2= yr-1, wheret1/2= 28 yr is the half-life loss rate from deposition isHerekDis the loss rate constant for deposition;kD=1/ D= day-1where D= 10 days is the lifetimeagainst deposition in the in the mass balance equations:2. Assuming that transfer of90Sr from the troposphere to the stratosphere (FTS) is negligible, the massbalance equation formS becomestddmSFTSFST LS =tddmTFSTFTS LT D =FSTkSTmS=FTSkTSmT=LSkdmS=LTkdmT=DkDmT=t ddmSkTSmTkSTkd+()mS =tddmTkSTmSkTSkdkD++()mT =tddmSkSTkd+()mS =6 Integrating this equation yields:Comparing to the observed decreasewithk = yr-1, we obtain:so thatkST= yr-1and hence S=1/kST= yr.

8 Note that radioactive decay has negligible effect onthe calculation of Sbecause the lifetime of90Sr against radioactive decay is long compared to theresidence time of air in the Total air mass must be at steady state between the troposphere and the stratosphere,and thereforeSince the pressure at a given altitude is proportional to the mass of atmosphere overhead,wherePsurface = 1000 hPa,Ptropopause = 150 hPa,Pstratopause = 1 hPa. We findmT /mS = and hence T = years. Compare to the lifetime D = 10 days of90Sr against deposition in the troposphere; the fractionof90Sr in the troposphere that is transferred back to the stratosphere is negligible, justifying ourassumption in question Letmrepresent the mass of HCFC in the troposphere. The HCFC loss rate by transfer to thestratosphere isLTS=m/ T, while the loss rate from oxidation isLox=m/ oxwhere oxis the lifetimeagainst oxidation.

9 The fractionfof emitted HCFC that penetrates the stratosphere isFor HCFC-123, ox = years andf = ; for HCFC-124, ox = years andf = 3 Interhemispheric exchangemSt()mS0()ekSTkd+() t=mSt()mS0()ek t=kSTkkd =mS'kSTmT'kTS= T SmT'mS'-------- =mT'mS'--------PsurfacePtropopause PtropopausePstratopause ---------------------------------------- --------------------=fLTSLTSLox+-------- ---------------1 T-----1 T-----1 ox-------+-------------------11 T ox-------+-----------------===71. We write mass balance equations formN andmS, and take the difference:In 1983,mN - mS = 7kg,E=15 kg; assuming steady state for(mN - mS)we obtain = 1/k = .96 yr2. Based on the equation:we see that the time scale for(mN-mS)to relax to steady state is 1/(2k+kd)= yr. The rise inEis slowrelative to this time scale so that(mN-mS)has the time to continually adjust to steady state 4 Long-range transport of acidityWe start from the puff model versions of the mass balance equations for SO2 and H2SO4:Integration of the mass balance equation for SO2 yields as solutionReplacing into the mass balance equation for H2SO4: (1)We seek the general solution to (1) as the sum of the general solution to the homogeneous equation (right-hand-side equal zero) and a particular solution to the full equation (1).

10 The general solution to thehomogeneous equation isdmNdt-----------EkmSkmN kcmN +=dmSdt----------kmNkmS kcmS =dmNm S()dt-----------------------------E2kkc+ ()mNmS () =2kkc+()EmNmS ()-------------------------=k12---EMNMS ----------------------kc ==dmNmS ()dt-----------------------------E2kkd+( )mNmS () =dSO2[]dt------------------k1SO2[] =dH2SO4[]dt--------------------------k1S O2[]k2H2SO4[] =SO2[]SO2[]oek1t =dH2SO4[]dt--------------------------k2H 2SO4[]+k1SO2[]oek1t =H2SO4[]Aek2t =8whereAis an integration constant. In addition, we can see from the form of (1) that it must admit asolution of the formBexp(-k1t)whereB= is a constant to be determined by substitution in the massbalance equation: we findB = [SO2]o/(k2-k1). The general solution to (1) is therefore:We derive the value of the integration constantAfrom the initial condition [H2SO4]o = 0 att= 0:and thus obtain the final solution for [H2SO4]:To obtain the SOLUTIONS as a function of distance downwind of the power plants we simply replacet=x/Uin the expressions for [SO2] and [H2SO4].