2010 Final Solutions - Harvard Universityeps5/exams/2010_Final_Solutions.pdf · Final Exam May 12, 2010 ... • The exam will last 3 hours. • Complete the problems directly on the

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Earth and Planetary Sciences 5 Final Exam

May 12, 2010 Name:________________________________________________ Teaching Fellow:_______________________________________

INSTRUCTIONS

• PUT YOUR NAME ON EACH PAGE. • The exam will last 3 hours. • Complete the problems directly on the exam. • Extra paper is available if needed. • Please show ALL your work so partial credit can be given! • Neatness is appreciated.

Scoring: Problem Score Multiple Choice _____ / 24 Southern California _____ / 24 Nitrogen Cycle _____ / 12 Ocean Carbon Chemistry _____ / 12 Stratospheric Chemistry _____ / 24 Tropospheric Chemistry _____ / 24 TOTAL SCORE _____ / 120  

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Section I: Multiple choice questions. Circle the best answer. (1 point each)

1. Greenhouse gases like CO2 warm the surface of the Earth by: a. Reflecting incoming shortwave radiation from the sun. b. Reflecting outgoing longwave radiation from the Earth. c. Absorbing and re-radiating incoming shortwave radiation from the sun. d. Absorbing and re-radiating outgoing longwave radiation from the Earth.

2. Radiation emitted by the Earth peaks at a different wavelength than radiation emitted by

the sun because: a. The Earth and the sun emit at different temperatures. b. The Earth and the sun have different gravitational acceleration. c. The Earth is a perfect blackbody and the sun isn’t. d. The Earth’s atmosphere has greenhouse gases that change the wavelength of

emission.

3. Which of the following describes the land-sea breeze during the daytime? a. Warmer temperatures over sea, greater scale height over land, flow from sea to land

aloft, flow from land to sea at the surface b. Warmer temperatures over land, smaller scale height over land, flow from sea to land

aloft, flow from land to sea at the surface c. Warmer temperatures over land, smaller scale height over sea, flow from land to

sea aloft, flow from sea to land at the surface d. Warmer temperatures over land, greater scale height over sea, flow from land to sea

aloft, flow from sea to land at the surface

4. The following figure shows the infrared emission spectrum of the Earth measured by a satellite over the Mediterranean Sea. Based on this figure, what is the approximate temperature at the surface of the Mediterranean? a. 220 K b. 260 K c. 280 K d. We can’t tell from this figure.

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5. An object with POSITIVE buoyancy (net force on it is up): I. Has a smaller volume than the volume of fluid it displaces (VOBJECT < VDISPLACED) II. Has a lower density than the density of fluid it displaces (ρOBJECT < ρDISPLACED) III. Has a smaller weight than the weight of fluid it displaces (ρOBJECT* VOBJECT *g < ρDISPLACED*VDISPLACED*g)

a. II only b. I and III only c. II and III only d. I, II, and III

6. If an atmosphere has a scale height of 7.4 km and a surface pressure of 1000 hPa, what is

the pressure at an altitude of 5.0 km? a. 230 hPa b. 510 hPa c. 1000 hPa d. 2000 hPa

7. Transfer of CO2 from the atmosphere to the ocean results in a decrease in the

concentration of dissolved ______ with a corresponding change of the ocean water to more ________ conditions. a. CO3

2-; acidic b. CO3

2-; basic c. HCO3

-; acidic d. HCO3

-; basic

8. Which of the following is closest to the natural pH of seawater? a. 4 b. 6 c. 8 d. 10

9. What is the dominant form of inorganic carbon in the ocean?

a. CO2(g) b. CO2(aq) c. CO3

2-

d. HCO3-

10. Which of the following is the correct definition of the equilibrium constant for the

equilibrium reaction HSO3- ↔ SO3

2- + H+? a. K = k[HSO3

-] b. K = k[SO3

2-][H+] c. K = [HSO3

-]/([SO32-][H2O])

d. K = ([SO32-][H+])/[HSO3

-]

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11. Roughly what percentage of CO2 emitted from fossil fuel combustion remains in the atmosphere? a. 1% b. 10% c. 50% d. 90%

12. Which of the following are true of the biosphere?

I. It provides a pathway for nitrogen fixation II. It can serve as a sink of CO2 III. It can serve as a source of CO2

a. I and II only b. I and III only c. II and III only d. I, II, and III

13. Which of the following constitute odd oxygen?

I. O II. O2

III. O3 a. I and II only b. I and III only c. II and III only d. I, II, and III

14. Which reactions are predominantly responsible for the production of odd oxygen in the

stratosphere? a. NOx catalytic cycling b. ClOx catalytic cycling from CFCs c. HOx catalytic cycling d. Chapman mechanism

15. What constitutes the major species of the family of NOx radicals?

a. NO & NO3 b. NO & NO2 c. NO, NO2, & HNO3 d. NO, NO2, NO3, & N2O

16. Absorption by ozone in the stratosphere prevents most radiation of this type from making

it to the surface: a. UV b. Visible c. Infrared d. Microwave

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17. Which of the following would not constitute an initiation step in a ClOx radical reaction chain? a. ClO + HO2 → HCl + O3 b. HCl + O → Cl + OH c. Cl2 + O → ClO + Cl d. ClOOCl + hν → ClOO + Cl

18. Which is the correct chronology of the Antarctic ozone hole?

a. ClNO3 and HCl reservoir formation; PSC destruction; Cl2 production; ClOx production; O3 destruction; PSC formation; Polar Sunrise

b. ClNO3 and HCl reservoir formation; PSC formation; Cl2 production; Polar Sunrise; ClOx production; O3 destruction; PSC destruction

c. ClNO3 and HCl reservoir destruction; PSC formation; Polar Sunrise; Cl2 production; ClOx production; O3 destruction; PSC formation

d. ClNO3 and HCl reservoir formation; Polar Sunrise; PSC destruction; Cl2 production; ClOx destruction; O3 destruction; PSC formation

19. A major sink of HOx radicals is the self reaction of HO2 to form hydrogen peroxide:

2HO2 → H2O2 + O2 ; k What is the rate of H2O2 production? a. 0.5 k [ HO2 ]2

b. k [ HO2 ]2

c. 2 k [ HO2 ]2

d. 2 k [ HO2 ]

20. Circle the gases that contribute to acidification of rainwater below pH 5 (you may select more than one). a. CH4 b. H2SO4 c. NH3 d. HNO3 e. OH

21. If we are emitting 16 x 1011 molec cm-2 s-1 of NOx and 2 x 1011 cm-2 s-1 atoms of

hydrocarbon C, in which air pollution regime are we located? a. NOx-limited b. HOx-limited c. VOC-limited d. Impossible to tell

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22. Which of the following correctly orders the EM spectrum from most energetic to least energetic photons? a. UV, IR, Visible b. UV, Visible, IR c. IR, UV, Visible d. IR, Visible, UV e. Visible, IR, UV f. Visible, UV, IR

23. Where would you expect the largest VOC source?

a. Phoenix, AZ b. Anchorage, AK c. Charlotte, NC d. Boston, MA

24. For the same mass concentration (g cm-3), which of the following penetrates more deeply

into human lungs? a. PM2.5 b. PM10 c. Equally bad

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Section II: Problems Show ALL your work so partial credit can be given!

I. Autumn in Southern California (24 Points) The Santa Ana winds are a persistent feature during the late fall and winter in Southern California. They are northeasterlies (winds from the northeast) caused by dense dry air moving by gravity from the Great Basin and Mohave Desert down to the coast. They are notoriously hot, dry, and strong, bringing extreme wildfire danger to the region. a) Imagine that when these hot desert winds reach the coast, they have a density of 1.15 kg m-3. If the coastal air they encounter at San Diego is 22ºC and 1013 mb, what is the density of the coastal air (in kg m-3)? Will the desert winds float on top of the coastal air, creating a stratified scenario, or sink beneath and therefore cause mixing? Assume the molecular mass of air is 28.97 g mol-1. (5 points) ρ = P M / kB T = 1.013e5 Pa • 28.97e-3 kg mol-1 / (6.022e23 molec mol-1) • (1.38e-23 J/K) • ( 22ºC + 273.15 )

= 1.20 kg m-3

The density of the desert winds is less than that of the marine air. Hence, they will float on top. b) The following is an atmospheric sounding from a weather balloon taken at the San Diego International Airport on Nov 9, 2009 at 4 pm PST. Were there any clouds or fog in San Diego at that time? (2 points)

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No, the temperature profile never meets the dew point, so no condensation ever occurred. It was completely clear, not even any fair weather cumulus clouds over the airport. c) Determine from the data the atmospheric lapse rates for the regions (i) 0-0.5 km, (ii) 0.5-0.65 km, and (iii) 0.65-1.5 km. Label them appropriately as stable, unstable, or neutral. (10 points) Reading from the graph, and using the equation ΓENV = -∆T / ∆z 0-0.5km: Γ0-0.5km = - ( 15ºC - 22ºC ) / (0.5km – 0km ) = 14 ºC km-1

This region is (highly) unstable as ΓENV >> ΓDRY

0.5-0.65km: Γ0.5-0.65km = - ( 18ºC - 15ºC ) / (0.65km – 0.5km ) = -20 ºC km-1

This region is (highly) stable as ΓENV << ΓDRY (Thermal inversion) 0.65-1.5: Γ0.65-1.5km = - ( 16ºC - 18ºC ) / (1.5km – 0.65km ) = 2.4ºC km-1

This region is stable as as ΓENV < ΓDRY

There are no neutral regions.  

d) To what altitude would an air parcel heated to 24ºC at the surface rise? Solve your answer graphically on the weather balloon plot by drawing an appropriate line for the parcel trajectory, and justify your choice of lapse rate. (7 points) Because the dew point is never met, no condensation occurs, so we follow the dry adiabatic lapse rate the entire way. We know at the surface, T1 = 24ºC and z1 = 0km. Choosing an arbitrary point to solve for, at z2 = 1.0 km, T2 = T1 - ΓDRY ( z2 – z1 ) = 24ºC – 9.8 ºC km-1 ( 1.0 km – 0km ) = 14.2ºC Drawing a line between (0km, 24ºC) and (1km, 14.2ºC), we see the parcel intersects the environmental lapse rate at ~0.65 km. If it were to rise higher, it would become more dense

than the surrounding air and sink back down. The parcel will rise to 0.65 km.

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II. The Nitrogen Cycle (12 points) The following figure shows a box model of the modern nitrogen cycle. Reservoir quantities are given in Tg N and fluxes are given in Tg N yr-1. Recall that 1 Tg = 1012 g.

a) Calculate the residence times of nitrogen in: 1. The land biota 2. The soil 3. The combined land reservoir (land biota + soil)

Why is the reservoir time longer for the combined land reservoir than for the individual soil and land biota reservoirs? (4 points) For all residence times, use τ = [N]/LN τ = (1•104 Tg N) / ( 2500 Tg N yr-1) = 4 yr τ = (7•104 Tg N) / ( 2300 Tg N yr-1 + 220 Tg N yr-1 + 20 Tg N yr-1) = 28 yr τ = (7•104 Tg N + 1•104 Tg N) / (220 Tg N yr-1 + 20 Tg N yr-1) = 330 yr

The residence time in the combined land reservoir is much longer than the residence time in either the biota or soil reservoirs because the rate of cycling between the land biota and the soil is much faster than the loss out of the combined reservoir. b) Now consider the ocean reservoirs. Are the individual ocean reservoirs in steady state? What about the combined ocean system (ocean biota + deep ocean)? (4 points) A reservoir is in steady state if Flux In = Flux Out For the ocean biota:

flux in = 20 Tg N yr-1 + 30 Tg N yr-1 + 1600 Tg N yr-1 = 1650 Tg N yr-1

flux out = 40 Tg N yr-1 + 1600 Tg N yr-1 = 1640 Tg N yr-1

NOT in steady state For the deep ocean:

flux in = 1600 Tg N yr-1

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flux out = 10 Tg N yr-1 + 1600 Tg N yr-1 = 1610 Tg N yr-1

NOT in steady state For the combined ocean system:

flux in = 20 Tg N yr-1 + 30 Tg N yr-1 = 50 Tg N yr-1

flux out = 40 Tg N yr-1 + 10 Tg N yr-1 = 50 Tg N yr-1

IN steady state c) The modern reservoir sizes shown in the figure above are the result of both natural processes and human activity. Over the last 100 years, human activity has added about 1.5•104 Tg N to the combined land reservoir and 750 Tg N to the combined ocean reservoir. By what percentage have these reservoirs increased? (4 points) Percent change is defined as 100 * (New Amount – Old Amount) / (Old Amount) Here we know the new amount for each reservoir (given in the figure), and the change, so we calculate the Old Amount = New Amount - Change Land reservoir: % = 100 * (1.5•104 Tg N) / (8•104 Tg N - 1.5•104 Tg N) = 23% Ocean reservoir: % = 100 * (750 Tg N) / (8•105 Tg N - 750 Tg N) = 0.1%

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III. Ocean Carbon Chemistry (12 Points) The chemical equilibrium of inorganic carbon species in ocean water may be described by the following three reactions:

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CO2(g)↔ CO2 • H2O α

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CO2 • H2O↔ HCO3− +H+ K1

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HCO3− ↔ CO3

2− +H+ K2

The equilibrium constant (α, K1, K2) for each of the chemical reactions is listed below:

a) What is the net reaction that occurs when CO2 is transferred from the atmosphere to the ocean? What would be the effect on atmospheric CO2 of adding carbonate ions (CO3

2-) (as Na2CO3, for example) to the ocean? (3 points) CO2(g) + CO3

2- 2 HCO3-

Adding carbonate ions (e.g., adding Na2CO3) would drive this reaction to the right. Each molecule of carbonate would react with a molecule of CO2 (in the process, forming two molecules of bicarbonate), drawing down atmospheric CO2. b) If the ocean temperature is 5°C, the pH is 8.1, and the concentration of carbonate is [CO3

2-] = 8.2•10-5 mol L-1, what is the concentration of bicarbonate (HCO3

-)? (3 points) Use the definition of the equilibrium constant: K2 = ([CO3

2-][H+])/[HCO3-]

[H+] = 10-pH = 10-8.1 = 7.9•10-9 mol L-1

[HCO3-] = ([CO3

2-][H+])/K2 = (8.2•10-5 mol L-1)(7.9•10-9 mol L-1)/(3.99•10-10 mol L-1) = 1.6•10-3 mol L-1

c) The Henry’s Law coefficient, α, is given by the relationship α = [CO2•H2O]/[CO2(g)]. If the temperature is still 5°C and the concentration of aqueous phase CO2 is [CO2•H2O] = 1.8•10-5 mol L-1 and the surface pressure is 1 atm, what is the concentration of gas phase CO2 in equilibrium with the water? Convert your answer to ppm. (3 points) [CO2(g)] = [CO2•H2O]/α = (1.8•10-5 mol L-1) / (5.25•10-2 mol L-1 atm-1) = 3.4•10-4 atm Convert to ppm: 3.4•10-4 atm * (1 ppm/10-6 atm) = 340 ppm d) Imagine the temperature of the water increases from 5°C to 10°C without anything else (pH, ion concentrations) changing. Would the concentration of gas phase CO2 increase or decrease?

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Qualitatively, what does this mean for the temperature dependence of ocean uptake of atmospheric CO2? (3 points) The Henry’s Law coefficient would decrease (see table). From the relationship provided in part (c), this means the concentration of gas phase CO2 would have to increase. Thus, the ocean can take up less atmospheric CO2 at warmer temperatures (CO2 is less soluble in water at warmer temperatures).

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IV. Stratospheric Chemistry (24 Points) a) Stratospheric ozone destruction through catalytic cycling is enhanced by the interaction of NOx, HOx, ClOx, and BrOx radicals with one another. From the following reactions, identify 4 catalytic cycles that destroy ozone and give the net reaction of the catalytic cycle. [Hint: You will need to include more than two reactions for most of the cycles. Make sure you balance the appropriate species.] (8 points)

(1) Br + O3 → BrO + O2 (2) BrCl + hν → Br + Cl (3) BrO + ClO → BrCl + O2 (4) BrO + ClO → ClOO + Br (5) BrO + HO2 → HOBr + O2 (6) Cl + CH4 → HCl + CH3 (7) Cl + O3 → ClO + O2 (8) ClO + HO2 → HOCl + O2 (9) ClO + NO → Cl + NO2 (10) ClOO + M → Cl + O2 (11) HOBr + hν → OH + Br (12) HOCl + hν → OH + Cl (13) NO2 + hν → NO + O (14) OH + NO2 + M → HNO3 + M (15) OH + O3 → HO2 + O2 (16) O + HO2 → O2 + OH

Cl + O3 → ClO + O2 OH + O3 → HO2 + O2 ClO + HO2 → HOCl + O2 HOCl + hν → OH + Cl net: 2O3 → 3O2  Br + O3 → BrO + O2 OH + O3 → HO2 + O2 BrO + HO2 → HOBr + O2 HOBr + hν → OH + Br net: 2O3 → 3O2 Br + O3 → BrO + O2 Cl + O3 → ClO + O2 BrO + ClO → BrCl + O2 BrCl + hν → Br + Cl net: 2O3 → 3O2 Br + O3 → BrO + O2 Cl + O3 → ClO + O2 BrO + ClO → ClOO + Br ClOO + M → Cl + O2 net: 2O3 → 3O2  

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OH + O3 → HO2 + O2 O + HO2 → O2 + OH net: O + O3 → 2O2  Cl + O3 → ClO + O2 OH + O3 → HO2 + O2 O + HO2 → O2 + OH ClO + NO → Cl + NO2 NO2 + hν → NO + O net: 2O3 → 2O2 b) Chlorine nitrate (ClONO2) is an important reservoir species in the stratosphere. It is formed by:

(1) ClO + NO2 + M → ClONO2 + M It is lost via photolysis:

(2) ClONO2 + hν → ClO + NO2 The rate constant for Reaction (2) is 5x10-5 s-1 in the middle stratosphere averaged over daylight hours. What is the lifetime of ClONO2 against photolysis in this region? What is the effect of this reaction on catalytic cycles affecting ozone? (4 points) τ = [ClONO2]/j[ClONO2] = 1/j = 1/(5x10-5 s-1) = 2x104 s = 5.6 h The photolysis of chlorine nitrate produces ClO and NO2 radicals which catalytically cycle and destroy ozone. c) What would be the effect on stratospheric ozone of increasing emissions of CFCs, and why? What would be the effect on stratospheric catalytic NOx cycling of increasing concentrations of N2O, and why? What would be the effect on stratospheric catalytic ClOx cycling of increasing concentrations of N2O, and why? (4 points) Increasing emissions of CFCs would lead to increased stratospheric concentrations of chlorine radicals and increased ozone destruction in the ClOx catalytic cycle.  Increasing concentrations of N2O would lead to increased stratospheric concentrations of NO and increased ozone destruction in the NOx catalytic cycle.  Increasing concentrations of N2O would lead to increased stratospheric concentrations of NO and which would tie up some of the ClOx in the ClNO3 reservoir, decreasing the loss of O3 due to catalytic cycling. d) Consider the following series of reactions which control the cycling between Cl and ClO in the ClOx chemical family:

(1) Cl + O3 → ClO + O2 k1 = 1.2x10-11 cm3 sec-1

(2) ClO + O → Cl + O2 k2 = 3.8x10-11 cm3 sec-1

(3) ClO + NO → Cl + NO2 k3 = 1.7x10-11 cm3 sec-1

At 50km, [O] = 2.3x109 molec cm-3, [O3] = 6.3x1010 molec cm-3, [NO] = 1.5x108 molec cm-3.

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What is the ratio of Cl to ClO at this altitude? (4 points) Steady state for Cl (or ClO): R1 = R2 + R3 k1[Cl][O3] = k2[ClO][O] + k3[ClO][NO] [Cl]/[ClO] = (k2[O] + k3[NO])/(k1[O3]) [Cl]/[ClO] = {(3.8x10-11 cm3 sec-1)*(2.3x109 molec cm-3) + (1.7x10-11 cm3 sec-1)*(1.5x108 molec cm-3)} /{(1.2x10-11 cm3 sec-1)*(6.3x1010 molec cm-3)} [Cl]/[ClO] = 0.12 e) Consider the following:

(1) Cl + O3 → ClO + O2 (2) Br + O3 → BrO + O3 (3) ClO + BrO → Cl + Br + O2

Reaction (3) is the rate limiting step and has rate constant k3

= 1x10-11 cm3 sec-1. What is the symbolic expression for ozone loss? Given number density of air 2.0x1018 molec cm-3 and mixing ratios [ClO] = 1 ppbv, [BrO] = 8 pptv, what is the loss rate of ozone? (4 points) [ClO] = (1/1x109)*(2.0x1018 molec cm-3) = 2.0x109 molec cm-3

[BrO] = (8/1x1012)*(2.0x1018 molec cm-3) = 1.6x107 molec cm-3

-d[O3]/dt = 2R3 = 2k3[ClO][BrO] = 2*(1x10-11 cm3 sec-1)*(2.0x109 molec cm-3)*(1.6x107 molec cm-3) -d[O3]/dt = 6.4x105 molec cm-3 s-1

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V. Tropospheric Chemistry (24 Points) a) The dominant sink of atmospheric methane (CH4) is reaction with OH CH4 + OH → CH3 + H2O ; k = 2.45 x 10-12 exp( -1775 / T ) Identify each of the four species as radical or non-radical. (3 points) CH4: Non-radical; even number of valence electrons OH: Radical; odd number of valence electrons CH3: Radical; odd number of valence electrons H2O: Non-radical; even number of valence electrons b) The rate constant at 298 K is k=6.3 x 10-15 cm3 molec-1 s-1, and present day concentrations of methane are 1.8 ppmv. If the pressure is 1000 hPa, the temperature is 298 K, and we assume an OH concentration of 1.0x106 molec cm-3, calculate the rate of methane loss in molec cm-3 s-1. (7 points) M = P / kB T = 1e5 Pa / ( 1.38 x 10-23 J K-1 • 298K ) = 2.43 x 1025 molec m-3

= 2.43 x 1019 molec cm-3

nCH4 = ( 1.8 x 10-6 ) ( 2.43 x 1019 molec cm-3 ) = 4.4 x 1013 molec CH4 cm-3

-dCH4/dt = k [CH4][OH] = ( 6.3 x 10-15 cm3 molec-1 s-1 ) ( 4.4 x 1013 molec cm-3 ) ( 1 x 106 molec cm-3 ) = 2.8 x 105 molec cm-3 s-1

c) The average mean global surface temperature for 2009 was 14.57ºC. Over the next 100 years, the IPCC report middle-of-the-road predictions suggest a temperature growth of about 1ºC. Assuming constant OH levels and constant CH4 emissions, would this represent a positive or negative feedback? Recall, methane is a very strong greenhouse gas. (7 points) Negative feedback; assuming constant OH and CH4, warming leads to faster rate constants, which leads to faster destruction of methane, leading to less absorption of outgoing longwave radiation, leading to less warming. d) Atmospheric measurements of the H2O2/HNO3 concentration ratio offer a simple diagnostic of whether ozone production in a polluted environment is NOx-limited or VOC-limited. Explain briefly why. (7 points) Recall the determination of whether a regime is NOx- or VOC-limited is determined by the relative loss pathway of the HOx cycling In the NOx-limited regime, the dominant loss mechanism forms H2O2. In the VOC-limited regime, the dominant loss mechanism forms HNO3. Therefore, the ratio of the two represents an estimate of the relative amounts of either loss pathway. Higher H2O2/HNO3 concentration ratios indicate a more NOx-limited regime.

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FORMULAS Ideal gas law: pV=NkT

p=ρkT/Ma p=nkT

Stefan-Boltzmann Law: Energy flux = σT4

(blackbody, energy at all wavelengths)

Solar Constant:

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FS =σTSUN

4 ⋅ 4πRSUN2

4πdEARTH −SUN2

Effective temperature:

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Teff =F(1− A)4σ

⎛

⎝ ⎜

⎞

⎠ ⎟ 1/ 4

Surface Temperature:

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Tg = (N +1)1/ 4Teff Einstein-Planck Equation:

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E = hυ = hc /λ Barometric Law:

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P(z) = P0e−z

H

Scale height:

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H =RTg

=kTmg

Dry Adiabatic Lapse Rate:

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Γdry = −ΔTΔz

=gCp

Angular velocity:

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Ω =2πt

Coriolis Force: FC = 2mvΩsin(λ)

Geostrophic Velocity:

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v =1

2Ωρ sinλΔPΔx

Beer’s Law:

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I = I0 exp(−δ) Photolysis Rate Constant:

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j =ϕσI UNITS AND CONVERSIONS 1 m = 100 cm = 1000 mm = 0.001 km 1 kg = 1000 g K = °C + 273.15 1 mole = 6.02•1023 molecules 1 mb = 100 N m-2

1 Pa = 1 N m-2 1 cm3 = 10-3 L 1 J = 1 N m 1 Watt = 1 J s-1

CONSTANTS Stefan-Boltzmann constant σ 5.67•10-8 W m-2 K-4 Earth’s radius REarth 6.38•106 m Radius of the Earth’s orbit around the Sun RSun-Earth 1.5x1011 m Solar constant for the Earth: FS 1380 W/m2 Acceleration due to gravity g 9.81 m s-2 Pressure at sea level Psurf 1.013•105 Pa Angular velocity for the Earth Ω 7.3•10-5 s-1 Boltzmann’s constant k 1.381•10-23 J K-1 Universal gas constant R 8.314 J mol-1 K-1

Avogadro’s number Av 6.02•1023 molecules mol-1

Dry adiabatic lapse rate for the Earth Γdry 9.8 K km-1

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