Undergraduates: Three quizzes (30 percent), Assignments (20 percent), final exam (50 percent). Graduates: Three quizzes (30 percent), Assignments (15 percent), project (10 percent), final exam (45 percent).
The library has a book by Michael McElroy. It is excellent. There is a blue 6 volume encyclopedia set in the reserve section of the librray that has a lot on atmospheric chemistry.
The project is due Nov 30. It should be an essay of about 5 - 6 pages (at least 1200 words), and you will also give a 10 - 15 minute talk on it to the class. The project should be oriented around some particular issue or controversy in which you give both sides.
Quiz 2: November 1
Quiz 3: November XX
Lecture Notes up to September 27. Chapters 1 - 3 and 6 from the text (but excluding Sections 2.4, 2.5) The required parts of Chapter 4 are Figures 4-12, 4-19, 4-22, 4-24, and the discussions of these Figures from the text. We discussed (6.18), but its derivation is not required. First four assignments. Required slides from first six chapters. All questions given for the first six Chapters of Previous Test Questions given above. However, we have not yet discussed the stratospheric dynamics so the material discussed at the start of Chapter 4 of the Previous Test Questions is not required.
The main material will be the content covered since Quiz 1: optical depth, aerosols, and stratospheric gas phase chemistry (i.e. not polar ozone loss). From the text: Section 7.6, Chapters 8 and 9, and Chapter 10 (but not 10.3 on Polar Ozone Loss). All class notes, slides, and assignment questions covering these topics. In terms of the old test questions, you would be potentially be responsible for questions dealing with Chapters 8, 9, and gas phase stratospheric chemistry from Chapter 10 (but not S6, S7, S8, S9, or S10).
Chapters 10 and 11 from the text, i.e. stratospheric and tropospheric chemistry. Required slides from Chapters 10 and 11. Assignments 5, 6, and 7. All lecture notes on Chapters 10 and 11. The main methodologies we have for analytically solving problems in atmospheric chemistry are box models and family style approximations, so these are also required.
Required material from the first three quizzes. Lecture notes and slides from the acid rain Chapter. Slides (but not lecture notes) from the Chapter 12 (Ozone Air Pollution). We did not do Assignment 8, but you should be familiar with how to answer a question like 13.1 or 13.4.
Measures of Atmospheric Composition; Required Slides: 1 - 7, 9 - 12, 14
Chapter 2 (Pressure); Required Slides: 1 - 7 only.
Chapter 3 (Models); Required Slides: all
Chapter 4 (Transport); Required Slides: 15, 19, 24
Chapter 6 (Biogeochemistry); Required Slides: 1 - 5, 10 - 28, 31 - 38.
(26) Shows the decrease in atmospheric CO2 over the last 50 million years due to increased weathering over the Himalayas (caused by collision of India with Asia). Anarctica was forested 50 million years ago.
(24) Shows that only a fraction of emitted CO2 (about half) stays in the atmosphere.
(27) For a fixed pH of 8.2, and a well mixed ocean, it would be expected that only 3% of inorganic carbon would be in the atmosphere. Figure 6-10 shows that this fraction for the preindustrial atmosphere was approximately (615/(840+9700+26000) = 0.017. I am not sure why this is a bit smaller than 3%, but perhaps because the numbers for carbon in the ocean include forms other than DIC, or the preindustrial value of [H+] was slightly lower. A more sophisticated calculation, in which the added CO2 lowers the pH of the ocean, would show that 28% of added CO2 should remain in the atmosphere. This is a much more realistic estimate, but still lower than observed, because it assumes that the ocean is in equilibrium with the atmosphere.
(28) The ocean mixed layer is the layer near the surface that is turbulently mixed by surface winds, and therefore comes into relatively fast steady state with atmospheric CO2. However, it is a tiny fraction of the ocean overall. The main route of the CO2 enhanced parts of the ocean in steady state with the current atmospheric CO2 is via deep water formation of dense (salty + cold) ocean water in the North Atlantic. But this process is slow (only a rate of 1.6/year relative to a deep ocean reservoir of 970), so this process would take more than 200 years to transport CO2 rich water to the deep ocean.
(31) Mainly showing how Ocean CO2 lags the increase in atmospheric CO2 (as you would expect since has "memory" of older atmosphere), and the increasing acidity of the ocean.
(32) Major processes of carbon transport in surface biomass.
(33) The main point here is that there is more carbon in living biomass and the soil than in the atmosphere, so that relatively modest changes in carbon storage on land can significantly affect atmospheric CO2.
(35) Atmospheric oxygen is going down as CO2 goes up. To some extent reversing the O2 buildup that occurred during the Carboniferous.
(36) Showing how simultaneous time series of CO2 and O2 can provide insight into the "missing sink" problem for CO2. The arrow for fossil fuel burning is based on an estimate of the fossil fuel mass burned during this period, plus an assumed C to H ratio for the fossil fuel that was burned. In problem 6.8, it is argued that fossil fuel burning should consume 1.4 moles of O2 for every mole of C that is burned. However, for this arrow, O2 goes down from -17 to -57 (40 decrease), while CO2 increases from 352 to 382 (30 increase), so in this paper, the author appears to be assuming that fossil fuel burning destroys 1.33 moles of O2 for every mole of C burned. Perhaps the C to H ratio of the assumed fossil fuel mix was a bit different. In any case, what you want to do is add some combination of an ocean uptake vector plus a land uptake vector to the tip of the fossil fuel burning vector to recover the observed vector. Since ocean uptake is O2 neutral, it is a pure CO2 reduction, whereas land uptake is a 1:1 reduction in CO2 and O2. Knowledge of the direction of these arrows helps us infer the specific combination of ocean and land uptake that is responsible for the observed CO2 + O2 time series.
(37) Problem 6.8 mentions that fossil fuel burning adds about 6.3 Pg/year to the atmosphere (in the 90's). This is much smaller than the rate of global photosynthesis (120 Pg/year), so that a small residual in net biome production can significantly affect the atmospheric CO2 budget.
Chapter 7 (Radiation and Climate); Required Slides: 25, 26, 27, 28
Chapter 8 (Aerosols); Required Slides: all except 11, 14, 30 - 32 (Slide 24 important)
(1) Air pollutants in the US. PM2.5 second leading cause of air pollution in excess of standards after ozone.
(2) Fine and Coarse Mode aerosols. Bimodal size distribution. Comparable in terms of mass. Different physical origins.
(3) PM2.5 can reach the lungs most efficiently. Smaller aerosols have a higher deposited fraction due to diffusion, and larger (coarse) due to impaction.
(4) Life expectancy versus PM2.5 (1978 - 1982). Life expectancy within a city correlates with the annual PM2.5 concentration. Obviously how close you live to a highway also makes a big difference within a city. Correlation is not causation, but this plot strongly suggests that PM2.5 does play a role in reducing life expectancy.
(5) Life expectancy versus PM2.5 (1997 - 2001). Significant improvement in urban air quality. Cars/factories cleaner.
(6) Annual mean PM2.5 from satellite measurements. Saharan dust, higher PM2.5 in eastern US, Europe, populated regions of India and China.
(7) Fine aerosol composition in North America. Higher organics in SE US, higher soil in MidWest arid/desert regions. Largest sulfate in Ohio/Pittsburgh. Higher nitrates in CA (likely from cars).
(8) SO2 emissions, sulphate aerosol concentration, and sulphate wet deposition in the US. SO2 has a relatively short chemical lifetime before being converted to a soluble form, so these patterns are all similar.
(9) Two questions
(10) Partitioning of SO2 in the acqueous phase. This is similar to the partioning of CO2 in water, but the curves shifted to lower pH since SO2 is more acidic in water. Rainwater pH generally 4 - 6. Acid Rain 4, Natural usually around 5.
(12) Oxidation of SO2 by H2O2. Just the resultant shown in red.
(13) Observed titration of sulfur dioxide (SO2) by hydrogen peroxide (H2O2) in a cloud. Slide 12 shows that, in the presence of cloud droplets, there is an effective acqueous phase mediated reaction whereby SO2 and H2O2 from the gas phase react to produce acqueous phase SO4 and 2 H+. This reaction is quite fast and often goes to completion (i.e. until one of the gas phase reactants is used up). In the process, the cloud droplets become more acidic, so that the rain these clouds produce will also be acidic. The gas phase mixing ratios of SO2 and H2O2 are often comparable. In a cloud, if SO2 is larger, and the reaction with H2O2 goes to completion, you would expect that there would be zero H2O2 left in the cloud and some SO2 to remain in the gas phase (depending on the acidity of the cloud droplets). Conversely if H2O2 is larger than SO2, you would expect there to be zero SO2 left in the gas phase with some non-zero H2O2. Since the reaction is fast enough, the species titrate each other, and can not coexist in the gas phase when cloud droplets are present. I think this is the main purpose of this figure.
(15) Very strong discrete sources associated with mining or power plants (or an active volcano). Don't weaker natural sources very well.
(16)
(17) Long term trends in US SO2 emissions. CAA means 1970 Clean Air Act.
(18) Decline of sulphate aerosol in the US. Models able to simulate this decline.
(19) Global SO2 emission trends
(20) SO2 emission trends from coal use in India
(21) Sulphate-nitrate-ammonium aerosol
(22) Global ammonia emissions
(23) Thermodynamic rules for SNA aerosol
(24) Three different regimes for SNA aerosol formation
(25) Aerosols contain organics.
(26) Primary and secondary organic aerosol.
(27) Fine particulate matter PM2.5 in the US.
(28) The main biogenic Volatile Organic Compounds (VOC's): sources of secondary organic aerosol (SOA).
(29) Two models for formation of SOA.
(33) Correlation of SOA with sulphate
(34) Long term trends of organics and sulphate aerosol in the Southeast
Chapter 10 (Stratospheric Chemistry)
Required Slides: all except 8 and 13. There are some comments below. We have directly covered much of this material in the lectures, so comments on most of the slides are not needed.
(8) Note that the water vapor mixing ratio is a minimum at the tropical tropopause. The stratospheric entry methane mixing ratio at the tropical tropopause is 1.6 ppmv, and as it is oxidized in the stratosphere, each CH4 produces two H2O. This is the main reason the water vapor mixing ratio of an air parcel increases with the length of time spent in the stratosphere, as shown here.
(9) Question 1 is straightforward. For 2, H2O2 formation followed by photolysis is not a HOx sink, but OH attack on H2O2 implies that H2O2 formation represents a loss of two HOx, so reduces ozone destruction from HOx.
(20) Note that NOy is also called total reactive nitrogen.
(21) N2O enters the stratosphere with a mixing ratio of about 320 ppbv. The N2O mixing ratio of an air parcel goes down with length of time spent in the stratosphere, as N2O is subject to UV photolysis and O1D attack.
(22) Based on the coincidence of the invention of the Haber process for producing artificial fertilizer, and the onset of the increase in N2O, you would say that artificial fertizers are almost certainly playing a role in the N2O increase.
(24) In class I have often used the image of the ozone bucket. This gives the magnitude of the various leaks (i.e. ozone destroying catayltic cycles) out the bottom of the bucket, as a function of height.
(25) Note that 100 hPa is roughly the tropopause. Like other tropospheric source gases, the CFC's are destroyed in the stratosphere at a rate that increases with altitude.
(28) This is an important figure which shows the conversion of the chlorine source gases to Cly in the stratosphere (the same as N2O is converted to NOy). HCl is the dominant chlorine reservoir at most altitudes.
(29) After the phase out of man made stable chlorine compunds in the early 1990's the less stable chlorine molecules have decreased more quickly (since shorter lifetimes).
(30) Question 1: HNO4 is a reservoir of both NOx and HOx, and as such, its formation would reduce decrease ozone destruction from both of these radical families. Question 2: As the stability goes down, the fraction of of the halide in the reservoir form (HF, HCl, HBr, HI) goes down, and it becomes better at destroying ozone.
(35) Important plot showing the seasonal cycle of Antarctic ozone depletion, which happens in their spring. Even though the solar minimum occurs June 21, there is a time lag to generate the temperature minimum (of about 4-6 weeks), then once the PSC's are produced by the cold temperatures, there is another lag to destroy ozone since the return of sunlight is needed to photolyze the chlorine dimer to destroy ozone rapidly. Therefore the ozone minimum (around Oct 1) lags the solar minimum by more than three months.
(37) The anticorrelation between ClO and ozone is evidence that ozone is being destroyed by ClOx. Note the filamentary structure produced by the wind shear around the polar vortex. In solid body rotation (like a record player), points stay at fixed relative distance, so rotation in the presence of a radial chemical gradient would not produce filaments. But the circular flow around the South Pole is not solid body rotation, so the wind shear acts on the chemical gradients across the boundary of the vortex to produce filaments. Think of using an egg beater to mix two different ingredients. The circular flow produced by the egg beater is not solid body rotation, so the two ingredients are mixed together. Despite this mixing, there is still come chemical isolation around the polar vortex (due to a PV gradient). The same considerations apply to a hurricane. Again, the rotation about the center is not in general solid body rotation at any height, and indeed at low levels, there has to be an inward spiralling flow to fuel the hurricane, and compensate for the divergent outflow in the upper troposphere.
(38) Note the growth in the size of the high ClO region as the polar night shrinks after June 15. In the polar night, ClO is low since the dimer does not photolyze. Outside the polar vortex, ClO is low since temperatures not low enough to make PSC's. So region of enhanced ClO like a doughnut in July and August.
(41) Important slide showing the difference in temperatures between the NP and SP winter. Note roughly 6 week temperature minimum lag in the Antarctic relative the solar minimum. Temperatures in the Arctic are too warm for pure ice PSC's, so do not get denitrification due to falling ice PSC's (with smal piece of NAT at the core).
(44) 4 main differences between Arctic and Antarctic polar ozone depletion. (1) Start with higher ozone columns in Arctic (420 DU as opposed to 270 DU), (2) Colder T's in Antarctic so larger volume of PSC exposed air and higher conversion of HCl + ClONO2 to ClOx, (3) Get pure ice PSC's in the Antarctic, and (4) get denitrification in the Antarctic.
(45) We discussed the mechanism in class why rising greenhouse gas concentrations cool the stratosphere.
(46) ODP stands for ozone depeletion potential. It depends on the longevity of the CFC, what fraction is destroyed in the troposphere, how many Cl's it has, etc.
(47) Stratospheric recovery should occur as Cly in the stratosphere goes down, but could be offset by cooling stratospheric temperatures.
(49) Question 1: this would mitigate ozone destruction since dimer formation would be a null cycle with respect to ozone. Question 2: see doughnut discussion above.
Chapter 11 (Tropospheric Chemistry); Required Slises: all except 13, 32, and 35.
(11) Question 1: You could say it would increase tropospheric OH by increasing the downward transport of stratospheric ozone into the troposphere, but the larger effect is likely that it would decrease the radiative flux with wavelength less than 320 nm, so decrease OH. Question 2: In general sunlight is just an accelerant of what is happening, so in polluted regions where ozone is being produced, it would tend to increase the rate of oxidation of hydrocarbons and produce more ozone, while in very clean regions (i.e. low CO, NO, and hydrocarbons) where ozone is being destroyed, it would tend to increase ozone destruction.
Chapter 12 (Urban Air Pollution); Required Slides: all except 16
Chapter 13 (Acid Rain); Required Slides: 1 - 6
(i) Draw a reaction map: chemicals connected by arrows representing the reactions.
(ii) Put a dashed box around the family.
(iii) Divide the reactions into those that stay in box and those that cross the box boundaries.
(iv) Reactions that stay in the box are the family partitioning reactions.
(v) Reactions that cross the boundaries of the box are family source/sink reactions.
(vi) Using the family partitioning reactions only, and steady state assumptions on individual family members, determine ratios between family members.
(vii) Impose steady state on the family as a whole using the family source/sink reactions.
(viii) to solve for individual concentrations, you usually have to implement the ratios from (iv) in the equation from (vii).
Students often get confused about why this works. How is it possible to ignore a reaction in one steady state expression but include it in another? This ends up working (i.e. giving highly accurate expressions for chemical concentrations), because reactions rates and concentrations often vary by many orders of magnitude, with the reactions creating and destroying family members being "slow" and the reaction rates creating and destroying individual family members often extremely "fast". In particular, radical species tend to interconvert very quickly while more stable species are created and destroyed much more slowly.
Assignments: Must be handed in IN PAPER. If I am not in my office, slide it under my door. If you are sick, it is OK to email me a copy of your assignment, but you must give me a paper copy as soon as possible. In general, I will not be printing or marking assignments handed in electronically.
Question 1
1.1 from Jacob. The text does not appear to give a formula for the saturated vapor pressure of water. You can use P(H20,sat) = P0 e[22.49 - (6142/T)] with T in Kelvin and P0 = 6.11 hPa. Fog will form when the relative humidity reaches 100%, i.e. that the ambient water vapor pressure is equal to the saturated water vapor pressure at that temperature.
Question 2
(i) 1.2 from Jacob. In a cloud the relative humidity is 100%, i.e. the water vapor pressure is equal to the saturated water vapor pressure at the cloud temperature. Note that when you use pressure in the ideal gas law you must convert hPa to Pa.
(ii) The pressure in the cloud is 800 hPa. What is the total number density in the cloud? We call this [M] or na. Note that a useful form of the ideal gas law is given in Exercice 1-1.
(iii) What is the mixing ratio of water vapor in the cloud?
(iv) What is the number density (or concentration) of water vapor in the cloud?
Question 3
1.3 from Jacob. For (1), express your answer in parts per million ppmv. For (2), express your answer in parts per billion ppbv. For (3), express your answer in both ozone molecules per m2, and Dobson units (DU). 1 DU is equal to 2.69 x 1020 molecules per meter squared. It is also the amount of ozone that would form a layer 0.01 mm thick at ``standard" surface temperature (273 K) and pressure (1013 hPa).
Question 1
3.2 from Jacob
1. Assume the mass in the box is m, and the density is ρ. You have to solve for the mass dm removed from the box in a given time dt, given a horizontal wind speed U. Solve for dm/dt and the coefficient in front of m is the first order loss rate. From this you get the residence time with respect to outward transport τ. (Answer: 5.8 days)
2. There are now two first order loss processes in the box. The fraction f that is exported is the amount that is exported by dynamics divided by the sum of the two processes.
Question 2
3.3 from Jacob.
1. For ms, there are two dynamical terms, and one chemical term. Refer to the first order loss rate due to radioactive decay as kr. For mt, there are two dynamical terms and two other terms (wet deposition and radioactive decay). Refer to the first order loss rate with respect to wet deposition as kw. Note that the question is only asking that you write down the equations. There is no need to attempt to solve for the k's. This is actually impossible at this point without an approximation.
Answer:
For stratospheric strontium: dmS/dt = kTS mT - (kST + kR) mS
For tropospheric strontium: dmT/dt = kST mS - (kTS + kR + kW) mT
Note that the question above says : do not attempt to solve for the k's. This means, for example, that you should include the dynamical transport of tropospheric strontium into the stratosphere at this point, even though later you can assume that this is zero.
2. You are given the total first order loss constant for strontium in the stratosphere. If transport from the troposphere can be ignored, it can be assumed that this k is equal to the sum of the k's from the two loss processes. The reason you can ignore transport from the troposphere is that mt is very small since washout is fast for water soluble compounds in the troposphere. It is almost impossible for any water soluble compound to go from the troposphere to the stratosphere. The first order loss rate for radioactive decay kr is indirectly given to you. You have to determine how to solve it from the half life. Note that the half life is not exactly equal to the lifetime.
3. You have to assume that the mass transport from the troposphere to the stratosphere is equal to the mass transport from the stratosphere to the troposphere. Note that, although this is quite accurate for the total mass (ignoring seasonal variations), it will almost never be true for a particular chemical species. Note that the dynamical residence time is the reciprocal of the outward transport first order loss rate. You must use the rule: the total mass of a layer is proportional to the pressure difference between the top and bottom. This only applies to the total mass, not to the mass of any specific compound. When answering the question, remember that the dominant loss process is the one with the largest first order loss rate (or fastest timescale). This was done in class.
4. This question is actually similar to the second part of questions 3.2. To solve you need the k's for outward transport and chemical loss (or can also solve using lifetimes). The fraction lost to a particular process is always the amount lost via that process, divided by the total loss. (I believe the fractions are 16% and 44%).
Question 1
3.7 from Jacob.
1. Use a 1 box model in which the box is the entire atmosphere (stratosphere plus troposphere). There are therefore no transport terms; only emission and chemical loss. Steady state means you can set dm/dt = 0.
2. You have to use the equation for dm/dt, the given E and first order loss rate, and the starting 1989 mass, to solve for the mass in 1996. After 1996, when the emission is zero, you have simple exponential decay. (I get 2050: 7 x 109 kg and 2100: 2 x 109 kg.)
3. No new math here. Just use the equations from part 2. (I get 2050: 9.5 x 109 kg and 2100: 5.7 x 109 kg.)
Question 2
6.1 parts (1,4,5) only.
For question 4, you could compare the amount of oxygen produced via hydrogen escape to the total mass of oxygen in the atmosphere. Or, you could compare it with the total mass of oxygen on earth (I guess determine via googling). Probably, the second comparison is more appropriate because there is lots of time for oxygen in the atmosphere to exchange with the solid rock reservoirs over the past 4.5 Billion years.
The easiest way to do question 5 is probably to infer it from the magnitude of the seasonal cycle in CO2 given in Fig 6-6.
Answers:
1. The amount of biomass in the tropical rainforests must be less than the total global biomass given in Fig 6-4 as 700 Pg of C. Since burning of 1 mole of C removes 1 mole of O from the atmosphere (R2), burning of the tropical rainforests can remove, at maximum, 700(16/12) = 933 Pg of O (multiplying by the molar mass ratio of oxygen to carbon). This is much smaller than the current mass of O in the atmosphere given in Table 6-4 as 1,200,000 Pg O. Therefore, burning of the tropical rainforest would reduce atmospheric oxygen by only a tiny relative amount. It is true that the reason we have lots of oxygen in our atmosphere is because of the growth of biomass, but this is mainly because of the biomass that was converted to fossil organic carbon (oil, gas, coal, etc.) millions of years ago.
4. An escape of 5.5 x 107 kg H per year corresponds to a loss of 2.43 x 1020 moles of H over the given earth lifetime. For every mole of H lost, we would gain half a mole of O. Using the molecular weight of O, this corresponds to a gain of 1.94 x 1018 kg of O over the earth lifetime. Figure 6-4 gives the earth's atmospheric O2 mass as 1.2 x 1018 kg of O. These are comparable, so I would say this is a significant O2 source.
5. O2 should be largest at the end of the growing season (i.e. time of maximum land biomass), so presumably early summer. Figure 6-6 suggests that the CO2 seasonal cycle is around 5 ppmv at Mauna Loa (it would tend to increase with latitude in the Northern Hemisphere). Since photosynthesis produces 1 O2 for every CO2 absorbed, the O2 seasonal cycle should also be about 5 ppmv.
Question 3
6.5 Part (3) only. The answers are roughly 20 % and 0.1% respectively. I have some suggestions for how to do this problem below, but all reasonable approaches accepted. I am not sure you will get these answers by following my suggestions below.
(1) To find the pre-industrial N mass of the land biota box, you would construct a mass balance equation for N in the land biota box, and assume that the N mass is in steady state with the (then) sources and sinks. You can assume that the preindustrial first order loss rate of N from the land biota box via decay is the same as now (from Fig 6-3). You obviously have to adjust the inputs via industrial fertilizer and biofixation from their current values, but keep the uptake from soil the same. Solve for the preindustrial N in the land biota box, and take the percent difference from the current value given in Fig 6-3.
(2) Note that the land biofixation rate in Fig 6-3 is given as 160, whereas it says it is currently 240 in the question. Don't worry about this apparent inconsistency. I think by "land biofixation" in the question they are actually referring to the sum of "industrial fertilizer + land biofixation".
(3) I would suspect that this way of solving for the preindustrial land biota N is an overestimate, since it is likely that the uptake from soil was smaller (due to the fact that the N in the soil reservoir was smaller). But I am not sure how this uptake can be adjusted to account for this.
(4) For the preindustrial N mass of the ocean biota, you would have no choice but to keep the input from upwelling, rivers, and biofixation the same. You would also have to assume that the first order loss rates of N from the ocean biota via denitrification and decay are the same as in the current ocean (but not the values of the fluxes themselves!). This leaves what to do about the input from rain (30 in the current atmosphere). I would say that if the total production of atmospheric fixed nitrogen decreases from 30 to 5, you can assume that the total decrease of 25 gives rise to decreases in rain input over land and ocean in proportion to their current values (i.e. that 30/(30+80) of this decrease is allocated to a decrease in N rain input to the ocean. You again assume steady state, and again solve for the preindustrial ocean biota mass of N given the known k's.
(5) The main thing is try and make reasonable arguments. I don't think there is a perfectly correct way of doing this question. Basically, if you have a box model, and you know all the inputs, and the first order loss rate k's, and assume steady state, the mass in the box is solvable, since it is the only unknown.
(6) It might seem a bit confusing that the removal of N from the ocean biota per year (1600) is larger than the N in the ocean (1000). But this just reflects the fact that organisms in the upper ocean containing N just sink to the deeper ocean relatively quickly (residence time of less than a year). Unfortunately for the ocean biota, there is no soil reservoir with which they can exchange and recycle nutrients multiple times. Surface oceans can therefore become starved of nutrients, except in places where there is upwelling from the deeper oceans. The relatively fast removal timescale is also partly why there hasn't been much increase in ocean biota N over the past century. (Also obviously since we don't apply industrial fertilizer to the ocean like we do the land, or increase nitrogen fixing crops over the ocean, and because increased N from combustion mostly falls over land.)
Question 1
6.2 parts 1 and 4 only.
Part 1. It turns out that this is a fairly complicated question, if you take pH feedbacks into account. You can answer in the simple way or the complicated way. Just looking for a valid argument.
6.2(1) : This question appears to have two answers. One answer is that by removing CO32- from sea water, the equilibrium will shift to replace the lost CO32- in reactions (R3) to (R5). The result would be a reduction in dissolved CO2, and a drawdown of CO2 from the atmosphere. However this process would also release more H+ and decrease ocean pH. Several students mentioned reaction (R6) in the text, in which if you remove CO32-, the quilibrium would shift the reaction in the reverse direction with the production of more CO2(g). The first answer assumes that the reduction in total dissolved inorganic carbon (DIC) with coral formation dominates, while the second answer assumes that the change in DIC partitioning associated with ocean acidification will dominate. I found the following on the web: "Because the precipitation of calcium carbonate results in the sequestering of carbon, it frequently has been thought that coral reefs functions as sinks of global atmospheric CO2. However, the precipitation of calcium carbonate is accompanied by a shift of pH that results in the release of CO2. This release of CO2 is less in buffered sea water than fresh water systems; nevertheless, coral reefs are sources, not sinks, of atmospheric carbon." In any case, not clear to me there is a definitive answer to this question without detailed calculations.
Part 4. As to whether this represents a positive or negative feedback, refer to the effect on the atmospheric CO2 budget only.
Question 2
6.8 (All parts)
1.1. We did this in class. First adjust the H2O to balance the H. Then adjust the O2 to balance the O. The point here is that the carbon to hydrogen ration of the fossil fuel type affects the amount of oxygen used up in fossil fuel combustion. For example, combustion of a methane molecule (CH4) would use up two O2 (one to produce the CO2 and one to produce the two H2O).
2.1. You use the known global fossil fuel combustion during this period plus the results from 1.1.
2.2. Should get about 8.8 ppmv increase in CO2 and 12.2 decrease in O2. This is an exercise in using the Keeling method.
6.8: It is reasonably easy to show that fossil fuel burning would have led to a 8.8 increase in CO2, and a 12.2 decrease in O2. Since only a 8.8 ppmv decrease in O2 was seen, this must mean 3.4 ppmv O2 was produced by photosynthesis (Dissolution of CO2 in oceans has no effect on O2). This would mean 3.4 ppmv removed from atmosphere by photosynthesis. So, of the 8.8 emitted CO2, 3.2 stayed in the atmosphere (36%), 3.4 removed by photosynthesis (39%), with the balance 25% being taken up by the oceans. :
Question 1
(i) Find the optical depth δ and fractional transmission of the atmosphere with overhead sun at 300 nm due to ozone absorption. Assume the ozone layer is 300 Dobson Units (1 DU = 2.69 x 1020 molecules/m2). The absorption of cross section of O3 at 300 nm is σ = 34.3 x 10-20 cm2. Note: A Dobson Unit is a column amount, not a concentration. The column amount is the integrated concentration over the depth of the atmosphere. You do not have to do a vertical integral of the product of the absorption constant times the ozone number density here, since the absorption cross section is independent of temperature (i.e. can be treated as a constant).
(ii) Find the optical depth δ and fractional transmission of the atmosphere due to ozone from (i) but with the sun at an angle of 60 degrees from the vertical (i.e. solar zenith angle = 60o.)
(iii) Find the optical depth δ and fractional transmission of the atmosphere with overhead sun at 220 nm due to oxygen absorption. Assume the surface pressure is 1000 hPa. The absorption cross section of O2 at 220 nm is σ = 4.46 x 10-24 cm2.
Answer:
(iii): Convert surface pressure of 1000 hPa to 100000 Pa, and then divide by g to get atmospheric column mass in kg/m2. Obtain atmospheric moles/m2 by dividing by the average molecular weight of dry air. Multiply this by 0.2 to get the fraction of atmospheric moles that are oxygen. Multiply by Avogadro's number to get the column molec O2/m2. Then multiply by the absorption cross-section. You should get optical depth δ = 19.6 and fractional transmission T = e-δ = 3.1 x 10-9. You can not assume a length L here. The atmosphere does not have a specific length. Also, there is no reason to use the surface pressure to calculate the atmospheric density at the surface.
Question 2
8.1 parts 1,2,3. For part 2, you have to set up a mass balance equation for Be-7 in the troposphere involving one emissions source (cosmic rays), two loss terms, and one transport source term. You can assume transport from the troposphere to the stratosphere is zero. Why? We already know that the lifetime of a molecule in the troposhere for transport to the stratosphere is around 8-10 years, which is much slower than the other processes mentioned. You also have to set up a mass balance equation for the stratosphere. And in this equation, you can also assume that transport from the troposphere to the stratosphere is again zero. We know in advance that this term is likely to be small since tropospheric aerosol washout in the troposphere will tend to make the concentration of any soluble species much smaller in the troposphere than the stratosphere (unless compensated by a larger source, which is not the case here). The first step in part 2 is to solve for the stratospheric mass of Be-7, then use in the mass balance equation for Be-7 in the troposphere. Assume steady state.
When you are writing up the assignment, please show the questions in the correct order. It is harder for me to mark if the order is random.
Question 1
10.2 (all parts)
Answer 1: 20 km: 0.0015 s, 45 km: 2.8 s.
Answer 2: 20 km: 1.5 x 10-5, 45 km: 0.028.
Answer 4: 20 km: 2 years, 45 km: 6 hours.
Question 2
10.3 (all 5 parts).
Answer 1.1: 6.6 x 10-7 s and 5.4 x 102 atoms per cm3
Answer 1.2: 2.3 s and 2.4 x 109 atoms per cm3
Answer 1.3: 18 hours
Answer 2 24 %
Question 1
10.4 from Jacob. Technically, for O3 loss, it should read Ox loss. The easiest way to solve (1) is through a topological approach: write down the reaction map (i.e. chemical species and reactions connecting them, and look for closed reaction loops involving the HOx family. Then determine whether a loop destroys Ox.) For (2), note that the definition of HOx here includes H.
Question 2
10.5 from Jacob (Parts 1, 2, 3 and 5 only)
Answer 1: 0.035 s and 30 s.
Answer 5: 35 minutes.
Question 3
10.10 from Jacob Part 3 only. For part 3.2, keep in mind: if NAT particles are forming, and HNO3 and H2O are being removed from the gas phase accordingly, in what direction in the plane are you effectively moving, given the relative abundances of HNO3 and H2O?
Question 4
Please have a look at Figure 3 of the paper below.
This paper is based on measurements of N2O and NOy from the ER-2 aircraft. To measure NOy, they have a special instrument that first converts all reactive nitrogen species to NO (This includes HNO3, NO2, ClONO2, HNO4, etc), and then measures the resulting total NO. There are two sinks of N2O in the stratosphere. It can react with O1D (see R19 from Chapter 10 of the text), or it can be photolyzed (see slide 21 of Chapter 10 slides). Keep in mind that the N2O mixing ratio is often used as a proxy for the "age" of an air parcel in the stratosphere (or equivalently the cumulative exposure of the air parcel to far UV light).
(i) Use the top plot of Figure 3 to estimate the fraction of N2O that is attacked by O1D in the stratosphere. Keep in mind: what is the NOy yield per N2O molecule attacked by O1D?
Answer 5: The main issue here is that each reaction of O1D with N2O produces two NO (i.e. two NOy). So the slope has to be divided by two to get the actual fractional attack of O1D on N2O. Correct answer is around 3.8%.
(ii) Deviations from the usual linear anticorrelation between N2O and NOy in the stratosphere are sometimes used to argue for the existence of irreversible denitrification in the stratosphere by ice forming around NAT particles, and then the particle falling into the troposphere. Point out in Figure 3 where you see evidence of this, and explain briefly why this interpretation might make sense.
Question 1
11.1 from Jacob
Question 2
11.2 from Jacob (all parts)
Question 3
11.3 from Jacob (Challenging) There is a missprint in 2a, CH3OOH is produced.
Question 1
12.1 from Jacob (to 3.2. i.e. do not do 4.1,4.2)
Question 2
13.1 from Jacob
This shows thunderstorm clouds in cross-section. The new younger cloud in the middle hasn't started to spread out yet, so doesn't have the large thin glaciated anvil at the top. The tropopause is the boundary between the red and white (in the tropics at about 15 km). The stratosphere is white because of the sulfate aerosol layer from the eruption of Mt. Pinatubo. This layer cooled the earth for several years after the eruption (1991 - 1993). The reason the layer persists in the stratosphere for several years is because clouds don't go into the stratosphere, so the little aerosol particles don't get rained out. (Would come down within a week as acid rain if near the surface.)