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 2
Quiz 3: November 30
Chapter 2 (Pressure): 6 and 7 only.
Chapter 3 (Models): all
Chapter 4 (Transport): 15, 19, 23, 24
Chapter 5 (The Continuity Equation): none
Chapter 6 (Biogeochemistry): all except 6 - 9, 29 - 30, 37, 39. Slide 36 is especially important.
Chapter 7 (Radiation and Climate): all except 5, 21.
Chapter 8 (Aerosols): all except 11 - 12 (but need for Acid rain chapter), 14, 28, 31 - 32, 35 - 37. (Slide 24 important)
Chapter 10 (Stratospheric Chemistry): all except 8 and 13.
Chapter 11 (Tropospheric Chemistry): all except 13, 32, and 35.
Chapter 12 (Urban Air Pollution): all except 16.
Chapter 13 (Acid Rain): 1 - 6.
Chapters 1-6 from the text. However, if there is a section from the text that was not discussed in class, it is unlikely it will be on the quiz. First two assignments, and lecture notes up to the end of Geochemical cycles (i.e. nothing on radiation). Required slides from first six chapters.
Chapters 7 - 10 from the text, except for sections 7.4.3, 7.5, 9.2, and only section 10.1 from Chapter 10. Pages 46 - 86 from the lectures notes. Required slides from Chapters 7, 8, 10. Assignments 4 and 5. In terms of the old test questions you would be potentially be responsible for questions dealing with Chapters 7, 8, 9, and S1 from Chapter 10. From last years quiz 2 below, all except question 2,3, and 7, which has not yet been covered.
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.
(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 ths 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 detsroyed much more slowly.
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 exp[22.49 - (6142/T)] with T in Kelvin and P0 = 6.11 hPa.
Question 2
1.2 from Jacob.
Question 3
1.3 from Jacob.
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.
2. There are now two first order loss processes in the box. The fraction f 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.
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).
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.
Question 3
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 cjemical 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.
3. No new math here. Just use the equations from part 2.
Question 1
6.1 parts (3,4,5) only
Question 2
6.2 parts (1,4)
Question 3
6.5 Part (3) only. I was able to get an answer roughly consistent with the book using the following assumptions: (1) The increases apply for 100 years; (2) the nitrogen lifetime in the land biota+soil reservoir is effectively infinite (actually about 320 years), so whatever was added has stayed; (3) increases in nitrogen fixation in the atmosphere due to fossil fuel are spread over the land and ocean in the same proportions as now; (4) increases in the biota+soil reservoir are distributed over the soil and biota reservoir in the same proportions as the current sizes of the reservoirs; (5) You can reproduce a very small impact on ocean biota by taking into account the fast removal of nitrogen from the ocean (about 0.6 year lifetime), though I haven't obtained 0.1%. The main thing is try and make reasonable arguments. I don't think there is a perfectly correct way of doing this. The above procedure would obviously yield an overestimate of the nitrogen in land biota + soil.
Question 4
6.8 (all parts)
Answer to Question 1
6.1(3) : Add 15CO2 + 15H2O -> 15(CH2O) + 15O2 to the reaction and you get net 15O2 production. It becomes an oxygen producing reaction since O2 was produced in the production of biomass (CH2O).
Answer to Question 2
6.2(1) : This question appears to have two answers. One answer is that by removing CO3(2-) from sea water, the equilibrium will shift to replace the lost CO3(2-) 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 CO3(2-), 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.
Answer to Question 4
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 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 E(20) molecules/m2). The absorption of cross section of O3 at 300 nm is 34.3E(-20) cm2. Note: A Dobsun Unit is a column amount, not a concentration. The column amount is the integrated concentration over the depth of the atmosphere.
(ii) Find the optical depth and fractional transmission of the atmosphere with overhead sun at 220 nm due to oxygen absorption. Assume the surface pressure is 1000 mb. The absorption of cross section of O2 at 220 nm is 4.46E(-24) cm2.
Question 2
7.2
Question 3
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.
Answer to Question 1
1(ii): Convert surface pressure to kg/m2. Convert to moles/m2 using average molec weight of dry air. Multiply by 0.2 since oxygen is about 20% of atm by molecular number. Multiply by Avogadro's number to get molec O2/m2. Then multiply by absorption cross-section. Should get optical depth around 17 and very small fractional transmission.
Answer to Question 3
8. (iii) Clouds that produce rain typically lift up air from near the surface where water vapor concentrations are highest. An air parcel will therefore have a shorter time before which it will be entrained into a rain producing cloud, than an air parcel in the upper troposphere.
Question 1
10.2 (all parts)
Question 2
10.3 (all parts)
Answer to Question 1
Question 1
10.4 from Jacob
Question 2
10.6 from Jacob (Parts 1 and 2 only)
Question 3
10.10 from Jacob
Question 1
11.1 from Jacob
Question 2
11.2 from Jacob
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.)