Equilibrium Thermodynamics

Transcription

1 Part I Equilibrium hermodynamics 1 Molecular hermodynamics Perhas the most basic equation in atmosheric thermodynamics is the ideal gas law = rr where is ressure, r is the air density, is temerature, and R is the gas constant for dry air. While the derivation of this equation takes some effort, it is very much worth it because it gives us a dee understanding of what ressure and temerature really are. Any work the atmoshere can do to move around air is derived from the internal energy of its comonent gases. How much of this energy does a unit volume of gas have? Consider the following diagram: Figure 1.1: A iston Say you are ushing on a iston filled with a gas. he ressure is simly = F/A (1.1) and the amount of work you would have to do to comress the gas is dw = F ( dx) (1.2) but so from (1.2) and (1.3) F = A (1.3) dw = Adx = dv (1.4) 5

2 where dv = Adx since you are ushing an area through a differential length to get a differential volume. Alternatively, we could switch the sign to get not the amount of work done on the gas but the amount of work done by the gas dw = dv (1.5) Where does the ressure come from? he ressure arises from the motion of the molecules of gas inside the iston banging against the iston wall. Each time they bang against the iston wall, the iston acts like a erfect reflector. If the iston does not move due to this constant banging, then there must be a force balancing the momentum associated with the banging of the molecules (note bold font to distinguish from ressure momentum is a vector), that is imarted to the iston wall with every collision. his force is F = d/ (1.6) We are assuming here that every molecule that hits the iston leaves with the same energy, i.e. the same seed v and mass m, and therefore the same momentum. he iston is a erfect reflector. But momentum is a vector! herefore, the momentum imarted to the iston by a single molecule is the change in momentum associated with coming in and bouncing out: D x = mv x ( mv x )=2mv x (1.7) What is Dt though? Suose we have N molecules in volume V, or n = N/V as the concentration 1. Only molecules within distance v x Dt hit the iston 2. he volume occuied by these molecules is DV = v x ADt 3. he number of molecules that hit the iston is DN = ndv = nv x ADt, where n = DN/DV is the molecular number concentration 4. he number of molecules that hit er unit time is DN/Dt = ndv /Dt = nv x A herefore, for all molecules, we combine number 4 and (1.7) to get So F x = DN D x Dt F x = nv x A 2mv x = DN Dt D x Now, taking (1.1), this leads to the exression for ressure exerted by all molecules in the x- direction = F x /A = 2nmv 2 x 6

3 But if we only want the one half of the x-direction that is in the direction of the iston then we need to divide by two to get the ressure on the iston = F x /A = nmv 2 x (1.8) here is a comlication here though, which is that molecules have an equilbrium distribution of seeds (un-normalized here) that obeys a Boltzmann distribution m 1/2 f (v x )= ex mv 2 2k x /2k For our uroses, we are interested in the average seed = nm v 2 x We could also multily and divide by two to write = 2n mv 2 x/2 (1.9) he term in brackets is obviously merely the average kinetic translational energy of each gas molecule in the x-direction u trans,x = mv 2 x/2 (1.10) (the dot refers to er molecule ). Since it will come u later, we will also note that, at equilibrium, the kinetic energy equals the otential energy of the molecule. We now have = 2n u trans,x (1.11) We also note the gaseous motion is obviously three-dimensional, and there is kinetic energy in each of these directions: v 2 x v 2 y v 2 z m 2 = m 2 = m 2 and although it is moderately difficult to rove, it is easy to believe that if we have three degrees of freedom (i.e. f = 3), that, like kinetic and otential energy, all degrees should share the energy equally, meaning each gets a third of the total v 2 x = v 2 /3 so so v 2 u trans = f u trans,x = 3 u trans,x = m 2 = 2 3 n u trans (1.12) his equation should start to be looking vaguely familiar. We should really be calling u trans the temerature of the gas molecule. However, hindsight is 20/20, and due to some historical misstes we have defined a temerature that is related to u trans by a constant factor u trans = f 2 k = 3 k (1.13) 2 7

4 where, k = J/K/molecule is Boltzmann s constant and 3k/2 = mv 2 /2. he average kinetic (or otential) energy er degree of freedom f is k /2. herefore, = nk (1.14) It is revealing here to consider that ressure has units of J/m 3, and therefore is an exression of energy density the number density of molecules n times k. What does k reresent? It is not the internal energy density er molecule, which for translational energy alone is u trans = 3 2k > k, and there is vibrational and rotational energy too, which makes the internal energy even bigger. Perhas the best way to think about k is that it is the total energy (kinetic lus otential) carried along the direction normal to a surface. Note k.e. = 1 2 k and.e = 1 2k. hus k is the total average energy er molecule that is available to do dv work. = nk is then the density of this available energy. We ll call this average energy µ 0 = k (1.15) It will be useful later. here are several ways to exress the ideal gas law, which relates ressure temerature and density in the atmoshere (among other laces). he most common formulations are V = NR where N is the number of kilomoles. In the atmosheric sciences, we tyically make some simlifications. We relace R by R = R /M, where M is the molecular weight of the gas in kg/kmol. In this case Alternatively, we can relace r by the secific volume a = 1/r: Main oints = rr (1.16) a = R (1.17) Pressure is a force over an area where the force is due to the change in momentum associated with molecules bouncing off their surroundings Pressure has units of energy density emerature has units of energy. he amount of energy k/2 is the amount of translational (or otential) energy er degree of freedom er molecule. here are three degrees of translational freedom so internal translational energy is 3k/2. Pressure has in effect two degrees of freedom due the molecular bounce he amount of energy k is the energy er molecule available to do dv work. Pressure is the density of energy available to do dv work on the surroundings 8

5 Question What if the volume of gas is an air arcel in a surrounding atmoshere? Do the equations above need to change in any way if the box is removed? 1.1 Atmosheric alication: water vaor and dry air he ressure exerted by a mixture of chemically inert gases is equal to the sum of artial ressures of the gases. he artial ressure is roortional to the molecular density. As we showed in Eqs and 1.16, in total = nk = rr But what if there is a mixture of gases, as in the atmoshere? Dalton s Law states that we can treat the gases as if they are indeendent from one another, making the good assumtion that all the molecules have the same temerature (or translational kinetic energy). In other words, we can just add u the molecular densities n i, or energy densities i, so that = ksn i = Sr i R i = S i i (1.18) So it is convenient to searate gases into their major constituents, which in the case of the atmoshere, is mostly dry air and moist air. For dry air (which is mostly nitrogen and oxygen) For water vaor we use the symbol e for ressure and d = r d R d (1.19) e = r v R v (1.20) he mean molar weight for dry air is M d =28.97 kg/kmol so R d = R /M d = 287Jdeg 1 kg 1, and for moist air M v = 18.0 kg/kmol so R v = R /M v = 461Jdeg 1 kg 1. hus the total ressure is = d + e =(r d R d + r v R v ) Note again, that we are saying that the water vaor molecules have the same temerature (kinetic energy) as the dry air molecules. We ll introduce an imortant arameter here e = R d R v = M v M d = (1.21) We define the mixing ratio (i.e. the ratio water vaor mass to dry air) in the atmoshere as Substitute the ideal gas law (1.19) and (1.20), we get w = m v m d (1.22) w = r v e/r v = r d ( e)/r d 9

6 e = e e w ' e e (1.23) An imortant adjustment that can be made to the ideal gas equation is to adjust for the added buoyancy that is associated with water vaor being lighter than dry air so that within a given volume, the density of air is given by r = m d + m v V = r d + r v = e R d + e R v or, with some algebra r = ale 1 R d ale e (1 e) = 1 R d (1 e) w e Rearranging = h 1 rr d (1 e) e w i = rr d v where v, defined this way, is the virtual temerature. It works out that to a good aroximation v ' e w = ( w) (1.24) e Figure 1.2: Moist versus dry air 10

7 his is a very useful quantity to know since the density of moist air is given by: r = = R d v R d ( w) At constant ressure, air that is either moist or hot will be less dense than air that isn t, and be more inclined to rise (Fig. 1.2). Question Dalton s law states that artial densities and ressures are additive. Can you rovide a comelling reason why temeratures are not additive as well? Why is there not a artial temerature? 11

8 2 Intensive and extensive variables Figure 2.1: Intensive (lower case) versus extensive (uer case) variables When considering hysical systems, it is often very useful to stay aware of whether the roerty being considered is intensive or extensive. An intensive variable is one that does not deend on the volume of the system, and an extensive variable is one that does. Intensive variables are sometimes called the bulk roerties of the system. For examle, consider the following exression of the ideal gas law = rr (2.1) Say we had a box with each of the roerties, r and, and we subdivided the box, it would not change the values of these roerties, so each of these roerties is intensive. However, r = m/v (2.2) Subdividing the box, into small boxes with smaller V would lead to a corresonding reduction in the mass in each subdivided box m. hus, m and V are extensive variables. he distinction is imortant, because often we want to know how much we have of a articular thing. he nature of the thing is defined by one or two of the intensive variables, r, and (note that from 2.1, we only need two to define the third). he amount we have is defined by the mass m. In atmosheric sciences, when we consider the ideal gas equation = rr where r is an intensive variable and mass m = rv is the extensive variable, where m normally refers to either the mass of dry air or the mass of water vaor. he convention used in Wallace and Hobbs, one that is commonly emloyed is to use the lower case for intensive quantities and the uer case for extensive quantities. hus, for examle, the internal energy er unit mass is given the symbol u and the internal energy is given the symbol U. In atmosheric sciences, we very often 12

9 thing about intensive quantities, because we reference them to a arcel of air in which the mass of the air (but not the volume) does not change with time. hus u = U m In the notes here, I will exress intensive variables that are referenced with resect to volume in bold lower case, i.e. the internal energy er unit volume is u. Where the intensive variable is er molecule it is exressed, e.g., as u. Main oints Intensive variables are indeendent of how much matter there is. Extensive variables deend on how much matter there is. Question his distinction between extensive variabiles and intensive variables is articularly imortant whenever we talk about flows, of air, radiation, or whatever. Why? 13

10 3 Heat, Work and the First Law of hermodynamics We can exress Eq for the ideal gas equation in terms of er unit mass rather than er molecule: = 2 3 ru trans (3.1) where u trans = 3 R (3.2) 2 is the translational energy er kilogram due to molecules whizzing around in three dimensions. here are other tyes of internal energy we can consider (and will later), but suffice to say we can talk about a total internal energy er kilogram u. Note that, like, u is an intensive variable, so the extensive roerty is U = mu where m is the mass of air. We call U, the total internal energy. Our question, is how do we change the total internal energy? Some outside agency adds energy to the system (e. g. condensation, absortion of sunlight or terrestrial radiation), through flows of some foreign material that does not define U. We call this energy heat he system changes its energy through flows of whatever material does define U (e. g. through exansion dv or lifting mgdz). We call this work. Whatever haens, there must be conservation of energy. We exress this as the First Law of hermodynamics DU = DQ DW (3.3) where, DQ is the heat added to the system, and DW is the work done by the system. Both heat and work can be either ositive or negative, so what is the difference between the two? What is most imortant to recognize is that heat and work are just names for energy. Our goal is foremost to kee an accurate account of the amount of energy that flows into and out of a system. hat said, there are different tyes of energy. he best way to distinguish heat from work is that heat is associated with with flows of a different tye of matter than is associated with internal energy U. So if U = mu, where m is the mass of the material we are interested in (say the air in a arcel of air), there is also an external energy associated with a different tye of material that we call heat (say hotons associated with warming sulight). For examle, radiative heating is associated with hotons. Radiative heating can heat a arcel of air comosed of molecules. Heating does not directly change the extensive variable of the air arcel m. In the atmoshere, we think about gases. For a gas, an easy way of visualizing work is to examine a iston Imagine there being a chemical exlosion in the iston that releases external energy DQ to the gas. he external material flow here is due to a redistribution of electrons in a chemical reaction. his flow of external energy or heating adds to the internal energy of the gas. An increase in the 14

11 dx A 1 2 dw = dv V1 dv V2 Figure 3.1: A iston at work internal energy of the gas is an increase in the amount of energy it has that is available to do work. As a consequence, the differential rate of doing work by the gas on the iston is but V = mr so and noting that x 1 dx/ = d lnx/ dw dw dw = A dx = dv = dv = mr V dv = mr d lnv he amount of work that is done by the gas in the iston is the amount of area under the curve. or, at constant temerature DW = DW = mr A similar derivative can be alied to the heating DQ = Z V2 V 1 Z t Z t dv d lnv 0 0 dq 0

12 hus, the change in internal energy is DU = Z t 0 dq 0 mr Z t 0 d lnv 0 0 (3.4) Heating adds to internal energy so that it can do work through exansion, which allows the internal energy to relax. An imortant oint to be aware of is that the total amount of internal energy we have DU is a consequence of a time integral of heating and doing work. he extensive roerty of how much we have (e.g. U or N) deends on ast flows. he resent is linked to the changes in the ast. Main oints otal energy is conserved, meaning it can only be shifted from one form to another When energy flows into a system, two things can haen. One is that the internal energy can go u. he other is that the system can do work on something else. Work, heating, and changes to internal energy haen over time. Current states evolve from ast states. Question When a system does work on something else, where does that energy go? Is one system s work another systems heat? 3.1 Atmosheric Alication: the general circulation Since in the atmoshere we treat most things er unit mass, we can rewrite the 1st law in terms of Jkg 1. du = dq dw (3.5) since a = V /m dw = da du = dq hus, u is an intensive variable, as it indeendent of the amount of air that is considered. Substituting = R /a, the change in internal energy is da (3.6) du = dq R a da = dq R d lna Du = Dq Dw = Dq R 16 Z t 0 d lna 0 (3.7)

13 Figure 3.2: Balance between incoming shortwave radiation and outgoing longwave radiation at the to of the atmoshere Within the context of Figure 3.2 and Eq. 3.4, we can think about why the atmoshere is able to do work. Due to an imbalance between incoming shortwave radiation and outgoing longwave radiation, there is net heating at the equator and there is a net cooling at the oles. Presumably these two things balance in total, so that the internal energy of the atmoshere as a whole does not change, so Du = 0 and Dq = R Z t 0 d lna 0 0 (3.8) his equation retty much sums u climate! Radiative heating Dq ermits work to be done through the exansive rocesses of convection and winds R t d lnv Cooling at the oles leads to work done on the oles through a shrinking. Meanwhile, the globally-averaged atmosheric temerature stays constant. Energy that is available to do work takes the form of meridional motions that we call the wind. Energy will not be created or destroyed, but it will be transorted from the equators to the oles because a high energy density is maintained by the imbalance at the equator and a low energy density is maintained at the oles. he one thing that is missing in this discussion is that air does not leave the oles once it gets there, so there must be a return circulation. 17

14 4 Internal energy in an ideal gas What if the internal energy is not constant when a gas is heated? In terms of mass, Eq can be exressed as u trans = 3 2 R where the number three reresents the number of degrees of freedom associated with the kinetic energy in the x, y, and z directions. By extension, the total internal energy er kilogram is u = f R (4.1) 2 where f is the number of degrees of freedom for a molecule. he total internal energy is not just the energy derived from the translational motions of the individual gas molecules, but also from internal motions by the molecules. Now this is central to atmosheric sciences because it allows us to exress how the temerature of gas is linked to how much we heat it (through, for examle, radiative absortion), i.e. the secific heat. Remember, heating doesn t change extensive quantities associated with how much of a articular thing we have (e.g mass or number of molecules). What it does change is the intensive roerty of the amount of internal energy er unit stuff, or equivalently, the temerature. But, by how much? From the First Law of thermodynamics (Eq. 3.6), du = dq Rearranging dq = du + dw = du d lna + R If a substance that is defined by the intensive roerty u is heated at constant secific volume a (e.g,. we add heat to a box filled with air, then d lna/ = 0, and the internal energy change is equal to du dq = a We would exect the temerature to go u in the box at some rate d/. Dividing both sides by d/, we get du dq = = c v d d a where c v is what we call the secific heat at constant volume (or, equivalently, constant a). Note that it is much more intuitive to think of it in inverse d = 1 dq c v Framed this way, c v tells us how much the temerature will rise for a given amount of heating. Now, we can substitute for u, our exression from Eq. 4.1, leading to q du c v = = = f d 2 R (4.2) a 18 dw

15 Figure 4.1: Heating at constant volume leads to a temerature and internal energy rise Just by insection, we can see that, if we add energy to a volume through heating, and the volume doesn t change, then the temerature will rise. But it will not rise by as much as might be exected because there are other degrees of freedom in the substance that get a share of the energy. f is at least three due to the translational modes of the molecules. If f is more then temerature must rise even less. If you were a French engineer in the 19th century, this would be rather imortant to know because increasing temerature is what is going to drive the iston, and you would want to know what the increase in temerature is for a given release of energy from burning coal. Since the engineers needed to figure out how much temerature (or internal energy rises), this led them to the concet of the secific heat at constant volume c v. From an atmosheric science standoint, radiative heating drives temerature erturbations, although it is not quite the same because the erturbations are not normally considered at constant volume (unless we are considering the lanet as a whole), but rather at constant ressure (see the section on enthaly). Still there is a fundamental roblem, as will be shown below, which is that there is basic disagreement between what classical hysics would redict the value of f to be and the value of c v that is actually measured exerimentally. In fact, this was a key oint that led 19th century hysicists to be concerned that the wonderful successes of the kinetic theory of gases weren t all that. In commenting on this roblem in 1869 Maxwell said in a lecture I have now ut before you what I consider to be the greatest difficulty yet encountered by the molecular theory, a conumdrum that aved the way for the quantum theory of matter. 1. Monotomic Molecule (examle Ar) 3 translational degrees of freedom f = 3,U tot = 3 2 R. Value of f for Ar imlied by observations at atmosheric temeratures f = Diatomic molecule (examles: N 2 and O 2 ) 19

16 3 translational degrees of freedom 2 rotational degrees of freedom (whole body) 2 vibrational degrees of freedom (P.E. lus K.E. (k/2 each) associated with change in bond length) 7 degrees of freedom f = 7,U tot = 7 2 R. Value of f for N 2 and O 2 imlied by observations at atmosheric temeratures f = 5. here is a big discreancy here! he classical theory works great for Argon but terrible for nitrogen and oxygen. Air is mostly a diatomic gas (N 2,O 2 ), so for er kilogram of gas u = 5 2 R = c v (4.3) where we would exect u = 7R /2 based on the kinetic theory of gases. A curious thing haens though which is that if the temerature of the gas goes u a lot, say to several thousand degrees, then f does indeed aroach a value of 7 as initially redicted. How can f be a function of temerature? Nitrogen is nitrogen is nitrogen, no matter the temerature. Right? Off hand, it makes no sense. One of the first successes of the quantum theory was to resolve this roblem. he value k/2 (like R /2) is a fine reresentation of the kinetic energy er indeendent mode rovided that k is not much less than the smallest ossible energy, the quantum energy hn, where n is the oscillation frequency for that mode! Here, h is the Planck constant. Vibrational modes are higher energy than rotational or translational modes. What haens is that if the natural frequency of a vibrational mode n is sufficiently high that k hn then these modes simly will not resond to an addition of energy to the gas. Effectively these vibrational modes are frozen because the bonds are too stiff. If the temerature gets sufficiently high, then these modes can be activated again. Otherwise, these degrees of freedom are unavailable at normal atmosheric temeratures. Our atmoshere is comosed rimarily of nitrogen and oxygen, both of which have stiff bonds with high natural vibration frequencies. hese vibrational modes are simly too high otential energy to interact with the translational modes we sense as temerature. hey are way u in the hills while the translational modes live in the valley. In any case, from Eq. 4.3, we can now rewrite our exression for the first law to be dq = c v a + a Heating leads to some combination of a temerature rise at constant volume and an exansion at constant temerature Main oints Heating doesn t all go into exansion work. Some goes into raising the temerature (4.4) Energy is shared among all available degrees of freedom. he more degrees of freedom the less temerature rises. Air has 5 degrees of freedom, three translational and two rotational. 20

17 Question Eq. 4.4 is not actually used very often in the atmosheric sciences to examine temerature changes in an air arcel. What is a limitation for alying the equation to the atmoshere that wouldn t have been a concern of most 19th century engineers and hysicists? 21

18 5 otal energy or the Enthaly So now we can think of there being two tyes of energy associated with the volume of a gas. here is the internal energy er unit mass u that is due to the rotational and translational motions of the molecules. Since each degree of freedom has a otential energy R /2, then we could generalize even further to say that the internal energy is the sum of all internal otentials of the system u = f  i=1 g int i (5.1) where g int i is each internal otential. here is also energy er unit mass associated with the ressure of the molecules on their surroundings a. he total energy is normally called the enthaly h = u + a (5.2) his ressure volume energy is sometimes referred to as an external otential. hus h = f  i=1 g int i + g ext (5.3) One way to think about this is that the external otential is the otential for flows that we can see, and the internal otential is for those we can t. For an ideal gas, since a = R h = c v + R or h =(c v + R) = f + 2 R (5.4) 2 where f is the number of degrees of freedom for the molecules. For air f = 5, so dh = 7 2 Rd (5.5) o see why we have introduced the enthaly, let s go back to the first law again (Eq. 3.7) dq = du+ dw = du+ R d lna (5.6) Now, reviously we evaluate ( q/ ) a to obtain the secific heat at constant volume c v (Eq. 4.2). What about the secific heat at constant ressure? From Eq. 5.6, taking the derivative with resect to temerature: q = u + R but, since a = R / then (d lna) =(d ln ), so q = lna u ln + R 22

19 Recognizing that u = c v and R d ln = Rd, we get q = c v + R (5.7) his is the same exression dervied in Eq. 5.4! hus dh = q (5.8) his is why we introduced the term enthaly. It is because heating at constant ressure increases the enthaly of a substance. his result is imortant because we can define the secific heat at constant ressure to be hus c = dh d = q (5.9) h =(c v + R) = c (5.10) herefore, it is easy to see that c = 7 2 R. For dry air, R = 287J/deg/kg, c v = 717J/deg/kg, so c = 1004J/deg/kg. For water vaor R = 461J/deg/kg, c v = 1463J/deg/kg, so c = 1952J/deg/kg. For air, the internal energy has five degrees of freedom and ressure two (due to the bounce). hus c : c v : R ' 7 : 5 : 2. his very imortant, because what it means that if we add energy to air, five arts of seven go into increasing the internal energy density, and two arts in seven go into increasing the ressure. he reason we have introduce enthaly as well as heating, even if they change at the same rate, is because enthaly is a roerty of the system (or a state variable) where as heat is not. You cannot oint to the heat of a substance but you can oint to its enthaly. Enthaly is one of the most imortant variables in atmosheric sciences, because we cannot control for the volume of an air arcel, but we can control for its ressure. Combining Eq with Eq. 3.6, and noting that d (a)= da + ad gives an alternative exression of the First Law dq = c a If there is heating the temerature rises and the ressure dros. Main oints he enthaly is the internal energy lus the energy the system exerts on its surroundings he energy exerted on the surroundings is a ressure volume energy (5.11) he secific heat is related to the number of degrees of freedom associated with the internal energy lus the number of degrees of freedom associated with the ressure he total number of degrees of freedom is = 7. herefore c : c v : R = 7:5:2 for air. 23

20 Questions Why are we generally more interested in the enthaly of an air arcel than its internal energy? 5.1 Atmosheric Alication: the dry and moist static energy In the atmoshere, we need to consider that there are other forms of internal energy that a arcel of air can have, other than just those associated with translational, vibrational and translational motions of a gas. A second form of internal energy is gravitational energy. In our atmoshere, air can move u and down. hus, add to the total internal energy the gravitational otential er unit mass u = Âg int i = c v + gz (5.12) i Adding a = R to this we get a revised enthaly called the dry static energy h d = gz + c v + R = gz + c (5.13) so heating at constant ressure yields q = dh d d = c + gdz (5.14) A second form of internal energy is that associated with the thermodynamic hase of a substance. he latent heat er unit mass associated with a change in hase from liquid to vaor is L v = J/kg at 0 degrees C. hus the energy associated with some fraction of dry air being in the vaor hase is L v w where w = m v /m d (Eq. 1.22). hus, adding the latent heat to the total internal energy, we get u = c v + gz + L v w (5.15) which leads us to the moist static energy (MSE in Wallace and Hobbs) so h m = c + gz + L v w (5.16) q = dh m d = c + gdz + L dw v (5.17) Of course, we could kee going and secify any number of things deending on the question we were interested. As shown in Fig. 3.2, heating can be related to the radiative flux divergence through 1 r F net z So the energy that is available to create changes in, z and w comes from the absortion and emission of radiation. he conversion of radiative flux divergence to a change in h m is the starting oint for radiative and climate rocesses. = q 24

21 6 Equilibrium solutions Very often we want to consider situations that are adiabatic, which strictly means that the internal energy and enthaly do not change because there is no heating. But this is a bit boring, and in fact it is never ossible (as will be shown). All objects radiate for examle. What we look at far more frequently is equilibrium conditions where energetic flows are in balance. Most of the textbook equations in atmosheric sciences are derived from this rincile. he starting oint is three exressions of the first law (Eq. 3.8), all basically equivalent. he first exression is or, if the work is done by an ideal gas. du = Dq Dt dw (6.1) And from Eq du = Dq Dt da dh = Dq Dt + a d So again, as in Eq. 5.8, if energy is added to a arcel of air at constant ressure, it s enthaly will go u by the same amount. 6.1 Case 1: Equilibrium solution for a materially closed system at constant entroy As will be shown, constant entroy s is a short hand for no net heating of the system, i.e. is is adiabatic, but work is being done on or by the system. In this case, from Eq. 6.1, we obtain u = s w s (6.2) (6.3) (6.4) or, from Eq. 5.1, Main oints Âi g int i = s w s (6.5) For a materially closed system, the adiabatic or equilibrium condition is that work done on a system raises its internal or otential energy. Work done by the system lowers it. Alternatively, a rise in the otential energy can be balanced by an ability to do work on something else. Potential energy is an ability to do work. 25

22 6.1.1 Atmosheric alication: the geootential What is the otential energy of the atmoshere? here are many ossible forms of otential energy but here we look at the gravitational otential. Remember from Eq. 1.2 that work is required to move a system a distance through a force field. Here, the force field is F g = mg, and the distance might be dz. hus, the work done on the system to increase its height, er unit mass is gdz. At constant temerature, from Eq. 5.12, du = gdz. hus, we can term a otential energy that is the amount of energy required to lift a mass m through a gravitation acceleration g through a distance dz. he intensive variable is the geootential, which has units of energy er mass F(z)= Z z 0 g(z)dz Basically, the geootential is the otential energy er unit mass that comes from lifting through a variable gravitational field at constant temerature (see able 3.1 in Wallace and Hobbs). For most uroses, we can assume gravity is a constant, so that F(z)=g or equivalently, we can take the derivative to get Z z 0 dz df( )=gdz (6.6) 6.2 Case 2: Equilibrium solution for a materially closed system at constant ressure Now we can consider a second system in which net heating of the system at constant ressure is zero. Energy may be entering the system through heating, and leaving the system through dissiation or cooling, but there is no net flux convergence or divergence. In other words, from Eq. 6.3: or, from Eq. 5.3, d  i dh = g int i q = 0 (6.7) + g ext i = q ransformations can still occur however, but they must obey the adiabatic condition above. Adiabatic here doesn t mean no heating, but rather no net heating. Main oints (6.8) For a materially closed system at constant ressure, the adiabatic or equilibrium condition is that energetic flows in must be balanced by energetic flows out, such that total energy er unit mass h (in whatever form) does not change. he different comonents of total energy can still vary, but there must be a balance between so that gains in one are offset by loss of another 26

23 Figure 6.1: An abstract illustration of where heating of a system is balanced by an equivalent rate of dissiation or cooling, so that all rimary system state variables stay fixed Atmosheric alication: the dry adiabatic lase rate For examle, if the air is dry, and we consider the dry static energy, a change in h d at equilibrium is given by (Eq. 5.14) dh d d = c + gdz = 0 A arcel that rises or falls without exchanging energy with its surroundings at a given ressure will have dh d / = 0 and will change its temerature at rate c = g or z = g (6.9) z c his is the dry adiabatic lase rate of about G d = g/c = 9.8 K/km. Of course we could do something similar for the moist static energy, but the derivation is more comlicated (and outlined in Wallace and Hobbs roblem 3.50). Note that c is not a function of ressure, and g is only a very very weak function of ressure (the atmoshere largely disaears by the time we care), so to a very close aroximation: d dz = g c 27

24 Figure 6.2: Heating at constant ressure leads to an enthaly or total energy rise 6.3 Case 3: Equilibrium in a materially closed system at constant temerature he final condition is constant temerature. A system doesn t have to be adiabatic to be in equilibrium. Alternatively, the system might be heated, but it sufficiently fast that the temerature, or energy er degree of freedom, stays constant. In this way heating balances working. From Eq. 6.1 q = w = a Substituting the ideal gas law and the relationshi between work and otential Âi g i lna = R or Âi g i ln = R or (6.10) (6.11) (6.12) Âi g i = a (6.13) 28

25 Figure 6.3: An abstract illustration of where heating of a system leads to working by a system while the internal energy or enthaly stay fixed so we have a range of relationshi now q w Âi g i lna = = = R Main oints ln = R a = = a (6.14) For a materially closed system at constant temerature, the adiabatic or equilibrium condition is that energetic flows in must be balanced by exansion work, such that total energy er unit mass h or u (in whatever form) does not change. If temerature does not change, then net heating leads to an increase in volume or a dro of ressure Atmosheric alication: the hydrostatic equation Since the total internal energy is given by  i g i = c + gz (Eq. 5.13), at constant temerature d  i g i = gdz. hus, from Eq since a = 1/r, we obtain the equation gdz a = ad a = rg z which suggests that ressure would increase with height. But, the atmoshere doesn t fall down. here must be a balancing ressure ushing uwards. No vertical motions requires that the total 29

26 Figure 6.4: Heating at constant temerature leads to an equivalent amount of working vertical ressure gradient at any location is zero. By changing sign we obtain the hydrostatic equation = rg (6.15) z Since density is always ositive, the air ressure must decrease with height Atmosheric alication: atmosheric thickness We can now substitute the ideal gas law. Assuming that the atmosheric layer is isothermal (i.e. constant temerature), then the density is (Eq. 1.24) hus, substituting we get r = (z) R = (z) R d v = (z)g (6.16) z R d v Suose we want to get the amount of otential energy er unit mass found between two layers, we are looking for a geootential difference F 2 F 1. his is useful, because air flows from high to low otential energy, so we can use this difference in two locations to figure out where the air will go. How do we find out this quantity? Re-arranging d = 30 gdz R d v

27 Since at constant temerature df = gdz (Eq. 6.6) and d/ = d ln Integrating between two levels Giving us Z z2 z 1 gdz = Z F2 R d v d ln = gdz (6.17) F 1 df = R d v Z 2 1 d ln = R d v Z 1 2 d ln DF = F 2 F 1 = R d v ln 1 2 (6.18) his is really imortant because differences in DF gives us the amount of energy er unity mass that is available to do thermodynamic work through atmosheric flows. It is one ossible form of the otential g we described earlier. Note that if we integrate Eq and change sign to account for the hydrostatic equation we can also obtain Eq DF = gdz = R d v ln 1 2 (6.19) Hurricanes are driven by a net heating of an atmosheric column, through a flux of energy from the ocean and a radiative cooling to sace. he heating ermits work to be done by way of an exansion of the column and a ressure dro. An exression that is related to Eq is to talk about the thickness of an atmosheric layer between two ressure surfaces. o get this we just divide Eq by gravity to get DZ = Z 2 Z 1 = R d v g ln 1 2 (6.20) It is easy to see from this equation a very intuitive result. A warm air column with higher v will be thicker than a cold air column, rovided that the column is defined as lying between two ressure surfaces 1 and 2. Warm air is less dense. 31

28 7 Quasi-equilibrium transitions Figure 7.1: An abstract illustration of transitions from one equilibrium state to another We have defined a two imortant equilibrium conditions. he first is one is heating at constant ressure and dq = dh (7.1) Assuming equilibrium, we were able to derive, for examle the dry adiabatic lase rate ( / z) = g/c. he second sense is that there is heating at constant temerature is balanced by working through a change in ressure at constant temerature. his leads to the following key result q = w = Âi g i = R lna ln = R = a = a From which we were able to show that heating increases the atmosheric thickness. Now we ask what will haen to the relationshi between heating and enthaly if the ressure is changed a little bit for the constant ressure condition (Eq. 7.2) or what will haen to the (7.2) 32

29 relationshi between heating and working if the temerature is changed a little bit for the constant temerature condition (Eq. 7.2). Effectively, we are asking what is the sensitivity of these equilibrium conditions to small changes. We are still looking for an equilibrium condition. However, we now want to know how the equilibrium solution is different at different temeratures. Effectively what we are seeking is an exression for quasi-equilibrium transitions. Conditions are changing slowly enough that something very close to equilibrium is always maintained. 7.1 Case 1: An ideal gas Suose for examle, there is no heating at constant ressure, then, from Eq. 7.1 and Eq. 5.11, c d ad = 0 But if ressure is constant then we only get the trivial solution that temerature is constant too. Not very interesting. But, let s integrate to see how a very small change in ressure would relate to a very small change in temerature at equilibrium: c ( 0 )=a ( 0 ) Now, we kee ressure 0 constant as we romised, and 0 too, but now we take the derivative of a small change with resect to temerature We end u with the exression c d ( 0 ) d = a d ( 0) d c = a d d (7.3) that could have obtained easily before, but how we got there matters as will become more clear below. Since a = R / d ln = c (7.4) d ln R or more generally from Eq. 5.9 d ln d ln = q = dh d(a) d (a) d Both the numerator and the denominator on the RHS are related to the number of degrees of freedom in the system. Eq. 7.5 is a bit of a master equation that we can take to almost any situation. It s really useful. (7.5) 33

30 7.2 Case 2: Phase changes But what about hase changes where changes are more abrut? What we have discussed thus far is the internal energy associated with sensible heat, i.e. the energy that can be associated with m < v 2 /2 > temerature. A second form of internal energy is that associated with the thermodynamic hase of a substance. It lays an extremely imortant role in the atmoshere because water is found in the liquid, ice and vaor hase, and raidly changes between all three, releasing or absorbing vast amounts of energy. From Eq. 7.2, the equilibrium solution is that the change in volume for a given heating is equal to inverse of the current ressure a = 1 q When we look at hase changes, we are concerned with the amount of heat that must be alied to a liquid to cause it go from a dense liquid hase to a gas that occuies much more sace er molecules. Because we are looking at a jum, since this is a hase change and the difference between a gas and a liquid is very large, we are interested in D differences. hus, the relationshi is no longer a differential, but rather Da = 1 Dq he amount of heat that must be alied to enable a hase change is the latent heat of vaorization Dq = L v At equilibrium, the ressure in the gas and liquid hase are equal. aking the ressure in the vaor hase, v a v = R v, so Da = L va v R v How does this equilibrium exression for constant temerature change if the temerature changes a small amount? aking the derivative with resect to temerature and keeing the ressure fixed dda d = La v R v 2 Here, we make a simlification. For secific volume, the difference between vaor and liquid is Da = a v a l = 1 r v 1 r l But the density of vaor is tyically tiny comared to that of liquid, by about a factor of 1000, so Da ' 1 r v = a v hus, the exression for the quasi-equilibrium transition to another temerature becomes da v d = L va v R v 2 34

31 or d lna v d ln = L v R v (7.6) Normally, this is exressed a little differently. Recognizing that d lna v = d ln, we get d ln d ln = L v R v (7.7) which again is similar to Eq. 7.5, excet d ln = Dh (7.8) d ln a v Main oints We are often not only interested in the equilibrium solution at constant ressure or temerature, but how the equilibrium solution changes as ressure or temerature change. he quasi-equilibrium solution for an ideal gas is related to the ratio of the gas constant to the secific heat at constant ressure he quasi-equilibrium solution for a hase change is related to the ratio of the the gas constant times the temerature to the latent heat Either exression is dimensionless Question What are the similarities between Eq. 7.5 and Eq. 7.7? Can you get to Eq. 7.7 directly from Eq. 7.5? 7.3 Atmosheric alication: otential temerature Integrating Eq. 7.4 from initial state i to final state f gives us R/c f f = i i We can use the above equation to derive an exression for the otential temerature q, which reresents the temerature a arcel would have if it were brought adiabatically from from (, ) to standard ressure 0, which is tyically set to 1000 mb. In this case q = R/c 0 35

32 he otential temerature is an imortant concet as it reresents a line of thermodynamic equilibrium for dry air in which there is no net exchange of energy with the environment through radiative absortion or emission, or through turbulent mixing, such that dh d = 0. In quasi-equilibrium transitions an air arcel may rise or sink, or move around horizontally, simultaneously changing both its temerature and ressure. But if there is no energy exchange, q is constant. Effectively, all arcels lying along a constant theta surface are in local thermodynamic equilibrium (Fig. 7.3). For examle, suose an air arcel rises from the surface at 1000 mb to 700 mb without mixing with its surroundings. If it s surface temerature is 25 C, what is its final temerature? f = i f i R/c = 269K Actually this equation can be related to the dry adiabat on a Skew log lot, if we consider from above that ln = R/c ln + const On the Skew log lot ln is the ordinate. Since not ln is the abscissa, adiabats are not straight but slightly curved from the lower right to uer left of the diagram Dry adiabats on a Figure 7.2: A Skew diagram for SLC Skew- are the same as lines of constant q. It is interesting to look at otential climatologically too since air motions tend to follow surfaces of constant otential temerature. his is often used in air quality modeling to see where air is going or coming from (look u HYSPLI for examle). 36

33 Figure 7.3: Zonal mean temerature and otential temeratures in the trooshere (From Houghton, 3rd Ed.). At least, the otential temerature is very useful for being a tracer of air motions in systems that are not changing raidly. What do we mean by raidly? Well, consider the following imortant timescales in the atmoshere. Free trooshere with no condensational rocesses 1 to 2 weeks is the timescale for radiative cooling to have an influence Synotic disturbances: timescale 1 to 2 weeks the same as the above, which is not a coincidence since radiative cooling and heating drive atmosheric motions Moist convection: timescales minutes to hours he imlication is that an assumtion of thermodynamic equilibrium, and motion along constant q surfaces is a useful guide over shorter time scales of erhas a few days, rovide there is no moist convection. Where would this work? Not the troics! 7.4 Atmosheric alication: saturation vaor ressure of water he famous Clausius Claeyron equation gives the sensitivity of the saturation (equilibrium) vaor ressure over a lane surface of ure water e S ( ). From Eq. 7.7, we have d lne S d ln = L v R v One would be temted to derive a solution for this equation as Lv 1 1 e S ( )=e S0 ( )ex 0 R v (7.9) (7.10) At freezing the latent heat is J kg 1 and R v = 461 J/K/kg. his is fine, but only if the difference between and 0 is small. he exression d ln /d ln is extraordinarily owerful, but 37

34 it must not be solved without care, because it is a sensitivity and other things that we have assumed to be constant might be changing with temerature also. he latent heat L v does vary somewhat by a few ercent over the normal range of atmosheric temeratures, and this does bad things to the accuracy of our nice exonential solution for e( ). A more accurate solution takes into account a solution for L v ( ). e s ( )=611.2ex( ) where, is in C. he reason for the slightly different form of the equation is that L v itself is a weak function of temerature e s (mb) (C) Figure 7.4: Clausius-Claeyron Equation for water vaor over liquid water Often you will hear the exression that at warm temeratures the air can hold more water vaor. Certainly from the C-C equation this seems to be correct, but semantically it is misleading. Nowhere in our discussion of the derivation have we needed to mention air at all. Rather, the C-C equation would hold even in the absence of dry air. hat said, in our atmoshere at least, having air resent is what enables the temerature to be as high as it is, since it is molecular collisions that enable radiative absortion to show u as a measurable temerature. 38

35 8 Equilibrium distributions Eq. 7.5 seems to imly that it is ossible to exerience a wide range on ressures and temeratures at equilibrium. his ends u being really imortant because it leads to distributions, something that is exlored throughout the sciences. It is very likely that you will send much of your career looking at distributions of one sort or another. It may hel immensely if you are able to understand their origins. here are two tyes of distributions that are commonly discussed, ower laws and exonential distributions. hese show u everywhere. Certainly throughout all of atmosheric sciences. But even your social networks will follow one of these two distributions. Note for examle that where we asssumed zero net heating at constant ressure, we obtained d ln = c d ln R his could be lotted on a log-log lot with on the x-axis and on the y-axis, and for all combinations of ressure and temerature there would be a straight line and the sloe would be c /R or 7/2 for dry air. Integrating would yield = c /R + const. which, of course, is the otential temerature exression. Alternatively, where we looked at zero at heating at constant temerature, and there was a big jum, we obtained the exression d ln = L v d ln R v his would not be suitable for a log-log lot because would be art of the sloe. But, we do get d ln d (R v ) = L v (R v ) 2 which suggests a log-linear lot. Integrating we get Lv = 0 ex ex R v 0 Lv R v which is an exonential distribution. Plotting log versus would yield a straight line with a sloe of L V /R v. his is great, but we can take this even further. What is d ln? Here, it hels to think of the ideal gas equation in the molecular form. We have already stated that = rr where R = g ext is a otential er unit mass. From Eq. 1.14, we could write the analogous exression in number form: = nk where µ ext = k. hus d ln d lnn µ ext d ln d ln µ ext n where n is the number density. he suggestion is that we now think of size distributions with a number concentration n on the y-axis and a otential energy available to do work µ ext on the x-axis. here are three basic tyes of size distributions to consider. 39

36 8.1 Log-log lot with a ositive sloe: Power Laws at constant temerature his tye of distribution describes quasi-equilibrium transitions where the jum is small, as we showed when deriving the otential temerature distribution from which can be written otherwise as d ln = c d ln R d lnn µ ext d ln µ ext n = dµtot dµ ext where µ tot = µ int + µ ext. he sloe is dµ tot /dµ ext, which for air is (c v + R)/R = 7/2. he general solution is n = n 0 (µ/µ 0 ) dµtot /dµ ext + const. (8.1) 8.2 Log-linear lot with a negative sloe. he Boltzmann distribution when jums are large For quasi-equilibrium transitions where the jum is big, we obtained the equation for hase changes But, we do get d ln = L v d (R v ) (R v ) 2 which can be generalized as d lnn µ ext dµ ext n = Dµtot µ ext (µ ext ) 2 the solution for the size distribution is n µ ext = n 0 ex Dµ tot µ ext µ ext he sloe on a log-linear lot of n versus Dµ tot is 1/µ ext. Exressed in terms of mass Dg tot g ext r g ext = r 0 ex What we are exressing here is how the number concentration changes when one otential changes and the other is fixed. One examle that has already been derived is the saturation vaor ressure (Eq. 7.10). g ext Atmosheric examle: the hysometric equation How does the density of air molecules at one level comare to the air density at another level? If g ext = R v is fixed and Dg tot = gdz is the otential that is changing, then gdz r = r 0 ex (8.4) R v 40 (8.2) (8.3)

37 his is the hysometric equation. Assuming the temerature of the atmoshere is constant, the density of the air dros of exonentially with height. Since we are assuming that temerature is constant, it follows too that Very often, one sees the exression = 0 ex = 0 ex gdz R v Dz H where H = R v /g is the Scale Height. he scale height is about 8 km at mid-latitudes Question Is the atmosheric temerature constant? Is it aroximately so, at least in the trooshere? 8.3 Log-Log lot with a negative sloe. Power Laws at constant ressure For internal distributions within a system that is at equilibrium (i.e. not one where there are quasiequilibrium transitions), we also get a ower law. Since = nµ ext, we can take the derivative with resect to time to obtain (8.5) (8.6) d = d (nµext ) n µ = µ ext ext + n (8.7) µ ext n If the ressure is constant and nothing is changing then d/ = 0 and he solution is d lnn µ ext d ln µ ext n = 1 (8.8) n = n 0 (µ/µ 0 ) 1 + const. his is a ower law. So, on a log-log lot of n versus µ ext the sloe is -1 or n µ 1/µ ext. here can be a large number of low otential things or a low number of high otential things. 41

38 8.3.1 Atmosheric examle: aerosol size distributions Figure 8.1: Junge aerosol size distribution (Junge, 1955) Normally, aerosol size distributions have what is known as a Jungian size distribution with a sloe of -3 on a log-log lot, rovided that the lot is n(r) where r is the aerosol radius. hus d lnn d lnr = 3 42

39 or n µ 1/r 3. Note, however, that the volume is given byv µ r 3. hus d lnn d ln µ = 1 d lnn 3 d lnr = 1 What does this tell us? Assuming that aerosols are formed by condensation, then we might assume (correctly) that the aerosol volume is related to an accumulated otential through the sum of all condensational work done. Question he equations for the ower law and the Boltzmann distribution should be the same in the limit of small jums. Can you show this? Main oints Power laws and exonential distributions are observed everywhere in nature. hey can be seen as a consequence of basic thermodynamic rinciles he distributions can be exressed as a function of many different variabiles, but most fundamentally as a number concentration versus some form of otential energy µ ext. Exonential distributions tend to show u when there are big jums Power laws tend to show u when there are small changes. he sloe is ositive for quasiequilibrium transitions and negative with a sloe of -1 at equilibrium. 9 Conservation of matter and energy From Eq. 6.7, we stated that the enthaly (or total energy) of an adiabatic arcel does not change with time. We also alluded to the fact that the system is materially closed, requiring that the amount of matter in the system is also constant. his could be written generally as Âm i = const. i Âg int i + g ext i = const. i 9.1 Atmosheric alication: Adiabatic liquid water content of a cloud he basic hysics we need to consider for understanding the formation of cloud in a arcel that rises and falls adiabatically is that heat and energy in the arcel are conserved Q = c + w = const. (Conservation of mass) h m = c + gz + L v w = const. (Conservation of energy) 43

40 where w is the water vaor mixing ratio and c is the liquid water mixing ratio. Q is the total water content, the quantity that is conserved if the arcel does not gain or lose mass. In a seudoadiabatic rocess, as the air exands, conservation of mass requires that the saturated mixing ratio decreases by an amount dw s as vaor is converted to condensed liquid. If the air remains saturated Q = c + w s (9.1) dq = 0 = dc + dw s (9.2) he liquid water content has units of g/m 3, so dc = dw s (9.3) LWC = rc (9.4) where r is the air density. yical values range from 0.05 g/m 3 in thin stratus to in excess of 1g/m 3 in cumulus towers. In a saturated adiabatic rocess dh m = 0 = c d + gdz + L v dw s (9.5) herefore L v dw s = c d + gdz dw s dz = c L v ale d dz + g c or, from Eq. 9.3 and 9.4 dw s dz = dlwc dz c L v (G d G s ) ' const. (9.6) = rc L v (G d G s ) ' const. (9.7) So we have just shown that LWC increases aroximately linearly with height. Lets try some tyical values for stratocumulus G s ' G d = r ' 1 c = 1004 L v = which gives us dlwc ' g/m 3 /km dz which is about right. A 200 m thick stratus cloud tyically has a cloud to value of LWC of 0.3 g/kg. 44

41 9.2 Atmosheric alication: Saturated adiabatic lase rate and the equivalent otential temerature But what is the moist adiabatic lase rate G s? Again, in a moist adiabatic rocess where mass is also conserved dh m = c d + gdz + L v dw s = 0 We can use this to derive the saturated adiabatic lase rate, G s d dz = g + L v dw s c c dz but w s is a function of temerature and ressure since w s ' ee s ( )/, where e s ( ) is given by the Clausius Claeyron equation (Eq. 7.9) hus ws dw s = dw s dz = ws de s d = L ve s R v 2 (9.8) ws d+ d dz + ws But the hydrostatic equation tells us that d/dz = rg. Also G s = d/dz dw s dz = ws ws rg G s (9.9) Equating this with Eq. 9.6, and solving for G s, we get G s = g c 1 rl ws 1 + L c ws d And recognizing w s ' ee s ( )/, and the C-C equation for e s ( ) we get G s = g c d dz 1 + Lw s R d (9.10) 1 + el2 w s c R d 2 It is often useful to refer to a thermodynamic chart for a first guess. he oint of learning them is to see how the essential ingredients of cloud hysics can be combined: conservation of energy and mass, the hydrostatic equation, and the Clausius-Claeyron equation for the saturation vaor ressure over a liquid surface. Another exression, that is useful, one that is derived in Wallace and Hobbs, is the Equivalent Potential emerature, which is conserved along moist adiabats. As before when we said that it is equivalent to say 45

42 Here and dh d = 0, dq = 0 R/c ale 0 ws L v q e = ex c ale ws L v q e = q ex c dh m = 0, dq e = 0 On the Skew- lot shown in Fig. 7.2, moist adiabats are the urle lines. Note that they do not cool with height as raidly as the orange dry adiabats. Note too that in the region around 600 mb where the dewoint and the temerature are almost the same, the temerature nearly arallels the nearest moist adiabat line. his suggests a well-mixed cloud layer, erhas altocumulus. 9.3 Atmosheric alication: Stability We are going to try to address the roblem of how a arcel of air rises within its environment when there is no exchange of energy or matter. he basic concetual idea is that a arcel starts somewhere in the atmoshere with some udraft velocity, erhas due to orograhic lifting or localized convergence of some sort. As the arcel rises its further acceleration or deceleration is due to its temerature relative to its environment. If the arcel is warmer than its environment, a buoyancy force will accelerate the arcel uwards towards regions of similar low environmental density aloft. However, as the arcel rises, it cools. In an idealized model in which the arcel does not mix with its surroundings, it cools with height at the dry adiabatic lase rate. Since the environment is cooling at slower rate with height, eventually the arcel is colder and less buoyant than its surroundings, and it begins to sink. What we will show is that an unstable dry atmoshere is characterized by d/dz = G > G d or dq/dz < 0. A stable dry atmoshere is characterized by G < G d or dq/dz > 0 he hysics is as follows: Within both the arcel and the environmental air, from the hydrostatic equation (Eq. 6.15) d dz = rg But the density r is different between the arcel and the environmental air. If r 0 and 0, and r and are the density and temerature of the arcel and the environment resectively. he uward ositive force exerienced by the arcel is r 0 r g er unit volume, or (r 0 r) g r 46

43 er unit mass. In other words, a arcel denser than its environment will accelerate downwards. Substituting the ideal gas equation, and assuming that the environmental temerature is close to the arcel temerature, we get a b = ( 0 ) g for the uward buoyancy force er mass (or acceleration). Figure 9.1: he eriod of the oscillations resulting from an initial erturbation is 2/N Actually, we should really be taking into account the fact that humidity also contributes to the buoyancy of the arcel, and use the virtual temerature v instead of the temerature (Eq. 1.24) v ' ( w) where w is the mixing ratio. So the buoyancy acceleration is then a b = ( v 0 v ) g his leads us to Eq in Wallace and Hobbs, since acceleration is the second time derivative of osition d 2 z 2 = ( v 0 (z) v (z)) g (9.11) v (z) We can imrove on this exression by noting that as the arcel goes u or down, its going to cool or warm at the dry adiabatic lase rate G d. hus v 0 v = v0 G d z 0 where z 0 = z z 0 is the dislacement from the equilibrium osition. Similarly, the environmental temerature at the same level is given by v = v0 Gz 0 47