First law of thermodynamics (Jan 12, 2016) page 1/7. Here are some comments on the material in Thompkins Chapter 1

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1 First law of thermodynamics (Jan 12, 2016) age 1/7 Here are some comments on the material in Thomkins Chater 1 1) Conservation of energy Adrian Thomkins (eq. 1.9) writes the first law as: du = d q d w (1) (although the bar notation isn t used by AT, it s more informative, since the d symbol indicates that only the variable u can actually have a differential. To understand why, it hels to think of the internal energy, which is a state variable (i.e. and instantaneous function of the system) as a noun and heating and working as verbs. You can t take derivative of heating and working because they are rocesses, not variables. You can, however, ask how much energy heating and working contribute to the system over a very short time interval, and write that contributions as d q = q t and d w = w t, where q t and w t are the heating and working rates, with units of J kg 1 s 1. When in doubt, you can do any thermodynamic derivation in terms of derivative instead of differentials, and write: or alternatively: du = q t w t (2) du = q t w t (3) Note that q and w have the units of a rate (J/kg/s), and are functions. Below, I ll use differentials to derive c v (see eq. (8) and derivatives to derive c ( equation (16) following Bohren and Albecht), so you can see the difference in aroach. Some definitions q is the rate of change of internal energy of the system as a consequence of a temerature difference between it and its surroundings and w ( J kg s 1 ) as the working rate: rate of change of internal energy of the system as a consequence of of a force exerted through a distance (ositive if work done by the system and in another text, (Q ( J)) similarly is defined as :... any sontaneous flow of energy from one object to another caused by a difference in temerature between objects Again note the use of flow this is a rocess, not a quantity. and work is also defined as a rocess or flow (W ( J)):... any other transfer of energy into or out of the system. You do work on a system whenever you ush on a iston, stir a cu of coffee, or run current through a resistor.

2 First law of thermodynamics (Jan 12, 2016) age 2/7 Soiler alert: If the above seems a little fuzzy, you can look u the statistical mechanics definition of heat and work. Bottom line is that heat is energy transfer between systems with no change to their energy levels (just changes to the number distribution of articles among those energy levels), where work involves some external change to the systems that does affect the value of the energy levels available to the the system. 2) Heat caacity We ve already calculated the internal energy for a monotomic gas in lecture 2: With 3 translational ( 1 2 mv2 x mv2 y mv2 z) degrees of freedom a monotomic or diatomic ideal gas has kinetic energy: u monotomic ( J kg 1 ) = # of molecules/kg KE = N A M d 3 kt 2 where M d is the mean molecular weight of air as before, and we have divided by M d to get the secific internal energy, u J kg 1. A diatomic gas has additional vibrational and rotational energy. There are two rotational degrees of freedom ( 1I 2 1ω I 2 2ω2) 2 (since rotation about the molecular axis has such a small moment of inertia (I 3 ) that it doesn t contribute significant energy) and two vibrational degrees of freedom (vibrational kinetic 1m ( 1 dl ) and otential energy 1 k(l l 2 0) 2 (see this Wikiedia discussion. As the web age at htt://en.wikiedia.org/wiki/heat caacity oints out, for light molecules like N 2 the vibrational energy is suressed (frozen out) at terrestrial temeratures because there is a big jum to the first vibrational energy level for light molecules. That leaves just the two rotational modes, contributing 2 N A 1 kt so that the total energy for an N M d 2 2 O 2 gas mixture in thermodynamic equilibrium is u diatomic ( J kg 1 ) = N A M d 5 kt 2 = c vt (5) Plugging in N A k=r =8314 J kg 1 K 1, and M d =28.97 kg kmol 1 I get c v =717 J kg 1 K 1, which agrees with the usual tabulated value. Also, take a look a this this link (scroll down to (Figure 4). For dry air at 300 K I get U = = J kg 1. This is about the same energy ossessed by a 1 kg weight moving at 656 km hr 1. 2.a) First ass: deriving c v using differentials The constant c v in (5) is is the heat caacity at constant volume. To see how this is related to the first law and the state variables ρ, T and, write the work done in time as w = dv, where v is the secific volume (i.e. 1/ρ), then (3) becomes (4) q t = du + dv (6)

3 First law of thermodynamics (Jan 12, 2016) age 3/7 Switching to finite differences and assuming constant volume (dv = 0): and taking the limit as t, T 0 q t t T = u T (7) c v = u (8) which is Thomkins 1.20 (why the artial derivative?). In words, c v is the amount of energy required to raise the temerature of the gas by unit temerature, given fixed volume. Two subtle but imortant oints: 1. c v is not a function of volume for an ideal gas, i.e.: 2 u v = 0 (9) 2. From (5) it s clear that u deends only on temerature, so we can change the artial to a total derivative in (8) and write or: c v = du (10) u = c v T (11) which (to reeat) is true no matter whether the volume is changing or not. 3) enthaly It s not often that we get to work with constant volume situations in the atmoshere so u is not conserved even when q = 0. Also the secific volume v is comaratively hard to measure. More useful than the internal energy in atmosheric thermodynamics is the secific enthaly, h ( J kg 1 ), defined as: h = u + v (12) Note that the v term in (12) looks like ressure-volume work, and can be interreted as the work required at constant ressure to make room for the system with internal energy u (Schroeder. 149). Using h we can rewrite the first law (du/ = q dv/) as This is Thomkins eq dh = q t + v d (13)

4 First law of thermodynamics (Jan 12, 2016) age 4/7 Taking d/ = 0 in (13), we can say that the heating rate q t is equal to the time rate of change of enthaly at constant ressure. Since evaoration and condensation occur at constant ressure we will find h useful when we deal with hase change. Now exand h in, T using calculus: and substitute (14) into (13) to get: dh = h q t = h + + h d ( h ) d v (14) (15) 3.a) Second ass: deriving c using derivatives If the ressure is constant, then d/=0, and (15) roduces the definition of the heat caacity at constant ressure: q t = c t at constant ressure (16) where: c = h Which is Thomkins eq If you don t like the hand-waving limits I took to get c v in terms of u in Section 2, equation (7) above, you can redo it in the same way as in (15) to get: (17) and taking constant volume (dv/ = 0): q t = u + ( u v + ) dv (18) c v = u q t = c v t (19) at constant volume (20) This derivation is a little longer than the one in Thomkins, but it doesn t rely on any confusing combinations state variables and rocesses like this one in Thomkins eq. 1.17: ( ) q c = (21) which tend to make mathematicians queasy, given what we ve already said about d q not being a erfect differential. P

5 First law of thermodynamics (Jan 12, 2016) age 5/7 3.b) Gibbs hase rule In the same way note from the definition of h in (12) that it also deends only on temerature for an ideal gas (why?). We can also show that c is roughly indeendent of temerature (and ressure) so (16) can be integrated to obtain: How are c v and c related? Recall the first law h c T (22) and substituting in ressure work: q t = du + w (23) q t = du + dv = c v + dv = c v + d(v) vd (24) But also note that the intensive form of the ideal gas law is: which imlies Where R d = 287 J K 1 kg 1 as before. This means: v = R d T (25) d(v) = R d (26) which imlies that q t = (c v + R d ) v d q t = c v d q t, = dh v d (27a) (27b) (27c) c = (c v + R d ) (28) Can you exlain hysically why c > c v? For dry air, c = 1004 J kg 1 K 1, while c v = 717 J kg 1 K 1. Note that I only need two indeendent coordinates (e.g. T and ) to secify this single-hase system as long as mass is conserved. That s because I ve got five quantities (, V, m, T and ρ) and three constraints: m = constant (29a) ρ = m/v (29b) = f(v, T ) equation of state (29c) This relationshi between indeendent arameters (degrees of freedom) and the number of hases/comonents in the system is known as the Gibb s hase rule, and we ll discuss it in more detail below.

6 First law of thermodynamics (Jan 12, 2016) age 6/7 4) Static energy and otential temerature Once again, recall the hydrostatic equation: d = ρgdz (30) Using (30) we can go back to the first law and show why enthaly is a very useful quantity, esecially to atmosheric modelers. Inserting (30) into (27c) gives: q t = dh + gdz (31) If there is no heating (i.e. the rocess is adiabatic then q = 0 and we can integrate (31) to get a conservation equation for the dry static energy h d : s d = h dry air + gz = constant (32) (comare Thomkins eq. 1.43,. 13). This conservation makes h d a useful tracer, since changes to h d have come come from some diabatic rocess (radiation, evaoration), and can t come from simly lifting or sinking air. Note also that gz is just the otential energy er kilogram. So as the arcel decreases or increases its altitude, energy is moving between otential and internal, while h d is staying the same. We can also take (32) aart to get an equation for the dry adiabatic lase rate: c dz dz = g (33a) = Γ d = g c 9.8 K km 1 (33b) What if we want to work with ressure instead of height? Then rewrite (27c) using the equation of state to get: q t = c R (note the lack of virtual temerature this is for dry air). Again, if q = 0, d (34) d c = R d (35a) T c d log T = R d d log (35b) Integrating both sides of (35b) from the surface, with (temerature, ressure) given by (θ, 0 ) to a lower ressure and lower (why?) temerature T gives Poisson s equation (Thomkins 1.37): ( ) ( ) c T log = log (36) R d θ 0

7 First law of thermodynamics (Jan 12, 2016) age 7/7 Equation (36) defines the otential temerature, θ which is conserved for adiabatic ascent/descent: θ = T ( ) Rd /c 0 = T ( ) (c cv)/c 0 = T ( ) γ 1 0 γ where γ = c /c v. Like s d, θ is constant for a dry adiabatic rocess. Problem: show in the same way as the above that another conservation relation exists for the secific volume, i.e.: (37) for adiabatic rocesses. v γ = constant (38)