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Atmospheric Thermodynamics J  J  J  J  J  J  J  J  J  J 

A Brief introduc�on Gas Laws: applica�ons for the atmosphere The Hydrosta�c Equa�on Defini�on of geopoten�al Scale Height and Hypsometric Equa�on : applica�ons First Law of Thermodynamics Adiaba�c Processes Water vapor in the air Stability The second Law of Thermodynamics : Applica�on of the concept of Entropy to the atmosphere

Masses in 1 mol of different substances have same ra�o, hence # molecules in one mol of any substance is CONSTANT Avogadro’s number ,NA. The value of NA is 6.022x1023per mole. Avogadro’s hypothesis, gases containing the same number of molecules occupy the same volumes at the same temperature and pressure. Therefore, the constant R in (3.1) for 1 mol is the same for all gases; it is called the universal gas constant (R*).The magnitude of R* is 8.3145J K-1mol-1. For 1 mol For n mols

Rd is R for 1kg of dry air

Apparent molecular weight dry air (all cons�tuents)

Boyle’s law,1which states if the temperature of a fixed mass of gas is held constant, the volume of the gas is inversely propor�onal to its pressure. Charles’ 1: for a fixed mass of gas at constant pressure, the volume of the gas is directly propor�onal to its absolute temperature. Charles’ 2: for a fixed mass of gas held within a fixed volume, the pressure of the gas is propor�onal to its absolute temperature. gram-molecular weight or a mole (abbreviated to mol), the molecular weight, M, of the substance expressed in grams.

Masses in 1 mol of different substances have same ra�o, hence # molecules in one mol of any substance is CONSTANT Avogadro’s number,NA. The value of NA is 6.022x1023per mole.

Apparent molecular weight of dry air Md J  Md can be also obtained by considering the main cons�tuents of the atmosphere and their propor�ons N2 x 0.78 + O2 * 0.21 + Ar * 0.01= 28.013*0.78+32.000*0.21 + 0.01*39.948= 28.969 or ~28.97 grams R* is the gas constant for 1 mol of ANY substance, the gas constant for 1 g of dry air is R*/Md, and for 1kg of dry air it is: Rd=1000 (R*/Md) = 1000 (8.3145/28.97)=287.0 J K-1kg-1 (11) Rd=Gas constant for dry air J  The ideal gas equa�on may be applied to individual gaseous components. For the water vapor: v v J  Where e is the pressure and αv is the specific volume of water vapor and Rv is the gas constant for 1kg of water vapor. J  Remember Mw water vapor: H2O ~ 18.016g = 18.016 x 10-3 kg J  Rv = 1000 R*/Mw (13) =1000 (8.3145)/18.016 = 461.51 J K-1kg-1

eα = R T

For one molecule gas constant (Boltzmans constant, k) For dry air

Gas Laws

=28.97 g

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Easy to conclude that… Mw water vapor: H2O ~ 18.016g = 18.016 x 10-3 kg Md Dry air: ~ 28.97 grams = 28.97 x 10-3 kg If Rv = 1000 (R*/Mw) and Rd=1000(R*/Md) Then, Rd/Rv = Mw/Md = ε = 0.622 J  Because air is a mixture of gases, it Obeys Dalton’s law of partial pressures, which states the total pressure exerted by a mixture of gases that do not interact chemically is equal to the sum of the partial pressure of the gases.



Tv ≡

T e 1 − (1 − ε ) p

Virtual Temperature

Rather than use a gas constant for moist air, the exact value of which would depend on the amount of water vapor in the air (which varies considerably), it is convenient to retain the gas constant for dry air (which we know how to measure) and use a ‘FICTITIOUS TEMPERATURE’ called VIRTUAL TEMPERATURE in the ideal gas equation. Since ε is positive and < 1 (=0.622), and e ≤ p, the denominator is positive and T (if e ≠ 0) In conclusion, The virtual temperature is the temperature that dry air would need to attain in order to have the same density as the moist air at the same pressure. Remember that moist air is less dense than dry air at the same temperature and pressure. Therefore, the virtual temperature is always greater than the actual temperature. However, even for very warm and moist air, the virtual temperature exceeds the actual temperature by only a few degrees

Hydrostatic Equation Column with unit cross-sectional area (say 1m2)

Consider “g” the gravity acceleration (m/s2)

pressure p + δp -δρ

δz

gpδz

Consider that δp is negative ( that is, z increases as p decreases)

pressure p In a hydrostatic equilibrium, and considering the limit as δz → 0

−

1 ∂p =g ρ ∂z

∂p = − ρg ∂z

or:

Geopoten�al J 

The geopotential Φ (phi) at any point in the Earth’s atmosphere is de�ined as the work that must be done against the Earth’s gravitational �ield to raise a mass of 1 kg from sea level to that point. In other words, Φ is the gravitational potential per Z unity of mass (Jkg-1, m2s-2)

dΦ ≡ gdz Using that gdz = −αdp

dΦ ≡ gdz = −αdp

dz

The geopotential Φ(z) at height z is thus given by:

z

Φ ( z ) = ∫ gdz 0

Hydrostatic Equation

(20)

(21)

Where Φ(0) at sea level (z=0) has, by convention, been taken as zero. Note that the geopotential at a particular point in the atmosphere depends ONLY ON THE HEIGHT of that point and NOT ON THE PATH

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Geopotential, Pressure and Temperature dependence

Geopoten�al Height Z

Z≡

J  Defined as

Φ( z ) 1 = go go

∫

z

0

gdz

(22)

Here we normalize geopotential to standard gravity

where go is the globally averaged acceleration due to gravity at the Earth’s surface (~ 9.81ms-2) Geopotential height is used as the vertical coordinate in most atmospheric applications in which energy plays an important role (e.g., in large-scale atmospheric motions). Values of z (actual height) and Z are almost the same in the lower atmosphere where go ~ g z(km) 0 1 10 100

Z(km) 0 1.00 9.99 98.47

g(ms-2) 9.81 9.80 9.77 9.50

Definition of Geopotential Height

Note the negative sign

Φ( z ) 1 = go go

p2

Z≡

z

∫ gdz 0

Φ 2 − Φ1 = − Rd ∫ Tv p1

dp p

It means that if we divide the right hand equation by go and get rid of the minus sign we have an equation for the Geopotential Height at p1 and p2! How to get rid of the minus sign?? Invert the limits of your integral and you are done!

Z 2 − Z1 = +

Rd go

∫

p1

p2

Tv

dp p

∂p

∂z

=−

p g Rd Tv

dΦ ≡ gdz = −αdp = − RT

The difference Z2 – Z1 is referred to as the geopotential thickness of the layer between pressure levels p2 and p1 and is dependent on the virtual temperature

dp dp = − Rd Tv p p

We see here that infinitesimal varia�ons in geopoten�al between two pressure levels are related to temperature and pressure. To know the Geopoten�al difference between Z1 and Z2 we need to integrate the above equa�on between two levels since temperature and pressure changes when we consider large varia�ons:

Φ2,Z2, p2

dp dΦ = − Rd Tv p

A few important things to think about:

Hydrostatic Equation Combined with Variation in the Geopotential Equation

∫

Φ2

Φ1

p2

dΦ = − ∫ Rd Tv p1

dp p

or p2

Φ1,Z1, p1

Φ 2 − Φ1 = − Rd ∫ Tv p1

dp p

So, why is this so important? Z 2 − Z1 = +

Rd go

∫

p1

p2

Tv

dp p

Since thickness is dependent on temperature, we can use this concept to observe warm and cold regions and related that to waves propagating in the atmosphere and how they change wind patterns…

Since thickness is dependent on temperature, we can use this concept to observe warm and cold regions and related that to waves propaga�ng in the atmosphere and how they change wind pa�erns…

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Now, going back to the thickness: Z 2 − Z1 = + H ∫

p1

p2

dp = H ln( p1 / p 2) Hypsometric Equation p

Or you may want to find p2 given p1 and the thickness:

⎡ (Z − Z1 )⎤ p2 = p1 exp⎢− 2 H ⎥⎦ ⎣ H≡

RT = 29.3T go

Exponential decrease of pressure with height

(Scale Height of the atmosphere)

Where T=255K =average T for troposphere and stratosphere

ln(p2 ) ln(p1 )

Radiosonde data will provide the temperature profile and we can obtain the mean Tv for that layer (Eq. 3.28)

Z 2 − Z1 = + Tv

Rd Tv ⎛ p1 ⎞ ln⎜⎜ ⎟⎟ go ⎝ p2 ⎠

-Pressure -Altitude -Geographical Position -Temperature -Relative Humidity (p,t) RH= partial pressure H2O (e) / saturated vapor pressure (e*(at some T)) -Wind (speed direction), rawinsonde (wind only)

Thickness and temperature J  The temperature of the atmosphere generally varies with height and the virtual temperature correc�on can not always be neglected. In this case, we have to find a way to resolve the integral below, since Tv is not always a “constant”

Z 2 − Z1 = +

Rd go

∫

p1

p2

Tv

dp p

In this case we have to obtain radiosonde data and resolve the integral

Thickness and Heights of Constant Pressure Surfaces J  Because pressure decreases monotonically with height, pressure surfaces (imaginary surfaces on which pressure is constant) never intersect. J  From the hypsometric Equa�on we learned that the thickness of the layer between any two pressure surfaces p2 and p1 is propor�onal to the mean virtual temperature Tv J  Therefore, as Tv increases the air between the two pressure levels expand and the layer becomes thicker

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Reduction of pressure to sea level J  Let the subscripts g and 0 refer to condi�ons at the ground and at sea level (Z=0), respec�vely. Then, for the layer between the Earth’s surface and sea level, the hypsometric equa�on assumes the form: ⎛p ⎞ Zg = + H ln⎜ o ⎟ Which can be resolved to obtain the sea-level pressure ⎜p ⎟ ⎝ g⎠ ⎛ go Z g ⎞ ⎛ Zg ⎞ ⎟⎟ po = pg exp⎜ ⎟ = pg exp⎜⎜ ⎝H⎠ ⎝ Rd Tv ⎠ If Zg is small, the scale height H can be evaluated from the ground temperature. Also, if Zg/H pB

9.81 m s ⎛ dT ⎞ g −⎜ = Γd = = 0.0098 K m −1 = 9.8 K km −1 ⎟= 1004 J K −1kg −1 ⎝ dz ⎠ c p Remember: variations in oC or K are equivalent

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Potential Temperature θ

Thermodynamic Diagrams

J  Is defined as the temperature that the parcel of air would have if it were expanded or compressed adiaba�cally from its exis�ng pressure and temperature to a standard pressure po (generally taken as 1000hPa) J  This concept is useful for many reasons. One of them is to compare masses of air from different al�tudes and from different regions too. We R /cp will se more about this later

On P-V diagram, compression (pdα> es:

ws = 0.622

es p

For the range of temperature observed in the Earth’s Atmosphere, p >> es:

ws = 0.622

es p

(63) Depends on temperature and pressure. Represented as dashed lines in the Skew-T lnp

Rela�ve humidity, RH, relates the ACTUAL amount of water vapor in the air to the maximum possible at the current temperature. RH = (specific humidity/satura�on specific humidity) X 100%

RH = 100

Saturation for cold air

If the air temperature increases, more water vapor can exist, and the ratio of the amount of water vapor in the air relative to saturation decreases.

Dew point Td LOW RH

w e ≅ 100 (64) ws es

More water vapor can exist in warm air than in cold air, so relative humidity depends on both the actual moisture content and the air temperature.

(63) Depends on temperature and pressure. Represented as dashed lines in the Skew-T lnp

High RH Saturation for warm air

Dew Point (Td)

Distribution of mean dew points across the U.S. in July.

temperature to which the air must be cooled at constant pressure to become saturated with respect to a plane surface of pure water. I

Dew is moisture condensed upon surfaces, specially during the night (why??) The dew point is the temperature to which the air must be cooled at constant pressure to become saturated with respect to a plane surface of pure water. In other words, dew point is the temperature at which satura�on mixing ra�o ws with respect to liquid water becomes equal to the actual mixing ra�on w.

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Lifting Condensation Level J  LCL is the pressure level to which moist air must be li�ed adiaba�cally to become saturated J  Mixing ra�o (w) and poten�al temperature (θ) remain constant (adiaba�c, no evap or condense) ws decreases un�l ws=w=Td. J  The LCL is intersec�on of w with ws for the parcel. So from A and Td at const p, follow parallel lines of θ and ws un�l they intersect . loca�on C J  NOTE: TD does not remain constant as pressure decreases because both the vapor and dry air expand J  If you know any moisture variable (TD, LCL, RH, w,…) and T and p you can determine the others

W=Ws

Ws decreasing Along dry adiabat

Wet bulb temperature (Tw) and Dew point (Td)

Wet-bulb temperature and dew point both involve cooling a hypothe�cal air parcel to satura�on, but there is a dis�nct difference. Td cool to satura�on while keeping moisture constant Tw temperature we obtain if we cool the air to satura�on by evapora�ng water into it , thus, moisture content has increased and the parcel will reach satura�on at a higher temperature.

Td < Tw < T

Wet / dry bulb temperature Dry bulb is commonly used temperature not affected by moisture of the air. Wet Bulb temperature is the lowest temperature to which a volume of air can be cooled at constant pressure by evaporating water into it. Wet bulb measured with Thermometer with moist cloth over end. Ambient air flow now evaporates from wet cloth (adiabatic evaporation) cooling it. Once it reaches a steady temperature, wet bulb is measured. Evaporation reduced if more moisture in air, at 100% RH Dry bulb=Wet bulb 1) FOLLOW THE SATURATION MIXING RATIO LINE UPWARD FROM THE DEWPOINT TEMPERATURE 2) FOLLOW THE DRY ADIABAT UPWARD FROM THE TEMPERATURE UNTIL IT CROSSES THE FIRST LINE 3) FOLLOW THE SATURATION ADIABAT DOWN TO THE ORIGINAL LEVEL. TEMPERATURE AT THIS POINT IS THE WET-BULB TEMPERATURE If a raindrop falls through a layer of air with a constant wet bulb temperature the raindrop will reach the wet bulb temperature

Latent Heat J  Latent Heat is the heat (actually Enthalpy) added or removed from a substance to produce a change of phase. J  When water evaporates to form water vapor, individual water vapor molecules must break free of a�rac�ve forces (hydrogen bonds)

–  The most energe�c molecules will break free, taking their energy with them –  Some kine�c energy converts to chemical and poten�al energy –  The average kine�c energy of the remaining liquid (TEMPERATURE) is reduced –  So evapora�on is associated with cooling

J  When water vapor condenses, individual water vapor molecules must convert some poten�al and chemical energy back to kine�c, so condensa�on heats

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Latent Heat J  Latent Heat is the heat (actually Enthalpy) added or removed from a substance to produce a change of phase. J  For water the latent heats are

–  Mel�ng (or freezing): Lm = 3.34x105 J kg-1 (at T = 0oC) –  Vaporiza�on (condensa�on): Lv = 22.5 x105 J kg-1 (at T = 100oC)

J  The temperature changes produced by freezing or condensing a gram of H2O in a kilogram of dry air (at 0oC) are: –  Lm/cp = (3.34x105 J kg-1 )/(1005 J K-1kg-1) = 332 K = 0.332 K (gm kg)-1 –  Lv/cp = (2.46x106 J kg-1 )/(1005 J K-1kg-1) = 2447 K = 2.447 K (gm kg)-1

–  So freezing a gram of water in a kilogram of dry air will warm it by about ⅓ of a degree –  And condensing a gram of water in the same kilogram of dry air will warm the air by 2½ degrees

Saturated Ascent J  Above the LCL a rising parcel ascends moist adiaba�cally J  Because the latent heat released by condensa�on compensates for some of the cooling due to expansion, the saturated adiaba�c lapse rate is less than the dry adiaba�c. Γd > Γs J  Typical values are: –  Γs ~ 4o K km-1 in very moist surface air –  Γs ~ 6 to 7o K km-1 for typical middle troposphere –  Γs approaches Γd = 9.8o K km-1 near the tropopause or in the Arc�c, where the air is very dry

Saturated Adiaba�c and Pseudoadiaba�c Processes J  As air rises adiaba�cally it cools un�l it becomes saturated, then condensa�on occurs J  Saturated (moist) adiaba�c process, condensa�on products remain in the rising parcel and latent heat does not pass through the boundaries J  If all of the condensa�on products immediately fall out of the air parcel (taking some heat), parcel is then said to undergo a pseudo (moist) adiaba�c process J  Heat of condensate very small, so these are virtually equivalent J  Pseudo-moist-adiaba�cally: All condensed moisture immediately precipitates from parcel. J  Moist adiaba�cally: All condensed moisture remains in parcel.

Saturated Ascent J  Satura�on adiabats show the path that a saturated air parcel follows as it rises pseudo moist adiaba�cally In skew T ln P satura�on adiabats are solid green lines

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Equivalent Potential Temperature θe equivalent potential temperature of a parcel of air when all the water vapor has condensed so that its saturation mixing ratio ws is zero. ⎛L w ⎞ θ e = θ exp ⎜⎜ v s ⎟⎟ ⎝ c pT ⎠

Normand’s Rule J 

2 J  1

Air is lifted pseudoadiabatically until all the vapor has condensed, released its latent heat, and fallen out. The air is then compressed dry adiabatically to the standard pressure of 1000hPa.

Equivalent potential temperature is a conserved quantity

Equivalent Potential Temperature, Continued J  Both equivalent poten�al temperature θe and wet bulb poten�al temperature θw are conserved J  When height coordinates (z) used instead of (p) we use the conserved quan�ty Moist Sta�c Energy MSE = (cpT + Φ) + Lvq = (cpT + gz) + Lq MSE = enthalpy + poten�al energy + latent heat content Takes the place of θe. The first two terms are called Dry Sta�c Energy (cpT + Φ) For dry adiaba�c ascent cpT -> Φ, or cpT ↓ and Φ↑ For saturated adiaba�c ascent cpT ↓, Φ ↑ , Lvq↓

Find LCL using poten�al temperature line(θ) that passes through the point located by the temperature and pressure of the air parcel,

J 

and the satura�on mixing ra�o line (ws) that passes through the point determined by the dew point and pressure of the air.

J 

Follow the equivalent poten�al temperature line (θe) (i.e.,the pseudoadiabat) that passes through the point located by the LCL to the original pressure of the air parcel, Tw

J 

Con�nue to 100hPa, θw

3 To calculate θe on Skew-T diagram. 1.  Go up a dry adiabat to the LCL 2.  Above the LCL follow the pseudoadiabat up un�l it approaches a dry adiabat. 3.  Follow that dry adiabat down to 1000 mb. That is the Equivalent Potenial temperature θe

Normand’s Rule: Tw can be determined by li�ing a parcel of air adiaba�cally to its LCL, then follow a moist adiabat from that temperature to to parcels actual pressure

Net effects of ascent J  When a parcel is lifted above LCL so that condensation occurs, latent heat gained is retained so that when it descends. –  Net increase in temperature and potential temperature. –  Decrease in moisture content –  No change in equivalent potential temperature or wet bulb potential temperature. (conserved for dry and saturated processes)

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Static Stability Recall, if parcel at O is li�ed it will follow Γd.

For (a) 𝚪𝚪A, parcel at A is colder (more dense) than surroundings (TB) and will sink back to O (stable), Γd - Γ = strength of restoring force

Stable 𝚪𝚪𝚪𝚪d If parcel at O lifted , will follow Γd. to A where it is warmer (less dense) than surroundings and will rise (unstable)

If the layer is stably stra�fied N is real N2 is posi�ve

TA>TB

Gravity waves

The parcel executes a buoyancy oscilla�on! The higher the Brunt –Väisälä frequency, the more stably stra�fied the atmosphere EG gravity waves

Condi�onal instability If the actual lapse rate of the atmosphere lies between the saturated adiaba�c lapse rate and the dry adiaba�c lapse rate, a parcel of air that is li�ed sufficiently far above its equilibrium level will become warmer than the ambient air.

From O Cools adiaba�cally to A (LCL) and is cooler than ambient air If li�ed further follows moist adiabat to B, Level of free convec�on (LFC) If li�ed beyond B becomes posi�vely buoyant If li�ing stops before B parcel is nega�vely buoyant and will return to O

The Second Law of thermodynamics is concerned with the maximum frac�on of a quan�ty of heat that can be converted into work. A cyclic process is a series of opera�ons by which the state of a substance (called the working substance) changes but the substance is finally returned to its original state in all respects. Final state=ini�al state



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Since the ini�al and final states of the working substance are the same in a cyclic process, and internal energy is a func�on of state, the internal energy of the working substance is unchanged in a cyclic process.

Net heat (Q) absorbed by the working substance in a cyclic process is not equal to zero (q is not a state variable) . The net heat supplied by the environment in a cyclic process is converted to mechanical WORK

Carnot Cycle J  A heat engine (or engine for short) is a device that does work through the agency of heat (Q). J  If during one cycle of an engine a quan�ty of heat Q1 is absorbed and heat Q2 is rejected, the amount of work done by the engine is Q1-Q2 and its efficiency is defined as

A working substance is said to undergo a reversible transforma�on if each state of the system is in equilibrium so that a reversal in the direc�on of an infinitesimal change returns the working substance and the environment to their original states.

Carnot Cycle Summary J  Originally developed to understand steam engines J  Adiaba�c compression in which the environment does work on the system, A to B J  Isothermal expansion, B to C, working substance doing work in contact with a warm reservoir extracts heat Q1 J  Adiaba�c expansion, C to D, also doing work J  Isothermal compression, D to A, working substance in contact with the cold reservoir gives up heat Q2

η=

Q1 − Q2 Q1

From A to B adiaba�c, q=0 Where 𝛾𝛾 = Cp/Cv For isothermal transforma�on B-C Boyles Law states

Q1

adiaba�c Q2

isothermal Combine these to find

Q1

Heat: from B to C, Q1 is absorbed from reservoir. Recall internal energy depends only on T and T does not change from B to C , so Q1 goes towards WORK and

->

Q2 Similarly , Q2 rejected to cold reservoir from D to A

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Q1

Q2

The ra�o of heat and temperature are the same!

Hence the the ra�o Qrev/T is the same

Q2/T2=Q1/T1

A-B and C-D are adiaba�c so no change in entropy. B-C (isothermal) entropy of the source is decreased by Q1/T1 and D-A, isothermal entropy of the sink is increased by Q2/T2 If Q2/T2=Q1/T1, no change in entropy in Carnot Cycle

Efficiency is at a theore�cal maximum for this Carnot Engine

Entropy

Entropy J  Isotherms are dis�nguished by differences in temperature J  Dry adiabats are dis�nguished by differences in θ J  Using the Carnot cycle as an example we know we can pass from adiabat to adiabat along isotherms, where heat (Qrev)is absorbed or rejected J  We also showed that the ra�o Qrev/T is the same no ma�er which isotherm is chosen J  Now we can use Qrev/T to measure the difference between two adiabats J  Qrev/T is called the difference in entropy (S) between two adiabats J  So the increase in entropy of a system between two states J  where dQrev is the quan�ty of heat that is added (reversibly) to the system at temperature T J  Entropy is a func�on of the state of the system, not the path taken

J  W=Q1 - Q2 is the heat converted into useful work. J  The ra�o of the useful work to total energy is η = (Q1 – Q2)/Q1 = (T1 − T2)/T1

J  The quan�ty Q/T, called Entropy (S), is constant in Reversible processes (Q1/T1 = Q2/T2), J  but increases in irreversible process as S=Q2/T2-Q1/T1 , which make up most of what really happens in the universe. J  Since the cold reservoir is always at T2 > 0 K, some waste heat must always be exhausted to the environment. J  For our engine, entropy increase S2-S1 a�er one cycle is given by the reduc�on of entropy of the Hot source and the increase of the cold sink

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Entropy and Poten�al Temperature J  Another way to write entropy is Tds = dq J  So that the 1st Law becomes: J  Tds = cpdT − αdp J  Or ds = cp(dT/T − (R/cp)dp/p) = cp dθ/θ J  Integrating s = cp lnθ + constant J  Thus, an adiabatic process in which potential temperature (θ) is constant is also an Isentropic process. J  Adiabats are frequently referred to as Isentropes

The Clausius – Clapeyron Equa�on J  The Clausius Clapeyron equa�on ( or the first latent heat equa�on) describes how the saturated vapor pressure above a liquid changes with temperature and also how the mel�ng point of a solid changes with pressure.

Temperature entropy diagram Can display the Carnot cycle as temperature vs entropy

Here adiabats are constant entropy lines So for the process ABCDA, the heat taken in Q1=XBCY the heat rejected Q2=XADY Q1-Q2=ABCD

Implica�ons for global warming J  CC equa�on implies that satura�on water vapor pressure changes approximately exponen�ally with temperature under typical atmospheric condi�ons, and hence the water-holding capacity of the atmosphere increases by about 7% for every 1 °C rise in temperature

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The second Law of Thermodynamics J  For a reversible transforma�on there is no change in entropy for in the universe (system + surroundings). If system receives heat (reversibly) the increase in entropy=the decrease in entropy of surroundings The concept of reversibility is an abstrac�on

J  In nature irreversible or spontaneous transforma�ons occur, and the entropy of the universe increases as a result of irreversible transforma�ons

J  Chapter 3 (Due 10/21/2015) 3.18 (b,d, f, h, l,r), 3.19, 3.20, 3.26, 3.27 (b)More fuel is required to li� a hot air balloon through an inversion than to li� it through a layer of the same depth that exhibits a steep temperature lapse rate. Other condi�ons being the same, more fuel is required to operate a hot air balloon on a hot day than on a cold day. (fuel warms balloon, density and buoyancy, Inversion warms, strong lapse cool) (d)The gas constant for moist air is greater than that for dry air. (cons�tuents) (f)Describe a procedure for conver�ng sta�on pressure to sea-level pressure. (Hypsometric) (h)If a low pressure system is colder than its surroundings, the amplitude of the depression in the geopoten�al height field increases with height.(Cold core low with warm core alo�) (l)A parcel of air cools when it is li�ed. Dry parcels cool more rapidly than moist parcels. (lapse rate, latent heat) (r)Which of the following pairs of quan��es are conserved when unsaturated air is li�ed: poten�al temperature and mixing ra�o, poten�al temperature and satura�on mixing ra�o, equivalent poten�al temperature and satura�on mixing ra�o?

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