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Module 3 Thermodynamics of the Dry Atmosphere

3.1 Introduction

Thermodynamics essentially deals with energy transformations in a system and its equilibrium states under such transformations. A system could be either open (no boundaries) or closed (boundaries). An open system exchanges energy and matter with its surroundings in attaining its equilibrium state, whereas a closed system has boundaries, which are impermeable to matter, but exchange of energy is possible as the system is not isolated. Thermodynamics plays a central role across a number of traditional disciplines of physics and chemistry because pressure and temperature are the key variables in the understanding the equilibrium states of a given system. The atmosphere and ocean are also complex systems where the exchanges of mass and energy are so intricate that their dynamics cannot be explained satisfactorily without appropriate consideration of thermodynamics. Thus, simulation of atmospheric phenomena ranging from cloud microphysics to large-scale atmospheric motions, involves the basic laws of dynamics and thermodynamics. The thermodynamic state of the atmosphere is greatly modified by the water vapour, which on transformation of phase adds heat to the system. Also, evaporation of liquid water is accompanied with the cooling of the system. The thermodynamics of the moist air is thus different from dry air under certain transformations with phase changes. We first derive expressions for lapse rate and stability relevant for the dry atmosphere using the first law of thermodynamics assuming atmosphere in hydrostatic equilibrium. Finally, the thermodynamics of moist air will be discussed.

3.2 Equation of State Using the equation of state, first of thermodynamics and the hydrostatic equilibrium, several expressions relevant for the dry atmosphere are derived. Air is a mixture of gases and all gases obey the same equation of state. The ideal gas law is expressed as

pV = mRT             (3.1) where V is the volume, p the pressure, m the mass, R the gas constant and T denotes the temperature. The Equation (3.1) can also be written as

p = ρRT  ,    ρ = mV                (3.2)

pα = RT   ,         α = 1ρ

(Specific volume) (3.3)

A gas is composed of molecules and the number of molecules can be calculated for its mass m contained in volume V by first calculating the number of moles n of the gas in the given

volume. Thus, we have n = mM

with M as the molecular weight of the given gas. The

Avogadro’s hypothesis states that gases containing the same number of molecules occupy the same volume at same temperature. That is, the number of molecules in one mole of any substance is constant and it is called the Avogadro Number (NA) NA = 6.022 x 1023 per mole (3.4)

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If we consider one mole of a gas then the gas constant R is same for one mole of any other gas; that is, R in eqn. (3.3) is same for one mole of all gases. Hence, it is called the universal gas constant R* = 8.3145 J K-1 mol-1. Thus, for n moles, the ideal gas reads as pV = nR*T (3.5) Note that R* is used in calculations with one mole which contains NA molecules of a gas. Since, NA is also a constant, the gas constant for one molecule is also a universal gas constant, known as the Boltzmann’s constant given by,

2310022.63145.8*×

==AN

Rk = 1.3807×10-23

Question 1: Write down the units of k. [Ans. JK-1 molec-1] If there are no molecules per unit volume, then from equation (3.5), we obtain p = n0kT (3.6) If pd and α d are respectively the pressure and specific volume of dry air then with Rd as the gas constant for 1 kg of dry air, the gas law becomes pdα d = RdT (3.7) The average molecular weight (Md ) of dry air is equal to 28.97 g mole

-1; so, one can find the gas constant for one gram of dry air as R* /Md and therefore the gas constant, Rd as

Rd =R*

Md /1000= 1000 R*

Md= 1000 8.3145

28.97= 287 J kg−1K −1

Question 2: Consider air as the mixture of gases made up almost entirely by nitrogen (N2), oxygen (O2) and argon (Ar) with concentration by volume as 78.08%, 20.95% and. 0.93%; calculate the average molecular weight of air.

3.3 Hydrostatic equation Consider a column of air in the vertical direction (z) with positive direction pointing upward. The weight of a slab at a height z of thickness δ z is ρgδ z , g is the acceleration due to gravity taken as constant. Now calculate the vertical force that act on this slab of air between z and z +δ z with atmospheric pressure p(z) and p(z +δ z) = p +δ p . Since pressure decreases with height, so the vertical force on unit cross-sectional area is −δ p . The hydrostatic balance requires that vertical force be balanced by the weight of the cross-section,

−δ p = gρδ z or ∂p∂z

= −gρ (3.8) Eqn. (3.8) is known as the hydrostatic equation and negative sign in this equation ensures that pressure decreases with height. Eq. (3.8) can be integrated from surface (z = 0) to a height z to obtain the following result

⎭⎬⎫

⎩⎨⎧−= ∫ H

dzpzpZ

Oo exp)( (3.9)

Fig. 3.1. Hydrostatic balance

z + δ z

z

Ground level

g

z= 0

−δ p

p(z)

p(z)+ δ p

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H is known as the scale height, and it is that altitude where pressure at surface reduces by a factor e−1 (e = 2.71828183) . H varies between 6 km at T = 210 K to 8.5 km at T = 290 K for the lower atmosphere.

Sea Level Pressure: If the mass of the earth were distributed uniformly over the globe with actual topography, the sea level would be 1.013 x 105 Pa or 1013.25 hPa, which is referred to as 1 atmosphere (or 1 atm).

Question 3: Show that in an atmosphere of uniform temperature (To ), p(z) = po e− z/H ; what

is the expression for H?

3.4 Geopotential The geopotential Φ at any point in the earth’s atmosphere is defined as the work done

to raise a mass of 1 kg from sea level to that point against the force of gravity. The geopotential at sea level Φ(0) = 0. In our calculations, we can therefore take, Φ = gz with z as the geometrical height in meters (units of geopotential J kg-1 or m2 s-2).

Φ Z( ) = gdzO

Z

∫ (3.10) The geopotential height is defined as,

Z = Φ(z)go

= 1go

gdzo

Z

∫ (3.11)

z(km) Z(km) g0 0 9.811 1 9.8010 9.99 9.77

g0 = 9.81ms−2 is the globally averaged value of g. In the lower atmosphere go ≈ g.

Geopotential thickness: From eqns. (3.10) and (3.8) with α = RdTp

, we have

dΦΦ1

Φ2

∫ = gdzZ1

Z2

∫ = − RdTdppp1

p2

∫ ⇒ Φ2 −Φ1 = −Rd Tdppp1

p2

∫ , which gives

Z2 − Z1 = Rdg0

T dppp2

p1

∫ or Z2 − Z1 =RdTg0

ln p1p2

⎛⎝⎜

⎞⎠⎟

(T = mean temperature) (3.12)

The difference, Z2 − Z1 is referred to as the geopotential thickness or simply the thickness of an atmospheric layer bounded by pressure levels p1 and p2. If [T ] is the average temperature

of this atmospheric layer, then with H = Rd[T ]go

≡ RdTgo

Z2 − Z1 = H lnp1p2

⎛⎝⎜

⎞⎠⎟

(3.13)

p2 = p1 exp −Z2 − Z1H

⎧⎨⎩

⎫⎬⎭

(3.14)

Eqn. (3.13) is known as the hypsometric equation in meteorology parlance.

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Exercise 1: Calculate the height (geopotential) of 1000 hPa and 900 hPa when atmospheric pressures at sea level is: (a) 1014 hPa and (b) 1020 hPa. Take H=8 km. (Ans.: 112m & 160m).

Question 4: Define T in (3.13). Ans. T =T dppp1

p2

∫dppp1

p2

∫ (weighted mean layer temperature).

Exercise 2: Calculate ΔZ of a layer between 500 hPa and 1000 hPa at a point in (i) tropics with T =15°C (Ans.: 5849 m); (ii) polar latitudes with T = −40°C (Ans.: 4732 m). An important conclusion from eq. (3.12) is that geopotential thickness is proportional to mean temperature of the layer between pressures p1 and p2 as seen in Fig. 3.2; when moisture is present in the layer T is replaced by T v (the mean virtual temperature. which will be defined later).

Fig. 3.2 The solid lines indicate isobars. Thickness of layer between isobars is larger in the warm area than over cold area. Panel (a): Winds in a warm core lows are strongest at the surface and decrease with height. Panel (b): Upper level lows sometimes do not extend down; so at lower levels, there is cold core (i.e. the high is topped by a low)

3.5 System of Units There are two system of units, viz., (i) MKS: Metre (m), Kilogramme (kg) and

Second (s); and (ii) cgs: centimetre (cm), gramme (g) and second (s).

Temperature: °C and °F; TF =95T + 32 , TF (°F) and T (°C)

Pressure: 1 bar = 105 Pa = 103 hPa; 1 millibar (mb) = 102 Pa Torricelli or mm Hg (torr); 1 torr = 133.322 Pa 1 atmosphere (atm) = 1.01325 bar = 1.01325 x 105 Pa = 760 torr Pound-per-square-inch: 1 psi = 6894.76 Pa

Derived units

Acceleration m s-2 cm s-2 Density kg m-3 g cm-3 Force Newton (N) dyne (dyn) N=kg m s-2 dyn = g cm s-2

Pressure Pascal (Pa) microbar (µbar) Pa = N m-2 µbar=dyn cm-2 Energy Joule (J) erg J=N m erg=dyn cm Specific energy J kg-1 = m2 s-2 erg g-1 = cm2 s-2 Power Watt (J s-1) erg s-1

760 mm of Hg at 0°C = 1 atm 1 cal = 4.1840 J

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3.6 Reduction of Pressure to Sea Level (PSL) In mountainous regions, the difference in surface pressure (ps) from one station to

another is largely due to difference in their elevation above the sea level. In order to determine the pressure change due to the passage of weather systems, the pressure are reduced to sea level (reference level). The sea level pressure is calculated using the hypsometric equation (3.14),

p2 = p1 exp −Z2 − Z1H

⎧⎨⎩

⎫⎬⎭

; ps = psl exp −Zs − Z1H

⎧⎨⎩

⎫⎬⎭

with p1 = psl and p2 = ps , Z1 = 0

psl = ps expZsH

⎧⎨⎩

⎫⎬⎭= ps exp

gZsRdT

⎧⎨⎩

⎫⎬⎭

(3.15)

3.7 Concept of a Parcel of Air We have assumed atmosphere in hydrostatic equilibrium. However, atmosphere is

constantly heated by solar radiation at the surface, so hot surface air will rise upwards, which may disturb this kind of equilibrium. When will atmosphere turn unstable? Here, the concept of an air parcel helps in determining the necessary conditions. We define an air parcel as a part of the atmosphere which is not different from other parts at the same level, but becomes a distinct portion once it is displaced from its original position. Its dimensions are sufficiently large in comparison to the mean-free path of air molecules but small enough to represent the average properties of the atmosphere of its location. Most importantly, a parcel is insulated from its surrounding during its displacements but instantaneously adjusts its pressure with the environmental pressure. Large expanses of air having consistent characteristics of temperature and moisture of regions where they originate are called “air masses”. The air mass retains its characteristics and does not mix even during large movements with the wind from places of its origin to other distant parts on the globe. The parcel movement is generally considered in the vertical direction, but air masses move horizontally over long distances. Parcels drawn from an air mass can give the characteristics and origin of air masses with the help of retro trajectories.

3.8 Adiabatic Lapse Rate Consider the vertical movement of a parcel at pressure p, temperature T and of specific

volume α (=1/ρ) in the atmosphere. Assume that atmosphere is in hydrostatic equilibrium; that is, the upward force due to pressure gradient in vertical is balanced by the weight of the parcel. The gravitational force and the buoyancy force are balanced; hence there is no need to consider them in the first law of thermodynamics. For a unit mass, the first law of thermodynamics states that a quantity dQ of heat added to the system is utilized in increasing the internal energy of the system and as the work done on the system. Stated mathematically, the First Law of Thermodynamics is given as

dQ = dU + dW (3.16) Internal energy increment: dU = CvdT   ; Work done by pressure: dW = p dα     dQ = Cv dT  + pdα (3.17) Cv is the specific heat of air at constant volume. Differentiation of the equation of state gives d(pα ) = d RdT( )   ⇒   p dα +αdp = RddT    (3.18) Using α = 1

ρ and Rd = Cp – Cv   in (3.18), we can write it in the following form

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pdα + dpρ

= RddT  = Cp − Cv( )dT . Substituting the value of pdα in (3.17), we get

dQ = CpdT − dpρ   ⇒  dQ = Cp  dT + gdz  (3.19)

If the parcel undergoes adiabatic expansion or compression during its movement in the vertical, there is no exchange of heat (i.e. heat neither enters or leaves the parcel during motion); hence dQ = 0 and (3.19) gives

dQ = CpdT + gdz = 0  ⇒ −dTdz

= gCp

= Γd (3.20)

Equation (3.20) gives the temperature decrease with height that can be calculated with g = 9.81ms−2 and Cp = 1005 J kg

−1 as

Γd =gCp

= 9.76 K km−1 (3.21)

This is the first important result in atmospheric thermodynamics. The adiabatic lapse rate Γd is the rate of change of temperature with height of a parcel of dry air when it adiabatically rises up or sinks down in a dry atmosphere. However, the profile of temperature from an

ascent of a radiosonde gives the actual lapse rate Γ = − dTdz

of the atmosphere. It varies

widely due to the presence of water vapour and the average value of Γ is 6-7 K km-1 in the troposphere; that is Γ < Γd .

Fig. 3.3 Atmospheric lapse rates: Panel (a):- The line marked OA is dry adiabat. The environmental temperature profile is the dotted line CXPYD. The part CX represents steep fall in temperature with height in the environment. Negative lapse rate corresponds to an adiabat sloping on the right; Panel (b):- Profile OA corresponds to dry adiabatic lapse rate; the dotted line PQ shows rise in temperatures with height (inversion or negative lapse rate); the dashed line CQ corresponds to “superadiabatic lapse rate” in the atmosphere.

The meaning of constant lapse rate: If the atmosphere is heated by contact with the earth’s surface and vertical motions set in, then heat will be distributed up in such a manner that the vertical temperature profile will have a constant gradient (Γd ) of ~10 K km

-1 with altitude.

Y

X

z

T

Γ = Γd Γ <Γd

Γ>Γd

O

A B

C

DΓ = 0

PQ

(a)

z

T

Γ = Γd

O

A

CP

Q

(b)

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Now, consider the stability of the atmosphere with respect to vertical motions with the

help of Fig. 3.3 that shows different vertical profiles of temperature (T vs z ) in the atmosphere. In Fig. 3.3a, different temperature profiles are given: profile marked OA is the dry adiabat (i.e. the temperature profile with constant lapse rate Γd ); profile CXYD is the actual temperature profile of the environment (actual lapse rate); and profile OB shows the lapse rate Γ greater than Γd , and dashed line profile shows that temperature of the atmosphere is uniform with height (isothermal layer) and Γ= 0. If a parcel initially at point X rises, then it will follow the dry adiabat line OA and would arrive at a location Q, where it is surrounded by ambient atmosphere with conditions at P on the profile CXYD. That is, parcel is warmer than the ambient atmosphere and shall continue to rise; hence the point X on profile CXYD is unstable. In other words, the parcels that are displaced from point X on the profile CXYD would never return to X. By similar argument point Y on profile CXYD is stable if it is displaced vertically; because parcels that are pushed up (down), being heavier (lighter) than the surrounding, shall return back to point Y. That is, atmosphere is stable above the point Y. Thus we have

(i) Atmosphere stable if − ∂T∂Z

< Γd (3.22)

(ii) Atmosphere unstable if − ∂T∂Z

> Γd (3.23)

3.9 Potential Temperature: The potential temperature θ of an air parcel is defined as the temperature that an air

parcel would have if it were expanded or compressed adiabatically from its existing pressure (p) and temperature (T) to a standard pressure p0 (generally taken as 1000 hPa). From the first law of Thermodynamics, we have the relation

dQ = CpdT − dpρ   (3.24)

Eqn. (3.24) can be written for an adiabatic transformation ( dQ = 0 ) as

dQ = CpdT − α dp = 0 ⇒ CpRd

dTT

− dpp

= 0 (3.25)

On integrating equation (3.25), we obtain

CpRd

dTTθ

T

∫ =dppp0

p

∫ ( θ = T0 at p = po ) ⇒CpRd

lnTθ= ln p

po

θ = T pop

⎛⎝⎜

⎞⎠⎟

Rd /Cp

(3.26)

The equation (3.26) is called the Poisson equation.

Define κ = RdCp

and γ =CpCv

then κ = (1− 1γ

) = 0.2856

In calculating the value of κ , we have used Rd = 287 J K−1kg−1 and Cp = 1005J K

−1kg−1 .

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Exercise 3: Starting with eq. (3.26), show that: Tγ

pγ −1= T0

γ

poγ −1 =

θγ

poγ −1 = const.

Thus, a new temperature θ can be defined with (3.26) which a parcel will have if brought adiabatically to a reference pressure po , taken as 1000 hPa. It must be noted that for an incompressible medium, temperature is the appropriate variable; but for the compressible atmosphere, it is the potential temperature θ. To study the dynamic evolution of the atmospheric flows θ is an appropriate variable, because it is a measure of the sum of potential and internal energy and it is conserved in adiabatic motions of the atmosphere. Quantities that remain invariant during any transformation are said to be conserved variables. The potential temperature is therefore an extremely important parameter in atmospheric thermodynamics. The conservation of θ simplifies the treatment of the dynamics of compressible fluids in the absence of heat sources and sinks, much like the dynamics of incompressible fluids that is rendered simpler (divergence free motions) because density remains constant. The conservation of θ is mathematically expressed as

dθdt

= 0 (3.27)

Exercise 3: A parcel of air has a temperature -51°C at 250 hPa level. What is its potential temperature? What temperature will the parcel have if it were brought into the cabin of a jet aircraft and compressed adiabatically to a cabin pressure of 850 hPa? [Hint: Start with T = (−51+ 273.15) K , p = 250hPa , p0 = 1000hPa ; calculate θ ; Parcel brought to cabin: T = 222.15 K , p = 250hPa , p0 = 850hPa ; calculate θ ]. Note: Taking the logarithm of (3.26), it becomes easier to differentiate (3.26); we obtain,

lnθ = lnT + RCpln po −

RCpln p . Now differentiate on both sides to get

1θdθdz

= 1TdTdz

− RCp1pdpdz

(3.28)

3.10 Static Stability In equation (3.28), if hydrostatic equation is used then we have,

1θdθdz

= 1TdTdz

− RCp1p(−gρ) ; ρ = p

RT

Tθdθdz

= dTdz

+ gCp

(3.29)

For an atmosphere with constant θ (i.e. an adiabatic atmosphere), the atmospheric lapse rate

dzdT− is obtained from (3.29) by setting dθ

dz= 0 , and we obtained the earlier result,

− dTdz

= gCp

= Γd (3.30)

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Hence, if potential temperature θ is a function of height, the atmospheric lapse rate

⎟⎠⎞⎜

⎝⎛ −=Γ

dzdT will differ from adiabatic lapse rate Γd and we have from (3.29),

Tθdθdz

= Γd − Γ (3.31)

If Γ < Γd , it implies that dzdθ is positive i.e. θ increase with height. Thus, an air parcel that

undergoes an adiabatic displacement from its equilibrium position will be positively buoyant when displaced vertically downward. The air parcel will be negatively buoyant (sinks) when displaced vertically upward from its equilibrium position. Such as atmosphere is said to be statically stable or stably stratified atmosphere 3.11 Buoyancy Oscillations

Adiabatic oscillations of a fluid parcel about its equilibrium position in a stably stratified atmosphere are called buoyancy oscillations. The frequency of such oscillations could be derived by vertically displacing a parcel upward or downward a small distance δz from its equilibrium position at a height z. The atmosphere is always considered in hydrostatic equilibrium. So any imbalance between the upward pressure gradient and its weight will result in vertical acceleration; but when there is a balance, we have

dp0dz

= −gρo => −1ρo

∂po∂z

− g = 0 (3.32 )

Here po and ρo are pressure and density of the environment. If p and ρ are the pressure and density of the parcel, any imbalance between the vertically acting pressure gradient force

and the weight of the parcel will result in vertical accelerations ( dwdt

) of the parcel. The

Newton’s second law gives the equation of motion of the parcel as,

dwdt

= − 1ρdpdz

− g or ρ dwdt

= − dpdz

− gρ (parcel) and w = ddt(δ z) ; hence

ρ d2 (δ z)dt 2

= − dpdz

− gρ . (3.33)

Since environment is in hydrostatic equilibrium, dpdz

in (3.33) is replaced by dp0dz

and we get

ρ d2 (δ z)dt 2

= − dp0dz

− gρ

Now replace dp0dz

by −gρo in the above equation using (3.32), and (3.33) becomes,

ρ d2 (δ z)dt 2

= g(ρ0− ρ) (3.34)

Note that the term g(ρ0− ρ) on the right hand side of (3.34) is the buoyancy force acting on

the parcel of unit volume. We can now write (3.34) as

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d2

dt 2(δ z) = g ρo − ρ

ρ (3.35)

As pointed out earlier that a parcel adjusts its pressure instantaneously with the surrounding pressure. Thus, for a parcel undergoing small displacements δz from its equilibrium position without disturbing its environment, we set p (parcel) = p0 (environ) . Hence, the pressure gradient

term in (3.33) is replaced by dpodz

in parcel method.

Fig. 3.4 Displacement of a parcel

Consider a parcel in Fig. 3.4 at a height z and it is displaced vertically by δ z to a new position z +δ z . As the parcel moves from its equilibrium position z to new position z +δ z , we can estimate the change in the thermodynamic variables T, θ and ρ. Initially, the parcel and environment are at the same temperature T, density and pressure, but when displaced its temperature is given by

Tp (z +δ z) = Tp (z)+dTdz

δ z or Tp (z +δ z) = T (z)− Γd δ z (3.36 a)

For the environment,

Tenv(z +δ z) = T (z)+dTdz

δ z or Tenv(z +δ z) = T (z)− Γδ z (3.36b)

The density ρ of parcel and ρo of environment can also be written at new position z1 . Now, one may derive the following expression that holds also at the new position of the parcel,

g ρo − ρρ

= gTp −TenvTenv

(3.37)

Using (3.36) and (3.37) in (3.35), we get

d 2

dt 2(δ z) = g

(Γ − Γd )δ z(T − Γδ z)

= g(Γ − Γd )δ z

T(1− Γ

Tδ z)−1 = g

(Γ − Γd )δ zT

1+ ΓTδ z + ...⎛⎝⎜

⎞⎠⎟

Neglecting term in (δ z)2 on the right hand side, we get the following oscillation equation for a parcel

d 2

dt 2(δ z) =−N 2 δ z ; N 2 = g

T(Γd − Γ). (3.38)

In eqn. (3.38), N 2 is the buoyancy frequency also called the Brunt-Väisälä frequency; N 2 is a measure of static stability of the atmosphere; for the lower atmosphere, the corresponding

period τ = 2πN

is a few minutes. The parcel undergoes oscillations when N 2 > 0 .

T T

Tp Tenv

z

z+δ z

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Exercise 4: If θ and θ0 are the potential temperatures of the parcel and the environment,

using (38), show that g ρo − ρρ

= gθ −θ0θ0

. The parcel undergoes adiabatic displacements,

hence θ(z +δ z) = θ0 (z) . Write for the environment, θo(z +δ z) = θo(z)+dθodz

δ z and show

that N 2 = gθ0

dθ0dz

= g d(lnθo )dz

.

The solution of (3.38) is given by δ z = A eiN t (3.39) If parcel oscillates about its mean position when N 2 > 0 , then for average tropospheric conditions,

N = 1.2 x 10-2 s-1 ⇒ τ = min7.82 =Nπ

If N = 0 , There will be no oscillations of the displaced parcel about its equilibrium position and parcel will be in neutral equilibrium (i.e. no accelerating force on parcel).

If N 2 < 0 , the potential temperature will decrease with height; which means the displacement of the parcel will increase exponential with time; in other words, parcels continue to move through the atmospheric column and the column is unstable. The above condition also lead us to write the conditions of static stability of the atmosphere

in terms of the potential temperature from the relation N 2 = gθ0

dθ0dz

. Thus, we have,

N 2 > 0 Atmosphere statically stable 0>dzd oθ

N 2 = 0 Statically neutral dθodz

= 0 (3.40)

N 2 < 0 Statically unstable 0<dzd oθ

An important note: On the synoptic scale, atmosphere is always stably stratified and when the stratification is unstable, convective overturning removes quickly the unstable regions that develop. The convective overturning essentially mixes the air in overlaying layers and it redefines the distribution of isentropes (surfaces of equal θ ) in the stratification. However, for moist atmosphere, situation is more complicated. If the temperature actually increases with height (temperature inversion) the atmosphere is gravitationally stable as warm air overlays the cold air. Temperature inversions inhibit mixing; and water vapour, aerosols and pollutants are trapped in the layer. During winter, temperature inversions occur during nights, therefore in the big cities, vehicular emissions will be trapped in this layer and there will be episodes of high pollution and smog after sunrise. However, it is worth mentioning that over global deserts, in day time hours, negative stability can exist in the air next to the ground with super-adiabatic lapse rates.