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Process II This will occur in two stages: (1) Decrease the pressure exerted by

the piston to 2.5 atm so that the speciﬁc volume will be 4 liters and

then (2) reduce the pressure to 1 atm with a speciﬁc volume of

10 liters. The work of expansion becomes the sum:

exp

¼ P

þ P

¼ 7:5 þ 6:0

¼ 13:5 Latm

The mechanical energy transfer from the system (dQ) is thus

13.5 Latm.

Process III Let the pressure exerted by the piston be continuously reduced such

that the pressure of the gas is inﬁnitesimally greater than that

exerted by the piston (otherwise no expansion would occur). Then

the pressure of the gas is essentially equal to that of the surround-

ings and since dw ¼ PdV,

exp

Pda ¼ R*T ln



23:03 Latm

In each of the above three cases, the system responded to the removal of external

weight to the piston by changing state and performing a net work on the surround-

ings while maintaining isothermal conditions within the system by absorbing heat

from the attached heat reservoir, the reservoir being part of the surroundings. In each

case, the total change in the system state was identical, although the energy absorbed

by the system from the surroundings, equal also to the work performed on the

surroundings by the process varied greatly.

Note that process III differed from the ﬁrst two processes in that it was an

equilibrium process, or one where the difference between the intensive driving

variable (pressure in this case) of the system differed only inﬁnitesimally from the

surroundings. The maximum work possible for the given change in system state

occurs in such an equilibrium process. All other processes, not in equilibrium,

produce less net work on the surroundings and thus also absorb less energy from

the heat reservoir!

It is important to note that, unlike the nonequilibrium processes, the net work for

the equilibrium process is dependent only on the initial and ﬁnal states of the system.

Hence if the equilibrium process is run in reverse, the same amount of work is

performed on the system as the system performed on the surroundings in the forward

direction. In the case of the nonequilibrium process, this is not the case and so the

amount of energy released to the surroundings for the reverse process is unequal to

that absorbed in the forward direction. Hence the system is irreversible. Hence we

can equate an irreversible process to one which is not in equilibrium.

196 PHYSICAL ATMOSPHERIC SCIENCE

For an isothermal reversible process involving an ideal gas, dU ¼0 and hence

dQ ¼ pdV ¼ dW

max

. Thus dQ is dependent only on the initial and ﬁnal states of the

system. Hence, for a reversible process, Q and W behave like stat e functions.Asa

consequence, since for either the irreversible or reversible processes, dQ ¼ dW ,it

follows that:

rev

¼ dW

max

ð42Þ

Since the dQ

rev

is related to a change in state, then the isothermal reversible result

described by process III must apply for the general isothermal case, i.e.,

rev

¼ R*d ln V ð43Þ

where the right-hand side of Eq. (43) is an exact differential. By this reasoning it is

concluded that the left-hand side of Eq. (43) must also be an exact differential of

some variable that we will deﬁne to be the entropy (S) of the system (joules per

kelvin). Since the change in entropy of the system is always an exact differential, its

value after any thermodynamic process is independent of the path of that process and

so entropy, itself, is also a state funct ion. We can write:

dS ¼

rev

ð44Þ

From Eq. (42) we can also show that

max

¼ TdS ð45Þ

which states that the maximum work that can be performed by a change in state is

equal to the temperature multiplied by the total change in entropy from the initial to

the ﬁnal state of the system. Since we showed that the net work performed on the

surroundings as a result of the isothermal process must be less than or equal to

max

, then it is implied that

dS >

ð46Þ

for an irreversible process. It is also reasoned that by virtue of the second law, and

obvious from the example described above, that a process where dS < dQ=T would

be a forbidden process. Since the entropy of the surroundings must also satisfy these

constraints, the entropy of the universe can either remain constant for a reversible

process or increase overall for an irreversible process. This is the mathematical

statement of the second law for the case of isothermal processes.

We must now expand this concept to a system not constrained to isothermal

conditions. To build a theory applicable to a general process, we consider the

following additional particular processes:

1. Adiabatic Reversible Expansion of an Ideal Gas For a reversible adiabatic

expansion, dQ ¼ 0 and so dS ¼ 0. This is an isentropic process. Since a

2 ATMOSPHERIC THERMODYNAMICS 197

reversible dry adiabatic process is one where y is conserved, a process with

constant y is also called an isent ropic process.

2. Heating of a Gas at Constant Volume: For a reversible process at constant

volume, the work term vanishes:

dS ¼

rev

¼ C

d ln T ð47Þ

3. Heating of an Ideal Gas at Constant Pressure: For a reversible process:

dS ¼

rev

¼ C

d ln T ð48Þ

The Carnot Cycle We demonstrated earlier that a net work can also be

performed on the surroundings as a system goes throug h a cyclic process. That

work, we showed, was equal to the area traced out on a P–V diagram by that cyclic

process. The maximum amount of work possible by any process was that performed

by a reversible process. We can call our system that performs work on the

surroundings as the working substance.

Now we ask, Where does the energy for the work performed originate? For a

cyclic process the initial and ﬁnal states of the working substance are the same so

any work performed over that process must come from the net heat absorbed by the

system from the surroundings during that process. We say net because the system

typically absorbs heat and also rejects heat to the surroundings. If the system

converted all of the heat absorbed to work without rejecting any we would say

that the system is 100% efﬁcient. That cannot happen typically. For instance, the

engine in a car burns gasoline to heat air within the cylinder that expands to do work.

But much of the heat given off from the gas actually ends up warming the engine and

ultimately the surroundings rather than maki ng the engine turn. The degree to which

the energy warms the surroundings compared to how much is used to turn the engine

and ultimately move the car (more energy is lost to friction in the engine and

transmission and so on) is a meas ure of how efﬁcient the car is.

For any engine, we can quantify its efﬁciency as:

Z 

Mechanical work performed by engine

Heat absorbed by engine



 Q

 q

ð49Þ

where Q

is the heat absorbed by the engine and Q

is the heat rejected by the engine

to the surroundings. In terms of speciﬁc heat:

Z ¼

 q

ð50Þ

The maximum efﬁciency that any engine can ever achieve would be that of an engine

198 PHYSICAL ATMOSPHERIC SCIENCE

mission, not perfectly insulated walls of the pistons, and so on would all be irrever-

sible parts of the cyclic process in a gasol ine engine. Now let’s imagine an engine

where the thermodynamic cyclic process is carried out perfectly reversibly. One

simple reversible thermodynamic process that ﬁts this bill is the Carnot cycle,

which powers an imaginary engine called a Carnot engine. No one can build

such a perfect frictionless engine, but we study it because it tells us what the

maximum efﬁciency possible is for cyclic process. Yes, even the Carnot engine is

not 100% efﬁcient and below we will see why.

First let’s describe the Carnot cycle. Simply put, the Carnot cycle is formed by

two adiabatic legs and two isothermal legs so that the net energy transfers can be

easily evaluated with simple formulations for adiabatic or isothermal processes.

Figure 5 shows such a process. The four legs of the process are labeled I to IV

beginning and ending at point A. To visualize this physically, we can imagine some

working substance contained within a cylinder having insulated walls and a conduct-

ing base, and a frictionless insulated beginning piston. The cycle is explained as:

I The cylinder is placed on a heat reservoir held at T ¼ T

. Beginning at

T ¼ T

, and V ¼ V

, the working substance is allowed to slowly expand

maintaining isothermal conditions.

II The cylinder is placed on an insulated stand and allowed to expand further to

volume V

and temperature T

. Since the working substance is totally

insulated for stage II, the process is adiabatic.

III The cylinder is taken from the stand and placed on a heat reservoir where

the working substance is held at T ¼ T

and slowly compressed (isother-

mally) to speciﬁc volume V

IV The cylinder is replaced on the insulated stand and the working substance is

slowly compressed (perhaps adding weight to piston slowly) causing the

Figure 5 Carnot cycle (from Iribarne and Godson, 1973).

2 ATMOSPHERIC THERMODYNAMICS 199

substance to warm adiabatically back to temperature T

and speciﬁc volume

This process is illustrated graphically in Figure 5. The cycle is reversible and

consists of two isotherms at temperatures T

and T

, where T

< T

and two adia-

bats, i.e., y

and y

The net work perfor med by the system is the area formed by the intersection of

the four curves. We can compute the work A and the heat Q for the four steps of the

process (direction indicated by arrows) in the following table:

ðIÞ DU

¼ 0 dQ

¼A

PdV¼ R*T

ln V

ðIIÞ dQ

¼ 0  A

¼DU

¼ C

ðT

 T

ðIIIÞ DU

III

¼ 0 dQ

¼A

III

¼R*T

ln V

ðIVÞ dQ

¼ 0  A

¼DU

¼C

ðT

 T

Since this is a cyclic process, the sum of the four terms Du is zero. The work

terms on the two adiabats cancel each other. The gas effectively absorbs the quantity

of heat Q

> 0 from the warmer reservoir and rejects it Q

< 0 in the colder

reservoir. The total work is then W ¼ A

þ A

III

¼ðQ

þ Q

Þ.

We can now relate q

and q

to the temperatures of the two heat reservoirs. From

Poisson’s equation,



R*=C



R*=C

; ð51Þ

and so:

; ð52Þ

which, substituting into the expressions for Q

and Q

gives

¼ 0; ð53Þ

or, alternatively



: ð54Þ

200 PHYSICAL ATMOSPHERIC SCIENCE

Now we can calculate the thermodynamic efﬁciency of the Carnot engine:

Z ¼

 T

: ð55Þ

Hence, the Carnot c ycle achieves its highest efﬁciencies when the temperature

difference between the two heat reservoirs is the greatest!

It is useful to generalize this result to the case of irreversible processes. To do so,

we consider the second law, and its implication that heat can only ﬂow from a

warmer toward a cooler temperature. As a ﬁrst step, it is noted that Eq. (55) is

valid for any reversible cycle performed between two heat sources T

and T

independent of the nature of the cycle and of the systems. This is a statement of

Carnot’s theorem.

Just as Eq. (55) holds for a reversible cycle, it can be shown that for any irre-

versible cycle, the second law requires that ðQ

ÞþðQ

Þ < 0, i.e., heat must

ﬂow from warm to cold and not vice versa. It can also be shown that any reversible

cycle can be decomposed into a number of Carnot cycles (reversible cycles between

two adiaba ts and two isotherms). Therefore, for any irreversible process it can be

shown that

 0 ð56Þ

which implies Eq. (46).

The Carnot cycle can also be viewed in reverse, in which case it is a refrigerating

machine. In that case a quantity Q

of heat is taken from a cold body (the cold

reservoir) and heat Q

is given to a hot reservoir. For this to happen, mecha nical

work must be performed on the system by the surroundings. In the case of a

refrigerator, an electric motor supplies the work.

In recent years, atmospheric scientists have discovered that certain types of

weather systems derive their energy from a Carnot cycle. In particular, the tropi cal

cyclone* is powered by a Carnot-like cycle created by the radia l circulation inward

to the storm center at the surface, up within the eye-wall clouds, outward at the storm

top, and then back downward far away from the storm center. State I, or the Carnot

cycle, is equivalent to the air moving inward toward the storm center along the ocean

surface at relatively constant temperature and toward low pressure in the storm

center. The heat reservoir is the ocean. In this case it heats not only through the

transfer of sensible heat but also through the transfer of latent heat. Stage II is found

in the eye wall, where the heated air rises moist adiabatically under falling tempera-

tures. Stage III occurs in the outﬂow at high levels away from the storm center. The

outﬂowing air maintains a constant temperature while slowly subsiding and pressure

*Tropical cyclones are known as ‘‘hurricanes’’ in the Atlantic and eastern Paciﬁc oceans, while in the

western Paciﬁc they are known as ‘‘typhoons.’’

2 ATMOSPHERIC THERMODYNAMICS 201

rising. The heat is lost to space by transfer through the emission of long-wave

radiation. Hence the cold plate is the radiation sink in space. Finally, stage IV

occurs as the air ﬁnally sinks back to the ground approximately moist adiaba tically.

This occurs mainly within the downdrafts of convective clouds in the periphery of

the storm. The result is a net gain in energy of the entire system, which is manifested

in the cyclone winds. The storm loses energy primarily to surface friction, which is

proportional to the speed of the winds. Therefore, the more energy released by the

Carnot cycle, the stronger the surface winds that come into equilibrium with the

Carnot cycle release. As with the discussion above, the efﬁciency of the cycle is

proportional to the temperature difference between storm top and the sea surface. As

a result, the strength attainable by a tropical cyclone is closely related to the sea

surface temperature.

Energy harnessed by the Carnot cycle is likely the basis for other types of weather

for which thermally driven convection is impor tant. These include a host of convec-

tive systems and possibly some aspects of Earth’s general circulation that derive their

energy from thermally direct circulations.

Restatement of First Law with Entropy The second law is a statement of

inequality regarding the limits of entropy behavior. When combined with the ﬁrst

law, the resulting relationships become inequalities:

Tds dU þ PdV

ð57Þ

TdS dH  VdP

ð58Þ

for the internal ene rgy and enthalpy forms, respectively. Alternatively, we may write

for the special case of a reversible process:

dU ¼ TdS PdVþ

and ð59Þ

dH ¼ TdSþ VdPþ

: ð60Þ

Note this differs from previous forms of the ﬁrst law in that we have an equation for

the parcel (or system) in terms of state function variables only. The difference

between the general Eqs. (57) and (58) and the special case Eqs. (59) and (60) is

a positive source term for entropy resulting from a nonequilibrium reaction that has

not yet been determined. In the world of dry atmospheric ther modynamics, we can

live with assuming the dry system is in equilibrium and so the second form is

sufﬁcient. However, when we begin to consider moist processes, the system is not

always in equilibrium, and so the sources for entropy from nonequilibrium processes

will have to be evaluated. The framework for the evaluation of these effects involves

the creation of free energy relations described in the next chapter.

202 PHYSICAL ATMOSPHERIC SCIENCE

Free Energy Functions In their present form, Eqs. (57) and (58) relate

thermodynamic functions that involve dependent variables that include the extensive

variable S. It is again convenient to make a variable transformation to convert the

dependence to variations in T instead of S. The transformation is made by deﬁning

the so-called free energy functions called the Helmholtz free energy (F) and the

Gibbs free energy (G) are deﬁned as:

F  U  TS

G  H  TS

In derivative form, they are written:

dF ¼ dU  TdS SdT

dG ¼ dH  TdS SdT

For our application to atmospheric problems, we will work primarily with the Gi bbs

free energy because of its reference to a constant pressure process.

Combining Eq. (61) with Eq. (60) we obtain

dG ¼SdTþ VdPþ

ð61Þ

where G

refers to Gibbs energies for each constituent of the system (k ) necessary to

form an equality of Eq. (58). The restrictions imposed by the second law and

reﬂected by those inequalities result in the requirement that

 0. Because

G ¼ GðT; V ; n

Þ is an exact differential, then

dG ¼



P;n

dT þ



T;n

dP þ



P;T ;n

ð62Þ

We can evaluate the potential internal energy and Gibbs free energy terms of Eq.

(61) as:



T;P



T;P

2 ATMOSPHERIC THERMODYNAMICS

203

where G

is the Gibbs free energy resulting from chemical potential. The terms of

Eq. (62) are then evaluated:



T;P



T;P



T;P

;



V ;n

¼S; and



T;n

¼P:

Concept of Equilibrium Equilibrium is not an absolute but is deﬁned in terms

of permitted processes. In statics, equilibrium is described as the state when the sum

of all forces is zero. This may be stated as a principle of work: If a system is

displaced minutely from an equilibrium state by a change in one of the system

properties, the sum of all the energy changes is zero. Neglecting internal potential

energies (A

), this leads to the statement that for a single component system of ideal

gas, undergoing a constant temperature, constant volume process, dF ¼ 0 while for

a constant temperature process and constant pressure process dG ¼ 0 at equilibrium.

Hence if one were to plot G (or F) as a function of system properties , equilibrium

will appear as a minimum along the G (or F) curve.

In general, we deﬁne a spontaneous process as one in which the system begins

out of equilibrium and moves toward equilibrium, resulting in an increase in entropy.

The rate of the process is undetermined thermodynamically. Consider the Gibbs free

energy for a constant temperature and constant pressure process. Since equilibrium

occurs at a minimum in Gibbs free energy, we can deﬁne:

A spontaneous process DG < 0

An equilibrium process DG ¼0

A forbidden process DG > 0

The Molar chemical potential of species k (m

) is deﬁned as:





T;p;n

: ð63Þ

For a system of only one component of a set on noninteracting components:

¼ G

ðmÞ¼H

 TS

ð64Þ

204 PHYSICAL ATMOSPHERIC SCIENCE

in which G

ðmÞ is the molar free energy of species k. We now write:

dG ¼SdTþ VdPþ

: ð65Þ

Hence the third term on the right-hand side (RHS) would represent free energies

resulting from the potential for interaction between all components in the system,

while the last term on the RHS represents the noninteracting free energy of each

component, for example, surface free energy.

Consider a system composed of a single ideal gas and held in a state of equili-

brium where dG ¼ 0. Ignoring internal potential energies, Eq. (65) becomes

dm ¼ SdT VdP: ð66Þ

Using the equation of state an d integrating, we obtain

m ¼ m

ðTÞþnR*T ln P; ð 67 Þ

where m

ðTÞ is the standard chemical potential at unit pressure (usually taken to be

1 atm). Hence, for an ideal gas with two constituents:

DG ¼W

max

¼ðm

 m

Þðn

 n

Þ¼nRT ln

: ð68Þ

We can also look at the chemical potential of a condensed phase, using the

concept of equilibrium. Consider liquid water in equilibrium with water vapor at a

temperature T and a partial pressure of e

. Applying the principle of virtual work, we

transfer dn moles of water from one phase to the other under the condition:

dn ¼ m

dn;

¼ m

þ R

T ln e

;

where m

and m

are the channel potentials of liquid and vapor respectively.

This is a general statement and demonstrates that the chemical potential of any

species is constant throughout the system (if the equilibrium constraints permit

transfer of the species). We can also see that for a nonequilibrium phase change:

DG ¼ m

 m

¼ R

T ln

; ð69Þ

where e

is the partial pressure of vapour not at equilibrium.

Hence if e

< e

, then DG < 0 for a process of condensation and hence that is a

forbidden process. By the same token, evaporation becomes a spontaneous process.

When m

¼ m

, the system is in equilibrium and at saturation. These concepts can be

2 ATMOSPHERIC THERMODYNAMICS 205