APPLICATIONS OF THE FIRST LAW

• Kinetic theory and internal energy: predicting the specific heats of dry air• Applications of the first law• Kirchoff’s theorem• Poisson’s equation• Potential temperature

KINETIC THEORY AND INTERNAL ENERGY

Without delving into the kinetic theory of gases, we can make use of one of its results in orderto obtain some insight into internal energy. each molecule possesses an amount of energy per degree of freedom equal towhere k is Boltzmann’s constant=1.381023 JK-1 . We can think of the number of degrees of freedom as the number of variables required tofully describe the motion of the molecules. So a monatomic molecule has 3 degrees of freedom (3 velocity components are required to specify its translational motion), a diatomicmolecule has 5 degrees of freedom (3 translational plus 2 angular velocities), while a triatomic molecule has 6 degrees of freedom (3 translational, 3 angular).

Dry air consists primarily of molecular Nitrogen (N2) and Oxygen (O2) and so isessentially diatomic with 5 degrees of freedom.

€

1

2kT

Hence the internal energy per mole is given by:

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1

2nkNAT =

1

2nR * T (4.1)

where n is the number of degrees of freedom and NA is Avogadro’s number equal toNA=6.021023 mol-1 . Note: Boltzmann’s constant may therefore be thought of as the universal gas constant per

molecule.

The internal energy per kilogram (i.e., the specific internal energy) is therefore given by:

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1

2n

R *T

M=

1

2nRT (4.2)

where M is the molecular weight of the gas. Thus, we would predict that for dry air,and since then . Moreover, since cp-cv=R, then we can also predict

that for dry air . Using R=287.05 Jkg-1K-1 , our prediction becomes for dry air:

cp=1005 Jkg-1K-1 . The table below shows that this prediction compares rather well withobservations.

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u =5

2RT

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cv =du

dT

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cv =5

2R

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c p =7

2R

-40oC +40oC

30 kPa 1004 1006

100 kPa 1006 1007

Table 4.1: Measured specific heat capacity at constant pressure, cp, for dry air.

POISSON’S EQUATIONS

Despite the fact that energy in the atmosphere/ocean system ultimately comes fromradiative heating, adiabatic processes in the atmosphere are of interest for several reasons.Often it is because real atmospheric processes occur quickly in comparison with the timescale for heat transfer, and so may be considered to be approximately adiabatic.

Alternatively, we may wish to make the adiabatic assumption simply because we are ignorantof the heat transfer and consequently must ignore it or give up. Poisson’s equations describerelationships between the state variables T, p, and for adiabatic processes.

When q=0 (adiabatic process), the first version of the first law may be written:

0=+ pdvdTcv (4.5)

Substituting for p from the ideal gas law, dividing by cvT, and using R=cp-cv we have, aftersome manipulation:

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d ln(Tvγ −1) = 0 (4.6)

where =cp/cv (=7/5 for an ideal gas). This implies that, for an ideal gas undergoing an adiabatic process:

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Tvγ −1 = constEq. (4.7) is one of three Poisson equations.

(4.7)

Eq. (4.7) can be used to explain a number of atmospheric phenomena. For example:

1. When you compress the air in a bicycle pump, v decreases and hence T increases and theair warms. (Sometimes the bicycle pump is noticeably warmer after use.)2. A similar explanation can be offered for the heating of subsiding (i.e., descending) air thatgives rise to a Chinook.3. Conversely, in a rising, expanding thermal, v increases and so T falls. This is a dynamicalcomponent that contributes to the decrease of temperature with height in the troposphere.(The other component is the height dependence of absorption of longwave radiation emitted bythe Earth.)

Starting with the second version of the first law, Eq. (3.8) and following a similar manipulation, one may arrive at the second Poisson equation for an adiabatic process in an ideal gas:

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Tp−κ = const (4.8)

where =R/cp (=2/7 for an ideal gas). Finally, by combining Eqs. (4.7) and (4.8), we arrive at the third Poisson equation:

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pvγ = const (4.9)

Where = cp / cv=7/5. Eq. (4.9) is the most familiar of Poisson’s equations, although they are all equivalent.

POTENTIAL TEMPERATURE

Meteorologists use Eq. (4.8) to define a quantity known as potential temperature (a thermodynamic concept that seems to be unique to meteorology). Suppose we start with aparcel of air in some arbitrary initial state specified by T, p. Let us move the air parcel adiabatically to a pressure of 100 kPa and call the temperature which it achieves thepotential temperature, . By using Eq. (4.8) in the initial and final states, it can be easily shownthat:

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≡T100

p

⎛

⎝ ⎜

⎞

⎠ ⎟

κ

(4.10)

Where p must be express in kPa (since we have used 100 kPa in the numerator). Because thepotential temperature of an air parcel is conserved under dry adiabatic processes, it may be usedas a tracer for air parcels.

Isentropic coordinates are ones in which potential temperature is used as the vertical coordinateinstead of height. In such a coordinate system, a parcel of dry air undergoing only adiabaticprocesses will always remain on the same coordinate surface. [NOTES: Why they are called isentropic coordinates will become clear when we discuss entropy and the Second Law of TD. We will also take into account the presence of water vapour. A similar concept isused in oceanography. These are isopycnal coordinates.]

The first law of thermodynamics may be expressed in yet a third version using potential temperature. Let us begin with the definition of potential temperature, Eq. (4.10), takelogarithms and then total differentials:

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d lnθ = d lnT −κd ln p (4.11)

Multiplying by cp and using the ideal gas law and the second version of the first law:

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c pd lnθ =δq

T(4.12)

This third version of the first law has several useful features. It is somewhat simpler than the previous two versions, it expresses the fact that the potential temperature is constant foradiabatic processes and varies for diabatic processes, and it seems to imply that the quantityon the right hand side is a function of state (since the potential temperature is a function ofstate). We will encounter q/T again soon in conjunction with the Second Law of Thermodynamics.

By considering Eq. (4.12) along with the first version of the first law for a cycle, and remembering that a function of state does not change in a cyclic process (e.g., ), it iseasy to show that

The integral on the right hand side is the net work done during the cycle. Hence this equationimplies that on a graph with axes cpln and T, the area enclosed by a cycle will equal thework done during the cycle. We will make use of this fact later when we consider theTephigram.

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du = 0∫

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c pTd lnθ∫ = pdv∫

THE SECOND LAW OF THERMODYNAMICS

• Thermodynamic surface for an ideal gas• Second Law• Entropy

THERMODYNAMIC SURFACE FOR AN IDEAL GAS

The ideal gas law is the equation of a surface in a three-dimensional space whosecoordinates are p, v, and T. This surface constitutes the ensemble of all states of an ideal gasthat are permitted by the ideal gas law. It is called the thermodynamic surface for that gas.Any changes in the gas’ state variables simply reflects movement on this surface. The followingdiagram illustrates this surface and on it are examples of isobaric, isothermal, isosteric, andadiabatic processes. Such processes may all be represented in the form:

(5.1)

Where n=0 is an isobar, n=1 is an isotherm, n= is an adiabat, and n is an isostere.

[NOTE: an isosteric process is one for which the specific volume remainsconstant. Such a process is also an isopycnic process because the density is constant. Processesfor which the total volume of the air parcel remains constant are called isochoric.]

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pv n = const

SECOND LAW

Sometimes the use of the first law will give rise to a seemingly impossible result. How do we know that it is impossible? Because it is entirely inconsistent with our collective experience. As an example, consider the cooling of bathwater. We will assume this to be an isobaric process, so that the first law is :

(5.2)

Let us imagine that 100 litres of water cool from 50oC to 20oC. If that heat is given to 10 kgof air in the bathroom (assuming the bathroom to be well insulated so that no heat is lost), whatwill be the final temperature of the air? If the warming of the air also occurs isobarically then the first law for the air is:

(5.3)

Equating the Q’s in Eqs. (5.2) and (5.3), we can solve for dTa using cp=1.0103 Jkg-1K-1

and cw=4.2103 Jkg-1K-1. The temperature change of the air is 1.3103 K!!!

We know this result is impossible, but it follows from the first law, so what is wrong? Clearly,in the final stages of the process as we have envisaged it, the air will be warmer than thebathwater. And yet we have implicitly assumed that heat will continue to flow from the waterinto the air to warm it. Experience tells us that heat simply does not flow spontaneously froma colder to a warmer body. This is called the Clausius formulation of the second law.

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q = cwdTw or δQ = mwcwdTw

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Q = mac pdTa

Rudolf Clausius (1822-1888) was a German physicist who established the foundations ofmodern thermodynamics in a seminal paper of 1850. He introduced the concepts of internalenergy and entropy (from the Greek for transformation). His great legacy to physics was theconcept of the irreversible increase of entropy (“Die Energie der Welt ist constant; dieEntropie strebt einen Maximum zu.” (1865)).

A more precise expression of the Clausius formulation is:

“It is impossible to construct a cyclicengine that will produce the sole effect of transferring

heat from a colder to a hotter reservoir.”

The term “sole effect” means that no work can be done. Of course heat pumps exist, so it ispossible to transfer heat from a colder to a warmer reservoir, but only by doing work.Oxford physicist P.W. Atkins in his book entitled “The Second Law” states it in yet another way (p. 9):

“…although the total quantity of energy must be conserved inany process….the distribution of that energy changes in an

irreversible manner.”

He also talks about a “fundamental tax,” stating that “Nature accepts the equivalence of heatand work, but demands a contribution whenever heat is converted into work.” (p. 21). Note,however, that there is no tax when work is converted into heat (for example, by friction).Since we may think of heat as disordered motion, and work as ordered motion, it would appearthat the second law also has something to say about a fundamental disymmetry between orderand disorder.

In order to formulate the second law mathematically, we will take another look at the bathtubproblem, invoking a cyclic engine to transfer the heat between the room air and the water in thetub, as in the following diagram.

We will focus our attention this time on the thermodynamics of the cyclic engine. Let ussuppose that TwTa . Let qa be the heat added to the engine by the room air, and let qw

be the heat added to the engine by the bath water. Our second law experience says that theformer will be positive and the latter negative (that is the air, being warmer, will give up heatto the engine and the bathtub water will gain heat from the engine). Moreover, the first lawrequires that the magnitudes of these heat transfers must be the same. Hence we may writeqa =q= -qw where we insist that the quantityq must be positive.

Let us now consider . Why we do this is not immediately obvious, although you shouldrecall from last lecture that q/T is a function of state. In this case:

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q

T∫

€

q

T∫ =

δqw

Tw

+δqa

Ta

= −δq

Tw

+δq

Ta

= δqTw − Ta

TwTa

⎛

⎝ ⎜

⎞

⎠ ⎟≤ 0

The inequality in Eq. (5.4) holds in general for real (irreversible) processes. The equalityin Eq. (5.4) will only hold as the two temperatures become equal. Heat transfer under such near-equilibrium conditions will be very slow and essentially reversible. If we defineentropy, s, as follows:

(5.5)

then, generally (dropping the line integral):

(5.6)

where the inequality holds for irreversible processes and the equality for reversibleprocesses. Clearly for adiabatic irreversible processes, the entropy must always increase,leading to the previously mentioned statement by Clausius. [NOTE: See also StephenHawking’s “A Brief History of Time” for a discussion of entropy and the arrow of time(Ch. 9).] For adiabatic reversible processes, entropy remains constant.

Formally, we can think of entropy as a function of state whose increase gives a measure of the energy of a system which has ceased to be available for work during a given process.

Note also from last lecture that ds=cpdln for reversible processes. Therefore, processes inwhich potential temperature is conserved are also isentropic processes.

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qrev

T≡ ds

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q

T≤ ds

Although it is important to be aware of the second law of thermodynamics, the field of atmospheric thermodynamics makes little use of it. We tend to assume that atmosphericprocesses occur reversibly.

However, since our atmosphere is in actuality an enormous heat engine that transports heat fromthe tropics to the poles, the concepts and ideas that arise from the second law are of some use.

SUMMARY OF REVERSIBLE AND IRREVERSIBLE PROCESSES:

From Zemansky and Dittman, “Macroscopic Physics”: “A reversible process is one that is performed in such a way that, at the conclusion of the process, both the system and the localsurroundings may be restored to their initial states, without producing any changes in therest of the universe.” For example, imagine a weight on a pulley. Suppose that the weight islowered by a small amount so that work is done and a transfer of heat takes place from thesystem to its surroundings. If the system can be restored to its original state, I.e., the weight islifted back up and the surroundings forced to part with the heat that they had gained duringthe lowering, then the original process is reversible.

A little thought should convince you that no real process is reversible. The abstraction of reversible processes, however, provides a clean theoretical foundation for the description ofreal world, irreversible processes.

What real world processes makes things irreversible?

Dissipative effects, such as viscosity, friction, inelasticity, electric resistance, and magnetichysteresis, etc.

Processes for which the conditions for mechanical, thermal, or chemical equilibrium, i.e., thermodynamic equilibrium are not satisfied.

In atmospheric and oceanic sciences, it is common to assume that there are no dissipative effects (particularly for flows at large spatial scales) and that motion of the atmosphere andoceans is isentropic. Viscous and frictional dissipation do, of course, occur (particularly inthe atmospheric and oceanic boundary layers) but a good understanding of the dynamics canbe obtained from inviscid theory.

APPLICATIONS OF THE SECOND LAW

• Carnot cycle• Applications of the second law• Combined first and second law• Helmholtz and Gibbs free energies

CARNOT CYCLE

The Carnot cycle is an ideal heat engine cycle that offers insights into other heat engines,including the atmosphere. It is a reversible cycle, performed by an ideal gas, that consists oftwo isothermal processes linked by two adiabatic processes, as illustrated below:

1

2

3

4

A

B

CD

Nicolas Leonard Sadi Carnot (1796-1832) led a short but interesting life. He was able topursue his scientific research as a result of his appointment to the army general staff. Hedied of scarlet fever followed by cholera, and all his papers were burned after his death.Only three major scientific works survive. His work on heat engines led to the ideal heatengine and cycle that now bear his name. Interestingly, he was able to make significantscientific advances while still believing in the erroneous caloric theory of heat, in whichheat is taken to be a function of state.

For each of the four processes in the Carnot cycle, we will consider the change in internalenergy, the heat added to the working gas and the work done by the gas (i.e., the componentsof the first law).

Isothermal expansion at temperature T1

du1=0 w1=q1

Adiabatic expansion between T1 and T3

du2=cv(T3-T1) q2=0 w2=-u2 ds2=0

Isothermal compression at T3

du3=0 w3=q3

Adiabatic compression between T3 and T1

du4=cv(T1-T3) q4=0 w4=-u4 ds4=0

Entire Carnot cycle du3=0 q=q1+q3 w= q1+ q3 ds=0

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q1 = pdvA

B

∫ = RT1 lnvB

vA

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q3 = pdvC

D

∫ = −RT3 lnvC

vD

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ds1 =δq1

T1

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ds3 =δq3

T3

Table 1: Thermodynamics of the Carnot cycleNotes: 1) q3 is negative because we consider q to be the heat added to the system. 2) The net work done in a Carnot cycle must equal the difference between the heat input and the heat exhausted. This work is also equal to the area contained within the Carnot cycle on a p-v diagram.3) In order to show that ds=0, we must show that vB/vA=vC/vD. This can be done by comparing the Poisson equations for the two adiabaticprocesses.

The efficiency of the Carnot cycle is defined to be the ratio of the mechanical work done to theheat absorbed from the hot reservoir. That is:

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E =δw

δq=

T1 − T3

T1

=1−T3

T1

(6.1)

It can be shown that this is the maximum efficiency of any cyclic engine working between thesame two temperatures. This is known as Carnot’s theorem. Note that the efficiency of theCarnot cycle can only be unity (i.e., 100%) if the cold reservoir (T3) has a temperature ofabsolute zero. In order to increase the efficiency of an engine, one would like to minimizeT3/T1.

EQUIVALENCE OF HEAT AND WORK, EFFICIENCY, AND THE FIRST LAW

Recall the first law: du=q- w. If we consider an engine cycle in a p-v diagram then the pathtaken (clockwise for a heat engine, counterclockwise for a refrigerator) may be expressed asthe line integral of the first law around the cycle:

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du = δq − δw∫∫∫Furthermore, since du is an exact differential its line integral vanishes, leaving us with:

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q = δw∫∫ (6.2)

i.e. there is an equivalence of net heat gained/lost and net work done by/on the system.

Efficiency, E, may also be considered to be the ratio of Work Out to Heat In.

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E =Work Out

Heat In

Consider the following idealized engine cycle:

The net heat input is QH=Q12+Q23. The net heat output is QC=Q34+Q41. The net heat in istherefore Q=QH-QC and we now know that this must be the net work done, i.e., W= QH-QC .The efficiency may then be expressed in terms of the heat as:

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E =QH − QC

QH

=1−QC

QH

Important Note: QH and QC here are MAGNITUDES of heat, and are therefore always positive.

Example: Suppose that the working fluid is an ideal gas.

From the first law we have that dU=Q- W. The incremental work done is related to pressureand volume by W=pdV. Also, the change in internal energy dU along any branch of the cycle isequal to dU=CvdT=(Cp-nR)dT. NOTE: we are not using molar values here, so Cp-Cv=nR.

Branch 12: Volume constant

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Q = dU = Cv (T2 − T1)1

2

∫1

2

∫ > 0 So heat is absorbed.

Branch 23: Pressure constantSince we are dealing with an ideal gas for which we can express the change involume in terms of the change in temperature as:

So the first law along this branch gives us: q=du+pdV=(cp-nR)dT+nRdT=cpdT from which we get that:

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pV = nRT

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dV =nR

pdT

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Q = Cp dT2

3

∫2

3

∫ = Cp (T3 − T2) > 0 So, again, heat is absorbed.

Branch 34: Volume constant

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Q = dU = Cv (T4 − T3)3

4

∫3

4

∫ < 0 So heat is rejected.

Branch 41: Pressure constant

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Q = Cp dT4

1

∫4

1

∫ = Cp (T4 − T1) < 0 So, again, heat is rejected.

The efficiency of this engine, E, is therefore:

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E =1−QC

QH

=1−Cv (T3 − T4 ) + Cp (T4 − T1)

Cp (T3 − T2) + Cv (T2 − T1)

Note that since we want the MAGNITUDE of QC, the signs in the numerator are such that itis a positive quantity.

What would an efficiency of 100% mean in this case? Well, we can substitute in T i=piVi/nRfor i=14 and show that in order for E to equal 1 we must have that:

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Cv (T3 − T4 ) + Cp (T4 − T1) = Cv ( p3V3 − p4V4 ) + Cp (p4V4 − p1V1) = 0

where we’ve cancelled out the nR’s. Now note that V3=V4, V1=V2, p1=p4, and p2=p3 to show

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V3Cv ( p3 − p4 ) + Cp p4 (V4 −V1) = 0

which can only be true if p3p4 and V4V1. In other words, 100% efficiency is only possibleif the cycle collapses to a single point on the p-V diagram, and no work is done!

Whoever said that doing no work is inefficient???????

In class example: Efficiency of the gasoline engine.

The atmosphere as a heat engine:

Our atmosphere is heated in the tropics (say Ttropics=313K) and cools at the poles (sayTpoles=233K). If we were to imagine our atmosphere as an enormous heat engine that convertsnet incoming heat into work (kinetic energy) then the efficiency of this engine is only 26%. In fact, the actual efficiency is much lower than this.

Sandström’s theorem (see Houghton, “The Physics of Atmospheres”) states that a steadycirculation in the atmosphere can be maintained only if the heat source is situated at a higherpressure than the heat sink. Some reflection on the Carnot cycle will reveal that the theoremmust be valid. Because the atmosphere of the Earth is transparent to solar radiation, therequirements of the theorem are satisfied for Earth. What about Venus?

APPLICATIONS OF THE SECOND LAW

• Impossible processes• Kelvin formulation of the Second Law• Combined First and Second Laws• Helmholtz and Gibbs Free Energies

1. Impossible Processes

Impossible processes are ones that violate the second law of thermodynamics, i.e., ones forwhich:

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ds <δq

T

(Recall that for adiabatic reversible processes ds=0. These processes are therefore calledisentropic and have many applications to the atmosphere and oceans. Adiabatic irreversibleprocesses have ds>0, since adiabatic processes have q=0.)

2. Kelvin Formulation of the Second Law

This may be postulated as follows: “It is impossible to construct a cyclic device that will producework and no other effect than the extraction of heat from a single heat source.” This result can bereadily deduced by considering an isothermal cyclic engine in contact with a single reservoir,and examining the consequences of the second law (which will demonstrate that, in fact, heatmust be lost from the engine to the reservoir, and hence that work must be done on the engine).

If it were not for the second law, as expressed by Kelvin, our energy worries would vanish.We could simply extract as much work as we needed from a very large heat source such as theoceans or the interior of the Earth.

3. Combined First and Second Laws

If we combine the first law in the form:

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du = δq − pdv or dh = δq + vdp (6.2)

with the second law in the form:

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Tds ≥ δq (6.3)

then the combined laws take the form:

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du ≤ Tds − pdv or dh ≤ Tds + vdp (6.4)

4. Helmholtz and Gibbs Free Energies

The Helmholtz and Gibbs free energies are functions of state. i.e., they are perfect differentials.They are defined, respectively, by:

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f ≡ u − Ts

g ≡ h − Ts (6.5)

Using them, the combined first and second laws (Eq. 6.4) may be written as:

where, again, the equality hold for reversible processes and the inequality holds forirreversible processes.

The Helmholtz free energy is particularly useful when considering chemical reactions thatoccur isothermally. You’ll note that the change of the Helmholtz energy during a reversible,isothermal process equals the work done on the system, and for a reversible, isothermal, andisochoric process the Helmholtz energy is conserved.

The Gibbs free energy is particularly useful when considering phase changes, which are isothermal and isobaric (and can be conceived of as occuring reversibly). Hence during suchphase changes the Gibbs free energy remains constant. The Gibbs energy can be used touniquely determine the triple point temperature and pressure of a substance, since g must bethe same for the solid, liquid, and vapour phases (2 equations, 2 unknowns (T,p)).For irreversible processes, the Gibbs free energy decreases as the substance approaches itsnew equilibrium. It is therefore clear that an equilibrium state is one at which the Gibbs energy is at a minimum.

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df ≤ −sdT − pdv

dg ≤ −sdT + vdp(6.6)

Hermann von Helmholtz (1821-1894) had become the patriarch ofGerman science by 1885. He began his career as an army doctor. As anMD and professor of physiology, he published a massive volume of workon physiological optics and acoustics. He also invented the ophthalmoscope. At the same time he began to publish work unrelatedto physiology--a paper on the hydrodynamics of vortex motion in 1858,and an analysis of the motion of violin strings in 1859, for example.In 1871, Helmholtz turned his attention full-time to physics, and madesignificant contributions in areas ranging from electrodynamics tothermodynamics and meteorology.

Josiah Willard Gibbs (1839-1903) was a Yale graduate and professor,who turned his attention to thermodynamics in the early 1870’s. Hecontributed to the use of geometrical methods and thermodynamicdiagrams. Perhaps his most significant contribution was to the understanding of thermodynamic equilibrium, which he viewed as anatural generalization of mechanical equilibrium, both being characterized by minimum energy. He also made a substantialcontribution to statistical mechanics.