equations of state - ntnu

CHAPTER 6

Equations of state

163 Workouts in Thermodynamics, © 2001 – 2013 Tore Haug-Warberg

Although thermodynamics builds on some basic, fundamental principlesof physical science, which form part of a well-defined mathematical theory, itis essentially an empirical science. This is particularly true of the many semi-empirical models that are used in engineering research. That is how it was inthe past, and that is how it will continue to be in the future, because with theexception of a handful of simplifications used in statistical mechanics1 there areno precise models, and there will always be the need for essentially physicalmodels with adjustable parameters.

A thermodynamic state description needs one or more functions of statethat define the asymptotes of the system (mixture) in a consistent manner. Ap-plying parameters to the functions gives engineers the flexibility they need,while it also ensures thermodynamic consistency. In this chapter we learnabout a variety of useful concepts, such as ideal gases, virial development,the Einstein–Debye phonon model and van der Waals theory, but first of allwe need to define the concept of an equation of state. Let us start with anycanonical differential, e.g. of internal energy

(6.1) dU = T dS − p dV +n∑

i=1µi dNi ,

defined using the canonical variables S ,V,Ni. If we know the fundamentalstate functions referred to as the equations of state of the system

T = T (S ,V, n) ,(6.2)p = p(S ,V, n) ,(6.3)µi = µi(S ,V, n) ,(6.4)

then Eq. 6.1 can be integrated using Euler’s 1st theorem to give U = TS −pV +

∑ni=1 µiNi, as shown in Paragraph 30 in Chapter 4. However, in practice

1 Above all crystalline phases, ideal gases and dilute electrolytes.

63

64 6. EQUATIONS OF STATE

this doesn’t work, because the relationships defined by 6.2–6.4 only exist asimplicit functions1. It is therefore more normal to start with

(6.5) (dG)T,p =n∑

i=1µi dNi

or alternatively

(6.6) (dA)T = −p dV +n∑

i=1µi dNi ,

which can be integrated in the same way if the state function(s) µi(T, p, n) orp(T,V, n) and µi(T,V, n) are known,2 but it is still difficult to accurately de-scribe the chemical potential of the compounds in a multi-component mixture.The exceptions to this are ideal (gas) mixtures for which we can obtain exactsolutions using statistical mechanics or, as we shall do in a moment, from in-tegrating the Gibbs–Duhem’s equation using just one additional assumption.By taking the ideal gas model as a reference, it is relatively easy to obtainthe residual functions Gr,p(T, p, n) =

∫(V(T, p, n) − V ıg) dp or Ar,v(T,V, n) =∫

(p(T,V, n) − pıg) dV, which describe the difference between a real mixtureand an equivalent ideal gas under the same physical conditions. If we combinethis with an expression for the energy of the reference gas, we get the Gibbsor Helmholtz energy of the mixture, and a full set of state functions can thenbe derived using differentiation. This leaves us with just V(T, p, n) or p(T,V,n) as the state functions.

The alternative to the above approach is to derive the thermodynamic func-tions in non-canonical coordinates:

U(T,V, n) rather than U(S ,V, n)H(T,V, n) rather than H(S , p, n)...

S (T, p, n) rather than S (H, p, n)

This is particularly useful if you only need to obtain U, H or S , but not Aor G. Typically, this is the case for physical state changes that occur at aconstant composition like for instance fluid flow in pipes, compressors andvalves. For systems of variable composition, where the chemical potential isimportant, it is more suitable to use A and G because µi = (∂A/∂Ni)T,V,Nj!i

=

(∂G/∂Ni)T,p,Nj!ican be obtained later using differentiation with respect to the

1 Only very rarely can entropy be written as an explicit (independent) variable in the statefunctions. 2 Many textbooks refer to p(T,V,n) as if there were no other state functions, butthat is wrong!

1. IDEAL GAS LAW 65

canonical variables. If we choose to disregard canonical theory for a mo-ment, then the chemical potential must be defined as µi = (∂H/∂Ni)T,p,Nj!i

−T (∂S/∂Ni)T,p,Nj!i

= hi − T si or a similar expression based on U, but this ap-proach normally involves more work.

1. Ideal gas law

The term ideal (perfect) gas refers to ideal, and from a physical point ofview hypothetical, descriptions of the conditions in a real gas. An ideal gas con-sists of individual particles (atoms, molecules or radicals) that move freely andindependently in space, without any kind of interaction. This is the ideal situa-tion because it involves ignoring the inter-molecular forces that exist betweenthe particles in a real gas. This includes all of the interactions caused by in-duced dipoles (van der Waals forces), dipole-dipole forces, dipole-quadrupoleforces, etc. In spite of this massive simplification, the total energy of the gasis not equal to zero, as the particles have their kinetic energy (translation), mo-ment of inertia (rotation), internal degrees if freedom (vibration, internal rota-tion, torsion, electronic excitation) and chemical binding energy. Furthermore,electron and nuclear spin can also play a role.

Ideal gases are only valid as a special case of the true thermodynamic statewhen pressure approaches zero, and can never be verified directly in a real(physical) system. Nevertheless, we must be allowed to consider the idealgas law one of the great scientific discoveries. Perhaps it is just as importantas Maxwell’s equations, which describe electromagnetism, although that is avery subjective judgement, as Maxwell’s equations are a complex set of fourlinked partial differential equations, whereas the ideal gas law is just a simplealgebraic equation. What they have in common, then, is not their mathematicalformulae, but rather their understanding of the underlying physics and of thepractical and theoretical implications.

1.1. Helmholtz potential.It is not a new idea that the world is composed of more or less distinct

material points, but the experiments that underpin the ideal gas law as we knowit today are more recent:

Name Observation YearBoyle ( PV )T,N = c1 1662Charles (V/T )p,N = c2 1787Dalton ( p/N )T,V = c3 1801Gay-Lussac (P/T )V,N = c4 1809Avogadro (V/N )T,p = c5 1811

Given what we know today, the above table accurately represents the scientificprogress, but at the time the scientists lacked our historical perspective, and we


should be cautious about attributing entirely rational motivations for their ob-servations. It was only in 1834 that Émile Clapeyron successfully formulatedthe ideal gas law in the simple form:

(6.7) pıg = NRTV .

The question we are going to look at is to what extent, and if so how, this lawcan be used to derive expressions for the Helmholtz energy of the gas. We cansee that the unit of the product p · V is the same as for A, which means thatthere may be a deeper relationship between pressure–volume and the energyfunctions.

The argument that follows is a great example of how physical and mathe-matical analysis can be successfully combined with experimental results (thanksto the insight of the old scientists, not the author). We will admittedly need tomake some choices along the way, which mean that the results are not entirelyconclusive, but this must be seen in the context of established practice for mea-suring and reporting basic thermodynamic functions.

The quickest way to obtain a useful result is to apply the Gibbs–DuhemEq. 4.23 described in Chapter 4. For a thermodynamic system with only onechemical component, the folowing applies:

s dτ + v dπ + dµ = 0 .

This differential has no general solution, as s and v are quite arbitrary functionsof the material state, but for a change of state that is assumed to be isothermal,then dτ = 0 and vıg = V/N = RT/p = −RT/π, giving us

(dµıg)T =RTπ dπ ,

or:

µıg(T, p) − µ(T ) = RTp∫

p

dππ = RT ln

(pp

).

It is also important to remember that thermodynamic energy functionsare first-order homogeneous functions to which Euler’s theorem applies. TheHelmholtz energy of a single-component system can therefore be written as theintegral A = −pV + µN. Substituting in pıg and µıg from the above equationsgives us:

Aıg = −NRT + Nµ(T ) + NRT ln(NRT

pV

).

We have thus derived the Helmholtz energy from the ideal gas law, using thefact that Euler’s theorem applies to the extensive properties of the gas, and

1. IDEAL GAS LAW 67

the Gibbs–Duhem equation, which is itself a direct result of Euler’s theorem.From Chapter 3 on Legendre transforms we know the general equations S =− (∂A/∂T )V,N and U = A + TS , which means that it also follows that:

U ıg = N

[µ − T

(∂µ∂T

)]− NRT = Nh − NRT .

In other words, the internal energy of the gas depends on the system’s sizeN and temperature T through the constant of integration1called µ. The lat-ter constant must be determined experimentally in each individual case, ortheoretically–experimentally if there are good simulation models for the chem-ical compound (based on spectrometric measurements). Our provisional con-clusion is therefore that we can calculate the Helmholtz energy of an ideal gaswith only one chemical component and its internal energy. From this the otherenergy functions can be obtained in a similar way using Legendre transforms,apart from one temperature dependent constant.

Our next challenge is to generalise this result into one that is similarly validfor a multi-component system. In order to do that, we must write the ideal gaslaw in its extended form

(6.8) pıg =∑i

NiRTV

so that we can identify the components that constitute the mixture. This is afunction of state in the form p(T,V, n), which gives us a good way to approachthe question of Helmholtz energy A(T,V, n), but in order to determine A usingthe Euler method of integration as shown in Chapter 4, we also need to knowthe equation of state µıg

i (T,V, n). This function cannot be obtained experimen-tally in the same way as pıg, and must therefore be derived theoretically. Oneway to do this is to apply the Gibbs–Duhem equation again:

S dτ + V dπ +∑i

Ni dµi = 0 .

The Gibbs–Duhem equation from Chapter 4 follows directly from the fact thatU is a homogeneous function to which Euler’s theorem applies when expressedusing the variables S , V and n. This means that there is an interdependencebetween the differentials of U which can be usefully exploited, and using theequation of state π = −pıg and the Gibbs–Duhem equation we will obtain a

1 The internal energy U ıg is inevitably independent of both pressure and volume: ideal gases area collection of free material points without any form of interaction. It is only the internal degreesof freedom of the particles that contribute to their energy, and these are also what determine theintegration constant µ.


possible solution for µi. Let us substitute for π = −pıg while keeping τ = Tconstant as shown below:

−V∑i

NiRT−V2 dV − V

∑i

RTV dNi +

∑i

Ni(dµi)ıgT= 0 ,

or∑i

(RT dVV − RT dNi

Ni+ (dµi)T ) = 0 .

The simplest solution for an expression with a sum equal to zero is to assumethat each individual term is equal to zero:

(dµi)ıgT

RT = − dVV +

dNi

Ni; ∀i ∈ [1, n] .

Thermodynamics is a phenomenological science, and we cannot be sure thatthe above expression correctly describes the gas mixture, but at least it is apossible solution. However, the result agrees with experimental observationsof dilute gases, which in the limit case p → 0 behave like a chaotic collectionof particles without any interaction, and where the macroscopic properties as-sociated with each particle are independent of the other particles. Interestingly,deriving the chemical potiential of a component of an ideal gas using statisticalmechanics gives us exactly the same answer1. This is a convincing demon-stration of the great power of formal thermodynamic analysis. Integrating thedifferential gives us

1RT (µıg

i (T,V, n) − µi (T )) = − ln VV+ ln Ni

Ni.

Note that the constant of integration µi , which will hereafter be called thestandard state of component i, is a function of temperature only. This meansthat for any given temperature T we must calculate or estimate a new valuefor µi . If the function of state were available in the form p(S ,V, n), then µiwould become a genuine constant of integration. In practice there are very fewsystems that can be described in this way, but radiation from black bodies, asdescribed in Paragraph 45, is one example.

The normal convention is to assume a pure-component standard state whereV/N

i = RT/p and where p = 1 bar, or, in older reference works, where p =

1 atm. The standard state µi should in that case be interpreted as representingµi (T, p), or more properly µıg

i(T, p,N1 = 0, . . . ,Ni = 1 mol, . . . ,Nn = 0).

1 Statistical mechanical theory is based on a microscopic description of the properties of parti-cles, and involves calculating their average values based on a relatively small number of axioms.Whereas at the macroscopic level we have to accept that our result is merely a possible result,the theoretician can assert that it is a correct result for a gas consisting of hypothetical particleswithout any physical extension or mutual interaction. In other words, a perfect gas.

1. IDEAL GAS LAW 69

Bearing in mind this fundamental understanding of what µi means, the chemi-cal potential of the component can be expressed as

(6.9) µıgi (T,V, n) = µi (T, p) + RT ln

(NiRTpV

).

Here we should particularly note that (dµıgi

)T,V,n = 0 must be true for all valuesof p. This is of course the case by definition, since T , V and n completelydetermine the thermodynamic state, but at the same time p acts as a parameterin the expression for µıg

i. It is therefore slightly unclear what happens if p

changes. The passive role of the standard pressure becomes more obvious ifwe partially differentiate µıg

iwith respect to p. This gives us (∂µi /∂p)T

−RT/p = 0, or put another way: if we choose to change p then we must alsocorrect µi to ensure that the total impact on µıg

iis zero.

It is important for our own self-esteem to acknowledge that in order toaccurately determine the standard state one needs information at a high levelincluding precise calorimetric measurements, advanced spectroscopy and quan-tum physics, or a combination of these things, and that these activities lie out-side our area of responsibility. We must therefore adapt our way of working tothe available thermodynamic databases, and not vice-versa, which means that itis important to use the data properly. Essentially, standard states are presentedin three different ways in the literature:

µi (T, p) = µi (T, p)(6.10)

µi (T, p) = ∆fhi (T ) − T si (T, p)(6.11)

µi (T, p) = ∆fhi (T) + (hi (T ) − hi (T)) − T si (T, p) .(6.12)

These equations follow from the definition G = U + pV − TS = H − TSfrom Chapter 3 applied to the standard state as defined in the above paragraph.Moreover, it is an experimental fact that all molecules, with the exception ofmonoatomic ones, have internal energy due to the bonding within the molecule.Normally this energy is reported as the enthalpy of formation ∆fh

i (T) of the

compound. All of the formation properties (enthalpy, volume, entropy, etc.) ofany given compound AaBb . . . Zz in an ideal gas state are defined analogouslyby

∆fhAaBb...Zz

= hıgAaBb...Zz

− ahαA − bhβB − . . . − zh

ζZ ,

where α, β and ζ indicate the phases (configurations) of the various elements.The phase can be liquid or solid, in which case the crystalline configurationis given, or a hypothetical ideal gas for gaseous elements. The chemical for-mula AaBb . . . Zz should here be seen as the elemental composition, i.e. not the


atomic composition, of the molecule. For example, hydrogen bromide wouldhave the formula (H2)0.5(Br2)0.5, giving the enthalpy of formation

∆fhHBr = h

ıgHBr − 0.5h

ıgH2− 0.5h

lıqBr2.

The reason for this discrepancy, with respect to the established practice forchemical equilibrium reactions, is that according to the IUPAC convention theenthalpy of a chemical element is defined as zero for the stable configuration

of the element1 at T = 298.15 K and p = 1 bar. Hence, in the above equationh

ıgH2= h

lıqBr2= 0 and the enthalpy of formation ∆fh

HBr ≡ h

ıgHBr is given relative to

this arbitrary reference. In particular, note that ∆fhBr2! 0, as the enthalpy ref-

erence by definition relates to bromine in the liquid phase and not for brominein the (ideal) gas phase.

This is the main principle that determines how the thermodynamic standardstate is reported, but there are, of course, exceptions to the rule: at least oneimportant database (JANAF) uses the more specific definition hi (T ) = 0 for theelements at all temperatures T , while another database perhaps uses T = 0 Kas its reference and yet another one may use gi (T, p) instead of si (T, p).There are therefore many ways of reporting the standard state, but for all ofthem it is true that

hi (T ) − hi (T) =T∫

T

cp,i(T ) dT ,(6.13)

si (T ) − si (T) =T∫

T

cp,i(T ) dln T .(6.14)

One of these equations (it doesn’t matter which) can be viewed as a definitionof c

p, while the other one follows from the total differential dh = T ds + v dp +∑µi dNi, which for constant pressure and composition gives us (dh)p,n = T ds.

The sections 2.1–2.4 explain a little bit about the fundamental physicalprinciples that, in their different ways, contribute to cv,i. In practice, all thermo-dynamic reference works are based on a large number of carefully recordedmeasurements supported by one or more theoretical calculations, which to-gether produce figures for ∆fh

i (T), si (T, p) and cp,i(T ). The enthalpy of

formation is generally measured using calorimetry, whereas the heat capacityof an ideal gas component essentially adheres to the superposition principle 2

cp,i = ctransv,i + crot

v,i + cvıbv,i + celec

v,i + R

1 Such as H2(g), Ar(g), S(rhombıc), Br2(lıq), Al(s), etc. 2 The gas constant R in the aboveequation is a result of the defintiion of H = U + pV , which for an ideal gas gives usHıg = U ıg + (pV)ıg = U ıg + NRT . If we then differentiate with respect to temperature weget cp,i = cv,i + R.

1. IDEAL GAS LAW 71

and can be calculated from spectrometric measurements. The JANAF tables1

are a good example of how far this development has gone. For relatively smallmolecules (fewer than 8–10 atoms) it is now possible to estimate c

p,i moreprecisely than using high-quality calorimetric measurements.

It is easy to lose one’s way in the details of these calculations. It is thereforeworth reiterating that using Euler’s theorem to integrate euqations 6.8 and 6.9always works, and that the Helmholtz energy of the gas can be expressed inthe following general form:

Aıg(T,V, n) = πıgV +∑i

Niµıgi

= −∑i

NiRT +∑i

Niµi (T, p)

︸!!!!!!!!!!!︷︷!!!!!!!!!!!︸G(T,p)

+∑i

NiRT ln(

NiRTpV

).

Note that, in this expression, the standard state term is a conventional Gibbsenergy function, because T and p have been selected as the state variables ofµi , rather than T and V. In some strange manner, something has gone wronghere, but what? It is clear that we no longer have a canonical description ofA(T,V, n), and that the unfortunate decision to combine T , V and n from theoriginal set of variables of pıg, with p from the standard state of µıg

i, is creating

unintended confusion.The consequences become even clearer if we take Aıg as our starting point

for differentiating U ıg = (∂(Aıg/T )/∂(1/T ))V,n or −S ıg = (∂Aıg/∂T )V,n. Thepartial derivative of Aıg is in both cases defined at constant volume, but howis it possible to implement this for µi (T, p)? Let us use U ıg as an example.We start with

Aıg

T= −

∑i

NiR +∑i

Niµ

i(T,p)T+

∑i

NiR ln( NiRTpV

)

followed by

d( Aıg

T )V,n =

∑i

Ni

(∂(µi /T )∂(1/T )

)

p

d( 1T ) +

(∂(µi /T )∂p

)

T

dp

−∑i

NiR[T d( 1

T ) + 1p

dp].

Finally we use the Gibbs–Helmholtz equations from Paragraph 28 on page 39together with (∂(µi /T )/∂p)T

= R/p from the equivalent discussion on page 69.The differential can now be expressed in the form

U ıg(T, n) =(d(Aıg/T )d(1/T )

)

V,n=

∑i

Nihi (T ) −

∑i

NiRT = Hıg − pıgV ,

1 JANAF thermochemical tables. 3rd edition. Part I,II. J. Phys. Chem. Ref. Data, Suppl., 14(1),1985.


which shows that U ıg is only a function of T and n because p and V cancelout during the derivation. The total derivative now has one degree of freedom,and therefore takes the same value as the partial derivative (∂(Aıg/T )/∂(1/T ))throughout the domain of definition. But why is it necessary to go via the totaldifferential? Well, because otherwise we risk writing (∂(µi /T )/∂(1/T ))

V,n =

ui . Firstly µi is a function of V and not of V , and secondly V varies with Twhen p has been chosen as the standard state. It is easy to make a mistakein those circumstances, and it is therefore safest to go via the total differential.The derivation also illustrates how risky it is to introduce non-canonical vari-ables for the thermodynamic potentials. In summary, the following equationsapply to ideal gas mixtures:

Aıg(T,V, n) =∑i

Niµi (T, p) +

∑i

NiRT[ln

(NiRTpV

)− 1

](6.15)

−S ıg(T,V, n) =∑i

Nisi (T, p) +

∑i

NiR ln(

NiRTpV

)(6.16)

U ıg(T, n) = Hıg(T, n) − NRT(6.17)

Hıg(T, n) =∑i

Nihi (T ) .(6.18)

1.2. Gibbs potential.Historicallly speaking, T, p, n has long been the favoured set of variables

for modelling all chemical systems, probably because temperature and pressureare particularly easy to control in a laboratory. In our case, it is possible tocalculate Gibbs energy by using the Euler method to integrate the chemicalpotential in Eq. 6.9, and substituting in the ideal gas law from Eq. 6.7

(6.19) µıgi

(T, p, n) = µi (T, p) + RT ln(

Ni pN p

),

which gives us:

(6.20) Gıg(T, p, n) =n∑

i=1µıg

iNi =

n∑i=1µi Ni + RT

n∑i=1

Ni ln(

Ni

N

)+ NRT ln

(pp

).

Alternatively, we can calculate the Gibbs energy as a Legendre transform ofHelmholtz energy (or vice–versa). This is undoubtedly possible, but since Gıg

and Aıg have different canonical variables, we also need to invert p(T,V, n) toV(T, p, n). If this is impossible, the result of the transformation will be an im-

plicit function1. For most equations of state, with the exception of ideal gases,the 2nd virial equation and certain other models, an explicit transformation

1 The specification V(p) has 3 solutions in the steam/liquid phases, and is not a function in theordinary sense of the word. On the other hand, p(V) is an unambiguous specification, whichmeans that there is no problem with doing a Legendre transform of Helmholtz energy to Gibbsenergy.

1. IDEAL GAS LAW 73

from T, p, n to T,V, n is impossible. This significantly complicates the practi-cal application of Legendre transforms.

1.3. Grand canonical potential.The ideal gas law is a simple model, which allows us to choose any system

variables that we like. For example, we can rewrite µ(T, p, n) or µ(T,V, n) asexpressions for n(T,V,µ). Very few models allow us to do this, and so wewill make use of the opportunity now that we have it. An alternative way ofexpressing Eq. 6.19 is:

(6.21) Nıgi

(T,V,µ) = pVRT

exp(µi−µi

RT

).

Substituted into the ideal gas law, Eq. 6.21 gives us an adequate equation ofstate for the grand canonical potential Ω(T,V,µ):

(6.22) pıg(T,µ) = RTV

n∑i=1

Nıgi = p

n∑i=1

exp(µi−µi

RT

).

The total differential of Ω for a one-component system was derived in Chap-ter 3, see Paragraph 24 on page 31. For an n-component system, we can statethat

(6.23) (dΩ)T,µ = −p dV .

The differential (dΩ)T,µ can be integrated directly, as long as the function ofstate p(T,V,µ) is expressed explicitly. Eq. 6.22 is explicit, and if we integrateusing Euler’s method at a given T and µ we obtain

(6.24) Ωıg(T,V,µ) = −pıgV = −pVn∑

i=1exp

(µi−µiRT

).

It may at first sight appear that p is a free variable in the above equations, butif we change p, then µi (T, p) varies accordingly, always keeping the value ofΩıg unchanged (show this). Also note that pıg(T, µi) is a function of only twovariables in Eq. 6.22, whereas Ωıg(T,V, µi) in Eq. 6.24 requires three variables.The intensive state is evidently given when n + 1 intensive variables have beendetermined, while the extensive state requires n+ 2 variables. This is a generalresult, which is not limited to ideal gases1. From Paragraph 35 in Chapter 4 we

1 In contrast to e.g. U ıg = U(T,n), which only applies to ideal gases.


know that the partial differentials of functions with only one extensive variablefulfil the requirements for homogeneity1:

∂T∂S

p,µ1,...,µn

=∂p∂V

T,µ1,...,µn

=∂µ1∂N1

T,p,µ2,...,µn

= . . . =∂µn

∂Nn

T,p,µ1,...,µn−1

= 0 .

The intensive variables are clearly independent of the size of the system, andthe state is therefore given when n+1 intensive variables have been determined,cf. Paragraph 39 on page 54. However, in order to determine the size of thesystem, we also need an extensive variable. In other words, a thermodynamicsystem requires n + 2 state variables, in spite of the fact that the intensive statecan be described with only n + 1 variables.

§ 40 Derive the partial derivatives of Ωıg(T,V, µ j) = −N ıgRT with respectto T , V and µi. Show that the results can be interpreted as −S ıg, −pıg and−N

ıgj

, without knowing the differentiation rules of the Legendre transform be-forehand.

Partial differentiation of Ω(T,V,µ). It’s a good idea to start by establishingsome shadow values before starting on the differentiation process itself. Insteadof differentiating pıgV in Eq. 6.24 we will use the fact that −Ωıg = pıgV =∑

i Nıgi RT = N ıgRT . If we differentiate N

ıgi (T,V,µ) in Eq. 6.21 we get:

(∂Nıgi

∂T

)V,µ

= Nıgi

RT si−R(µi−µi )(RT )2 − N

ıgi

T =N

ıgi

RT

(sıgi− R

),

(∂Nıgi

∂V

)T,µ

=N

ıgi

V ,

(∂Nıgi

∂µ j

)T,V,µk! j

=δi j N

ıgi

RT .

Here we have made use of the relationship −si = − (∂µi /∂T )p,n, which also

means that −si = − (∂µi /∂T )V,µ

. Think about this result for a moment. Apartfrom this fundamental observation, the simplification of RT si −R(µi−µi ) in thefirst partial differential equation relies on an understanding of certain principlesof thermodynamics. In the general case, µi = hi − T si. In the special case of

1 Zero is zero, most people probably think, but the partial differentials have different units inthis case, and the similarity only applies to the measure (the absolute value) of these quantities.

2. INTRA-MOLECULAR DEGREES OF FREEDOM 75

ideal gases, hıgi= hi , and hence T si − (µi − µi ) = T s

ıgi

. We are now ready toderive the partial differentials of Ωıg

(∂Ωıg

∂T

)V,µ

= −RTn∑

i=1

(∂Nıgi

∂T

)V,µ− NR = −

n∑i=1

Ni(sıgi− R) − NR = −S ıg ,

(∂Ωıg

∂V

)T,µ

= −RTn∑

i=1

(∂Nıgi

∂V

)T,µ

= −RTn∑

i=1

Nıgi

V= −pıg ,

(∂Ωıg

∂µ j

)T,V,µk! j

= −RTn∑

i=1

(∂Nıgi

∂µ j

)T,V,µk! j

= −n∑

i=1N

ıgi δi j = −N

ıgj

2. Intra-molecular degrees of freedom

The ideal gas law was established in several stages over the course of the17th-19th centuries, as an experimental law for the so-called permanent gases(air, N2, H2, etc.). We now know that there is a direct link between the macro-scopic characteristics of the gas and the microscopic properties of the individ-ual molecules. The related discipline is called statistical mechanics or statis-tical thermodynamics. It would be far too big a job to discuss this theory indetail, but we will take the time to look at some of the details that underpin ourunderstanding of ideal gases. Without further introduction, let us assume thatthe relationship between the particles’ microstates and Helmholtz energy is

ANRT= − ln Q ,

where Q is the total partition function of the system1,2. For a system with nparticles in defined locations, Q = qn, but if the particles are not in a fixedlocation (as in a gas), then Q = 1

n! qn. The molecular partition function q

approximates to the product of q ≈ qtransqvıbqrot · · · over the (assumed) numberof independent terms.

In principle, this is all we need to know about the gas, but in order to calcu-late the necessary properties, we must first know the quantum mechanical de-scription of each individual input to the model. We can only barely scratch thesurface of quantum mechanics, and the descriptions given in this section havetherefore been kept as simple as possible. However, before we start lookingat microstates, we will first establish the fundamental macroscopic thermody-namic equations that apply to ideal gases.

1 D. F. Lawden. Principles of Thermodynamics and Statistical Mechanics. John Wiley & Sons,1987. 2 Robert P. H. Gasser and W. Graham Richards. An Introduction to Statistical Thermo-

dynamics. World Scientific Publishing Co. Pte. Ltd., 2nd edition, 1995.


2.1. Translation.A particle-in-a-box has three identical degrees of translational freedom.

Even though the particle can in principle move freely within the control volume,its kinetic energy is quantified (only standing waves are permitted). In a cubicbox, each direction will have its own quantum number jx, jy and jz, which isgiven the same unit of energy ε = h2/8ml2. The associated partition function1

can be written

qtrans =∞∑

jx=1

∞∑jy=1

∞∑jz=1

exp(−( j2x+ j2y+ j2z )ε

kT

).

For high quantum numbers j2 = j2x + j2y + j2z ) 0 is a virtually continuousfunction over (an eighth of) the surface of a sphere with a radius of j. Thesurface area of the octant of the sphere is π j2/2, which leads us to the integral

limkT)ε

qtrans = π2

∞∫

0j2 exp

(− j2εkT

)d j

= π2

(kTε

)3/2 ∞∫

0x2e−x2

dx

︸!!!!!!!︷︷!!!!!!!︸√π/4

, x2 =j2εkT ,

→ (2πmkT )3/2Vh3 .

When going from summation to integration, it has been assumed that kT ) εx.Combining this with j2 ) 0 gives us j2ε/kT < 1 over much of the area ofintegration, which ensures that the integral is a good approximation of the sum.

§ 41 Calculate the typical translational temperature ε/k = h2/8ml2k for thelight gases hydrogen, helium and neon. Assume that the lengths of the sides ofthe box are ≥ 1 nm. Comment on the answer.

Particle-in-a-box. The table below gives the values for ε/k and the standardboiling

Mw

g/molε/kK

Tb

K

H2 2.0159 1.19 20.39He 4.0026 0.60 4.224Ne 20.179 0.12 27.09

point of these three low-boiling-point com-pounds.The length of the sides of the boxis l = 1 nm (which is right at the limit ofthe range of validity of classical thermo-dynamics). Except in the case of He, ε/kis reassuringly far below the boiling point

1 Per Chr. Hemmer. Statistisk mekanikk. Tapir Forlag, 1970. In Norwegian.


of the compounds, and in practice all gases have a virtually continuous spec-trum of translational energies1.

For an ensemble of n particles-in-a-box without fixed locations, the totalpartition function becomes Qtrans = 1

n! (qtrans)n. This results in a function for

communal entropy that only depends on the system’s volume and the mass andnumber of particles, while the geometry and bonds are irrelevant:

Qtrans =(qtrans)n

n!n→∞→ en(qtrans)n

nn

ε,kT−→

(en

)n [V(2πmkT )3/2

h3

]n.

Let us substitute in n = NAN, m = NA-1Mw, k = NA

-1R and ln e = 1. Differenti-ating gives us a set of canonical equations of state for Helmholtz energy:

ptransVNRT = 1 ,µtrans

RT = 1 − ln[V(2πMwRT )3/2

NNA4h3

].

Two of thermodynamic properties that can be derived from this are ctransv/R

= 3/2 and standard entropy

strans(1 atm)R = −1.164855 . . . + 5

2 ln T[K] +

32 ln Mw

[g/mol] ,

which is also known as the Sackur2–Tetrode3 equation.

§ 42 Use the Sackur–Tetrode equation to estimate s298.15 K,1 bar for all ofthe noble gases. Compare your answer with experimental values.

Sackur–Tetrode. A convincing comparison of measured and calculatedvalues is shown in Table 6.1. These calculations are more precise than experi-mental results, but the theory cannot be tested on multi-atomic molecules, be-cause the internal degrees of freedom of the molecules also contribute energy.This topic will be further discussed in the following sections on vibrationaland rotational degrees of freedom.

1 Note that ε ∝ l-2. If the length of the box is changed to e.g. 10 nm, then ε/k will be reducedby a factor of 100. This means that even He behaves in the classical manner in a macroscopic

control volume. 2 O. Sackur. Ann. Physik, 36:398, 1911. 3 H. Tetrode. Ann. Physik, 38:434,1912. Correction in 39 (1912) 255.


Table 6.1 Comparison of experimental standard entropies with the Sackur–Tetrodeequation (the margin of error for R, NA, h and Mw is less than 0.01 cal/molK).

Mw

g/mols

cal/K mol† strans

cal/K molMw

g/mols

cal/K mol† strans

cal/K mol

He 4.0026 - 30.11 Kr 83.80 39.17 ± 0.1 39.18Ne 20.179 35.01 ± 0.1 34.93 Xe 131.29 40.70 ± 0.3 40.52Ar 39.948 36.95 ± 0.2 36.97 Rn 222.0 - 42.08

† Gilbert Newton Lewis and Merle Randall. Thermodynamics. McGraw-Hill Book Company,Inc., 2nd edition, 1961.

2.2. Vibration.The atoms in a molecule are held together by forces which, when subjected

to small perturbations, allow the atoms to vibrate as if their centres of wereconnected by elastic springs. The partition function for a harmonic oscillatorof this kind, with a characteristic frequency ν, quantum number j and energyfactor ε = hν = hω/2π = !ω, is

qvıb =∞∑j=0

exp[−( j+ 1

2 )εkT

]= exp

(−ε

2kT

) ∞∑j=0

exp(− jε

kT

).

The summation can be written as a geometric series qvıb =√

u∑∞

j=0 uj =√

u/

(1 − u) where u = e−x = exp(−ε/kT ). If we divide the numerator and denom-inator by

√e−x, we get

qvıb =√

u1−u→

√e−x

1−e−x =1√

ex−√

e−x.

We can obtain a canonical equation of state if we substitute 2 sinh(x/2) =√

ex

−√

e−x. When inserted into x = ε/kT this produces

µvıb

RT = − ln qvıb = ln[2 sinh

(hν

2kT

)].

Differentiating twice with respect to temperature gives us

(6.25)cvıb

v

R =( hν

2kT )2

sinh2( hν2kT ) .

§ 43 The heat capacity of a harmonic oscillator has an upper temperaturelimit of limT→∞ cvıb

v= R. Show this.

Harmonic oscillator. The power series expansion of sinh(x) = x + 13! x3 +

O(x5) applied to kT ) hν gives:

limkT)hν

cvıbv

R = limx,1

x2

(x+ 13! x3 ··· )2 = 1 .


A harmonic oscillator has two equally important energy contributions (kineticand potential energy), which gives us cv → R/2 + R/2 = R as the upper tem-perature limit. In a multi-atomic molecule with many vibration frequencies,the contributions of each individual oscillator must be summed (linearly) inaccordance with the equipartition principle.

The table below shows the characteristic vibrational temperatures ε/k =hν/k = hc/λk for a few di- and multi-atomic molecules. Unlike the contribu-tions from translation and rotation, the vibrational state is only fully developedat high temperatures:

g1,2,3,···λ-1

1,2,3,···cm-1

ε/2kK

H2 1 4162 ca. 3000CH4 3, 2, 1, 1 1306, 1534, 2916, 3019 900–2200HCl 1 2885 ca. 2100H2O 1, 1, 1 1595, 3657, 3756 1100–2700

One of the striking features of the oscillator equations is that the ground state(with the quantum number j = 0) makes a positive contribution to the totalenergy. If we ignore the ground state, the partition function simplifies to

qEinstein → 11−exp( −εkT ) =

11−e−x .

The heat capacity can then be expressed as Einstein’s1 equation:

(6.26) cEinsteinv = x2ex

(ex−1)2 .

Our intuition tells us that the Eqs. 6.25 and 6.26 are equivalent, because theground state only makes a constant (temperature-independent) energy contri-bution. The explanation for this is by no means obvious at first glance, but ifwe compare the equations, we can see that it is the case (show this).

2.3. Rotation.A rotating system with a fixed distribution of mass is called a rigid ro-

tor. The moment of inertia of the rotor determines how much kinetic energy itstores at any given frequency. In quantum mechanics, the moment of inertiaof isolated atoms is zero, whereas diatomic (linear) molecules have degeneratemoments of inertia about two of their axes of rotation and a zero moment of in-ertia about the third axis. In multi-atomic (non-linear) molecules, the momentsof inertia about each axis of rotation are different, unless the symmetry of themolecule dictates otherwise. The partition function for a linear molecule with a

1 Albert Einstein, 1879–1955. German physicist.


moment of inertia I, quantum number j and energy factor ε = !2/2I = h2/8π2Ican be expressed

qrot,2D = 1σ

∞∑j=0

(2 j + 1) exp[− j( j+1)ε

kT

],

where σ corrects for any rotational symmetry of the molecule (σH2O = 2,σNH3 = 3, σCH4 = 12). There is no analytical limit to the above summa-tion, but expanding the power series1 gives us qrot = kT

ε (1 + ε3kT+ 1

15 ( εkT

)2 +4

315 ( εkT )3 + · · · ) provided that 0.3kT > ε. The upper temperature limit can becalculated using integration to obtain an approximation of the partition sum, asshown below:

limkT)ε

qrot,2D = 1σ

∞∫

j=0(2 j + 1) exp

[− j( j+1)εkT

]d j

= − kTσε e− j( j+1) εkT

∣∣∣∞0

→ 8π2IkTσh2 .

The rotational behaviour of multi-atomic (non-linear) molecules is more com-plex than that of linear molecules, and we will therefore simply state the parti-tion function without further explanation2:

limkT)ε

qrot,3D =(8π2IkT )3/2

σh3 .

M =∑i

Mi

(yT

i yiI − yiyTi

)∼ QΛQT .

Here yi = xi − x represents the molecular coordinates of each of the atoms imeasured in relation to the molecule’s centre of mass x =

∑i M-1

w Mixi.

§ 44 Determine the moment of inertia of an H2O molecule if the bondangle is 104.5 and the O–H distance is 0.958 Å.

Moment of inertia. Let us number the atoms H(1), H(2) and O(3). Theirrespsective atomic weights are M1 = 1.00794 g mol-1, M2 = 1.00794 g mol-1

and M3 = 15.9994 g mol-1. The molecule can be oriented in an infinite number

1 Gilbert Newton Lewis and Merle Randall. Thermodynamics. McGraw-Hill Book Company,Inc., 2nd edition, 1961. 2 Note the close analogy between qrot,3D and qtrans. Both of the termsinclude a kinetic energy contribution, and so there are clear similarties.


of ways, and the coordinates given below are simply based on an arbitrarychoice (origin at O(3) and H(1) along the x axis):

x1

[Å]=

−.95800

,

x2

[Å]=

0.958 · sin(14.5·2π360 ) = 0.240

0.958 · cos(14.5·2π360 ) = 0.927

0

, x3

[Å]=

000

.

From this information, the inertia tensor follows

M[g mol-1 Å

2]=

0.818 −.262 0−.262 0.954 0

0 0 1.772

.

Diagonalising M " QΛQT gives us the eigenvalues λ1 = 0.615, λ2 = 1.156and λ3 = 1.772, which have the same units as M itself. The moment of inertiacan now be calculated as I =

3√0.615 · 1.156 · 1.772 = 1.080 g mol-1 Å2=

1.794 · 10-47 kg m2.

Finally, the canonical equation of state describing the rotation of non-linearmolecules can be written

µrot,3D = −RT ln (8π2IRT )3/2

σNA3h3 ,

as long as we understand that I = 3√λ1λ2λ3 represents the molar moment ofinertia. Most molecules, with the exception of hydrogen (isotopes), thereforehave a relatively high moment of inertia, which means that the upper tempera-ture limit is as low as T " 50 K:

Ref.1 σ λi·1047

kg m2I·1047

kg m2ε/kK

Tb

K

H2 2 0.460 0.460 0 0.460 87.6 20.39CH4 12 5.313 5.313 5.313 5.313 7.6 111.66HCl 1 2.641 2.641 0 2.641 15.3 188.15H2O 2 1.022 1.920 2.942 1.794 22.5 373.15

This gives us the classical heat capacity values of R and 3/2R for linear andnon-linear molecules respectively.

1 G. M. Barrow. Physical Chemistry. McGraw-Hill, 3rd edition, 1979.


2.4. Other intra-molecular degrees of freedom.Vibration and rotation are the “classical” contributions to the intra-molecular

energy function. For simple molecules, these contributions can be added to-gether, provided that the amplitude of vibration is small and doesn’t disturb themolecule’s centre of mass (harmonic oscillator). However, the general descrip-tion of the molecule is far more complex, and must take into account anhar-monic vibration or internal rotational and torsional fluctuations. In hydrocar-bons, e.g. CH3, groups will rotate freely around the bond axis, whereas doublebonds such as CH2 = CH2 only allow torsional fluctuations. Internal rota-tion can also be partially prevented by sufficiently low temperatures (staggeredrotation). An important theoretical limit in this context is the equipartition prin-ciple, which states that a fully developed energy degree of freedom contributes12R to cv (vibration results in two degrees of freedom, consisting of potentialand kinetic energy).

Electron excitation is a phenomenon that typically occurs at high tempera-tures, and the outer electrons in a stable molecule will (gradually) be excited attemperatures # 3000 K. In radicals, excitation can occur at " 1000 K, becausetheir electrons are more weakly bonded.

The spin of elementary particles can also contribute to the energy stateof a molecule. In the case of hydrogen, nuclear spin (cf. parahydrogen andorthohydrogen) plays a key role at temperatures " 400 K. Similarly, electronspin is significant if the molecule has unpaired electrons (cf. NO, O2 and NO2).

3. Photons

Thermal radiation (in a closed control volume) can be treated as a thermo-dynamic system, provided that we’re not interested in the spectral distribution1

of the radiation. In this case it is actually possible to set up the canonical equa-tions of state2 for internal energy

prad =14

3

√3

4b

(SV

)4,

T rad =3√

3S4bV,

hvor

b = 8π5k4

15h3c3 = 7.56591 . . . · 10-16 Jm3 K4 .

Note that the number of photons present in the radiation space is not included inthe equations (determined by S and V). In this respect eletromagnetic radiationdeviates from the classical notion that the particle number of a system can be

1 An energy distribution function which has frequency as a free variable. 2 Herbert Callen.Thermodynamics and an Introduction to Thermostatistics. Wiley, 2nd edition, 1985.

4. PHONONS 83

treated as a free variable because it is countable. An unexpected consequenceof this is that the chemical potential of the photon must be set at zero (thisfollows from a Lagrange optimisation of the equilibrium state). Using Euler’smethod to do a simplified integration with respect to only S and V gives us theinternal energy of the radiation space:

U rad(S ,V) = T radS − pradV =34

3√

3S 4

4bV.

By eliminating S we can obtain the Stefan1–Boltzmann2 law of radiation U rad =

bVT 4. Alternatively, we can replace T with p, which gives us U rad = 3pV .

§ 45 One of the most important fission reactions in a uranium bomb canbe written 1

0 n+23592 U→ 142

56 Ba+9136 Kr+1

0 n. The energy released is 3.5 · 10-11 Jper fission event. Calculate T rad and prad if 1% of the atoms take part in thefission reaction.

Atomic bomb. The density of the 235 uranium isotope is 19.1 g cm-3 andits molar mass is 235.04 g mol-1. If 1% of the mass is involved in the reaction,the energy density will be u = V -1U = 0.01 · 6.0221 · 1023 · 3.5 · 10-11 · 19.1 ·106/235.04 = 1.71 · 1016 J m-3. From u = bT 4 = 3p we can obtain T rad =

6.9 · 107 K and prad = 5.7 · 1010 bar. These figures are representative for theuranium bomb Little Boy, which was dropped on Hiroshima on 6 August 1945.

4. Phonons

The Debye3 model of vibration is based on a periodic crystal lattice. Ateach lattice point there is a particle (atom, group of atoms or molecule) oscil-lating in the force field of the neighbouring particles. The partition functionadheres to the Einstein model on page 79, but in such a way that the frequencycan vary for each individual particle. Together all of the particles vibrate withina wide spectrum of frequencies (called phonons), right up to the critical fre-quency ωmax, which is an indirect measure of the total number of degrees offreedom in the system:

ωmax∫

0g(ω) dω = 3n .

The function g(ω) indicates the multiplicity for each frequency, and the nextimportant assumption in the Debye model is g(ω) ∝ ω2. This is not, of course,entirely accurate, but it is nevertheless a much better assumption than saying

1 Jozef Stefan alias Joseph Stefan, 1835–1993. Slovenian physicist and poet. 2 Ludwig Boltz-mann, 1844–1906. Austrian physicist. 3 Petrus Josephus Wilhelmus Debije, alias Peter JosephWilliam Debye, 1884–1966. Dutch–American physicist.


that all of the particles vibrate with the same frequency, as is assumed in theEinstein model. The explanation for this1 is complicated, so we will not dwellon it here, but the result is incredibly simple

g(ω) = 9nω3

maxω2 .

The partition function can now be expressed in a modified Einstein model

QDebye =ωmax∏ω

1[1−exp( −!ωkT )

]g(ω) ,

where the ω represent all of the discrete frequencies in the crystal. At the limitvalue n → ∞ there will be a continuum of frequencies, and the Helmholtzenergy can be expressed as

limn→∞

ADebye = 9NRTω3

max

ωmax∫

0ω2 ln

[1 − exp

(−!ωkT

)]dω .

The canonical equation of state can be obtained in the normal way throughdifferentiation. When expressed in the dimensionless form it makes sense tointroduce a characteristic vibrational temperature θD = !ωmax/k also referredto as the Debye temperature. The chemical potential then becomes

µDebye

RT = 9(

TθD

)3 θD/T∫

0x2 ln

(1 − e−x) dx ,

where x = !ω/kT = θD/T . If we differentiate the partition function twice withrespect to temperature, it gives us

cDebyev

R= 9

(TθD

)3 θD/T∫

0

x4ex

(ex−1)2 dx ,

but note that these expressions are two independent approximations of the parti-tion function. This introduces a thermodynamic inconsistency where c

Debyev

!

−T ∂2µDebye/∂T∂T .There is no analytical solution to the above integral, but at the upper tem-

perature limit θD/T → 0 and expanding the power series ex = 1 + x + O(x2)gives us

limT→∞

cDebyev

R = 9(

TθD

)3 θD/T∫

0

x4(1+x+··· )(x+··· )2 dx . 9

(TθD

)3 θD/T∫

0x2 dx = 3 .

1 Herbert Callen. Thermodynamics and an Introduction to Thermostatistics. Wiley, 2nd edition,1985.

4. PHONONS 85

This result shows that each atom in the lattice oscillates in 3 independent di-rections if the temperature is sufficiently high. Many metals (Pb, Sn, Ag, …)are essentially in this state at room temperature. The limit value is also re-ferred to as the Dulong1–Petit2 law. The table below shows the original datameasured by the two men in 1819. Recalculating their data to our presentvalues for the atomic masses and heat capacity of water gives the limit valuecp/R = 3.020 ± 0.0043 while the theoretical estimate stated above is 3. Thismust said to be an extraordinary good fit — not least so when considering thatthe measurements were done almost 200 years ago.

Table 6.2 Experimental heat capacities of 12 metals plus sulphur measured by Dulongand Petit4. It is maybe a little weird that sulphur is on this list but according to the

classic thoughts of the time the element was closely related to the metals.

cp† Mw

‡ cp · Mw cp† Mw

‡ cp · Mw

Bi 0.0288 13.30 0.3830 Pb 0.0293 12.95 0.3794Au 0.0298 12.43 0.3704 Pt 0.0314 11.16 0.3740Sn 0.0514 7.35 0.3779 Ag 0.0557 6.75 0.3759Zn 0.0927 4.03 0.3736 Te 0.0912 4.03 0.3675Cu 0.0949 3.957 0.3755 Ni 0.1035 3.69 0.3819Fe 0.1100 3.392 0.3731 Co 0.1498 2.46 0.3686S 0.1880 2.011 0.3780 - - - -

† Relative heat capacity. ‡ Relative atomic mass. The reference values are the heat capacityof liquid water and the atomic mass of oxygen respectively.

For the lower temperature limit, the ratio θD/T → ∞ and the integral is aconstant. This result has a practical application in the proportional relationship

limT→0

cDebyev

R ∝ T 3

when determining the entropy of a substance by calorimetry. Instead of thehighly time-consuming and difficult process of measuring near 0 K, all youneed to do is measure the heat capacity down to 15 − 20 K. Often this is suffi-cient to determine the factor of proportionality in the above equation, allowingyou to extrapolate the measurements down to 0 K.

1 Pierre Louis Dulong, 1785–1838. French chemist. 2 Aléxis Thérèse Petit, 1791–1820.French physicist. 3 The value of cp ·Mw for each of the substances is multiplied with the conver-sion factor Mw,O · cp,H2O/(R ·Mw,H2O) where Mw,O = 15.9994 g mol-1, cp,H2O = 75.327 J mol-1 K-1,R = 8.3145 J mol-1 K-1 and Mw,H2O = 18.0153 g mol-1 before the average value and the standarddeviation is calculated in the usual way as x = N-1 ∑

xi and σ =√

N-1 ∑(xi − x)2.


5. Free electrons

The bonding electrons in a metal belong to a group of particles calledfermions, named after the physicist Enrico Fermi1. One of the characteris-tics of fermions is that they cannot share the same one-particle state (the Pauli2

exclusion principle). In a metal, the electron energy levels will therefore befilled up from the lowest level up to the Fermi level equivalent to the chemicalpotential (µ) of the electrons in the conduction band. The internal energy ofthese electrons can be expressed

Uelec

NRT =uRT +

π2

4 ·kTµ+ . . .

The chemical potential of electrons is a relatively abstract concept, and an alter-native to this value is the Fermi temperature TF =

µk , which for normal metals

is approximately 105 K. Differentiating Uelec gives us the lower temperaturelimit for the electronic contribution to the heat capacity of metals:

limT→0

celecv

R= π

2T2TF∝ T .

This is in contrast to crystalline phases with covalent bonds where limT→0(cv)∝ T 3 due to the phonon contribution. The practical implication of this is thatthe heat capacity of metals is dominated by the fermion contribution at suffi-ciently low temperatures (T " 5 K).

6. Virial theorem of dilute gases

The virial equation relates to ideal gases. There is an equivalent powerseries expansion for osmotic pressure in dilute solutions, but it is harder toderive. Expand the power series p/RT in molar density ρ = N/V:

(6.27) pvırVNRT=∞∑

k=1Bk(T )

(NV

)k−1= 1 + B2(T ) N

V+ · · ·

This is the normal way to express the virial development of gases, but it is alsopossible to invert the series from p(ρ) to ρ(p) as shown below. Here the serieshas been truncated after the second term, which means that the index 2 can beleft off the symbol B:

pV2.vır

NRT = 1 + BNV + · · · . 1 + Bp

RT + · · ·

Multiplying both sides by NRT/p gives the volume-explicit form V2.vır =

NRT/p + BN, which reveals that B is a correction factor for the molar vol-ume of the gas. Experimentally B = limp→0(V − RT/p) is used, but this limit

1 Enrico Fermi, 1901–1954. Italian–American physicist. 2 Wolfgang Ernst Pauli, 1900–1958.Austrian physicist.

6. VIRIAL THEOREM OF DILUTE GASES 87

tells us nothing about how B varies with temperature. In practice B < 0 forT < TB ≈ 6Tc, where TB is known as the Boyle1 temperature. Statistical me-chanics explains the theoretical relationship between B and the associated pairpotential φ(r) for the interaction between two molecules in the gas phase:

(6.28) B(T ) = 2πNA

∞∫

0

(1 − e

−φ(r)kT

)r2 dr .

This equation holds very well for non-polar molecules with virtually perfectspherical symmetry, but for rod-shaped molecules and molecules with perma-nent dipole moments integration must be done with respect to the (two) inter-molecular angles of orientation as well as the inter-molecular distance. Thisresults in an integral that looks like the one in 6.28, but where φ(r) representsthe mean pair potential of the molecules.

σ r

φHard spheres. Let us assume that φ = 0 for r > σ and

φ = ∞ for r ≤ σ. That means viewing the molecules ashard spheres, with no pairwise interaction. Substitutingthis into the integral for B gives us

Bhs = NA

σ∫

0r2 dr = 2πNAσ

3

3

The hard sphere potential produces a reasonable approximation of a real systemat high temperatures, but it doesn’t explain why B is temperature-dependent.

−ε σ r

φPotential well. Let us assume that φ = 0 for r > ασ,

φ = −ε for σ < r ≤ ασ and φ = ∞ for r ≤ σ. Thatmeans viewing the molecules as hard spheres with pair-wise interaction within the potential well.Substituting thissimple potential into the integral for B (where α > 1),

demonstrates that B(T ) is a strongly temperature-dependent function with thehard-sphere potential as its upper temperature limit:

Bsw = Bhs + 2πNA

ασ∫

σ

(1 − e

εkT

)r2 dr

= Bhs[1 + (α3 − 1)(1 − e

εkT )

].

Another connection between ε, TB and α is revealed if you solve the equationBsw(TB) = 0:

εkTB= ln α3

α3−1 ⇔ α3 = eε

kTB

eε

kTB −1

1 Robert Boyle, 1627–1691. English chemist and philosopher.


Figure 6.1 The second virial co-efficient of methane calculatedusing the hard sphere, square-well and Lennard-Jones 6:12-potential where σ = 3.85 Åand ε = 1.96 · 10-21 J. BothBsw and BLJ satisfy the lowertemperature limit limT→0 B =

−∞, whilst the upper tempera-ture limit limT→∞ B = 0+ is onlyfulfilled for BLJ (this can only beseen once you reach T # 104 K).

0 200 400 600 800-400

-300

-200

-100

0

100

BL

J[c

m3

mol

-1]

T [K]

TB

Bhs

Bsw

BLJ

−ε σ r

φLennard-Jones1. Let us assume that φ = 4ε

[(σ

r)12 −

(σr

)6] for r > 0. This version is often referred to as the6:12 potential. The exponent 6 has a theoretical basis inthe London2 theory of dispersion forces involving instan-taneous dipoles, whilst the exponent 12 is empirical.BLJ

cannot be determined analytically, and numerical integration is therefore re-quired. This is simple in theory, but not quite so straightforward in practice.In order to prevent numerical problems, the lower limit is estimated using ahard-sphere contribution and the upper limit using a power series expansion,see the Matlab programme H:1.5. A typical temperature progression is shownin Figure 6.1 along with the experimental data for methane3.

7. Van der Waals equation

For a given vector x consisting of the mole fractions x1, x2, . . . , xn the vander Waals4 equation of state can be expressed as

(6.29) pVdW(T, v, x) = RTv−b(x) −

a(x)v2 ,

where a(x) =∑

i

∑j ai j xix j tells us something about the attractive forces be-

tween the molecules and b(x) =∑

i bi xi represents the hard-sphere volume.This means that there must be a relationship between the van der Waals equa-

1 John Edward Lennard-Jones, 1894–1954. English mathematician and physicist. 2 Fritz Lon-don, 1900–1954. German–american physicist. 3 J. H. Dymond and E. B. Smith. The Virial

Coefficients of Pure Gases and Mixtures. Oxford University Press, 1980. 4 Johannes Diderikvan der Waals, 1837–1923. Dutch physicist.

7. VAN DER WAALS EQUATION 89

tion and the virial equation in Section 6. We can demonstrate this relationshipby expanding the power series pVdWV/NRT at the limit V → ∞ (ideal gas):

pVdWVNRT =

11−b( N

V ) −a

RT

(NV

)

= 1 + b(

NV

)+ b2

(NV

)2+ · · · + bn

(NV

)n− a

RT

(NV

).

If we compare the coefficients in this series with the ones in Eq. 6.27, we findthat Bvdw

2 = b − a/RT , Bvdw3 = b2, Bvdw

4 = b3 etc. Only Bvdw2 bears any resem-

blance to real systems; the other virial coefficients are completely unrelated toreality. In spite of the massive simplification, the van der Waals equation ofstate is often used as a starting point for more complicated equations of state,see Chapter 12. It is beyond the scope of this chapter to discuss all of the im-plications of the van der Waals equation here, but we will at the very least noteone important property: the parameters aii and bi can be obtained by measur-ing the critical point of each individual compound experimentally. This doesnot mean that the equation is valid under near-critical conditions (it isn’t), butthese parametric values produce universal behaviour which is very valuable.

§ 46 Show that the parameters b = RTc/8pc, vc = 3b and a = 27(RTc)2/64pc meet the criteria for a mechanical–critical point defined by (∂p/∂v)T,x =

0, (∂2 p/∂v∂v)T,x = 0 and p(Tc, vc) = pc.

The values given can obviously be verified by direct substitution, but wewill choose a different approach. If you multiply both sides of the equation byv2(v − b) you can rewrite the equation of state 6.29 in the form p(v − b)v2 =

RTv2 − a(v − b), or as

(6.30) v3 − v2(b + RTp

)+ v a

p− ab

p= 0 .

This shows that (at constant pressure) the van der Waals equation can be treatedas a cubic equation with volume as a free variable. From the problem statement,we know that the pV isotherm has a degenerate extreme point1 as its criticalpoint. This implies that the equation of state must have three coincident roots(v − vc)3 = 0, or in its expanded form:

(6.31) v3 − v23vc + v3v2c − v3

c = 0 .

1 Both a maximum and a minimum (a horizontal point of inflection).


Critical points. The two Eqs. 6.30 and 6.31 have the same algebraic struc-ture, which allows us to compare the coefficients term by term. This gives riseto the following set of equations:

3vc = b + RTc

pc,

3v2c =

apc,

v3c =

abpc,

which must be solved with respect to a, b and vc. Alternatively we could havesolved the equations with respect to e.g. a, b and R, but traditionally vc isfavoured because it is subject to a relatively large degree of experimental un-certainty. However, this means that the van der Waals equation of state has toofew degrees of freedom to accurately reproduce all measurements of pc, Tc andvc. The solution to the above set of equations can be expressed

38 =

vc pc

RTc,(6.32)

b = 13 vc ,(6.33)

a = 3pcv2c ,(6.34)

which can in turn be rewritten in the form used in the problem statement. Weshould note that the critical compressibility factor zVdW

c = pcvc/RTc =38 be-

haves as a universal constant because vc is a dependent variable in the solution.Experimentally this is no more than a rough approximation as 0.25 < zc < 0.35for most molecular compounds.

§ 47 Pressure is an intensive property (homogeneous function of degreezero), which implies that p(T, v, x) = p(T,V, n). Rewrite the van der WaalsEq. 6.29 in this form.

VDW in the extensive form. Dimensional analysis shows that if v→ V andx → Nx in Eq. 6.29, then homogeneity will be preserved. Consequently, theequation of state can be written in the form

(6.35) pVdW(T,V, n) = NRTV−Nb(x) −

N2a(x)V2

§ 48 Show that, in its reduced form, the van der Waals equation can bewritten as pr = 8Tr/ (3vr − 1) − 3/v2

r . Draw the function for Tr ∈ 0.75, 1, 1.5.Identify all of the asymptotes to the graph. Specify the physical range of thefunction.

8. MURNAGHAN’S EQUATION 91

Corresponding state. Rewrite the van der Waals equation in its reducedform by substituting Eqs. 6.33 and 6.34 into Eq. 6.29:

(6.36) pVdW = RT

v−13 vc

− 3pcv2c

v2 .

Then divide by pc and define reduced pressure as pr = p/pc and reduced vol-ume as vr = v/vc:

(6.37) pVdWr =

(RT

pcvc

)

vr−13− 3

v2r.

Finally substitute in R/(pcvc) = 1/(zcTc) = 8/(3Tc) from Eq. 6.32 and definereduced temperature as Tr = T/Tc. This gives us the van der Waals equationas formulated in the problem statement:

(6.38) pVdWr = 8Tr

3vr−1 −3v2

r.

vr

pr

c.p.

The physical range is defined by vr ∈⟨

13 ,∞

⟩.

Note that the equation contains no arbitrary modelparameters. We call this a universal equation, andhere we will take the opportunity to mention theprinciple of corresponding states, which states thatrelated substances behave virtually alike at reducedstates. For the lowest isotherm (Tr = 0.75) thereare up to three possible solutions for pr = pr(vr).The fluid is thermodynamically unstable in the mid-

range where (∂pr/∂vr) > 0. This results in a phase transition for the system asdescribed in Chapter 13 on vapour–liquid equilibria.

8. Murnaghan’s equation

According to Murnaghan1’s equation, isobaric expansivity α and isother-mal compressibility β are two independent state functions of temperature andpressure respectively. The equation is most useful for pure solids and stoichio-metric compounds. To derive the chemical potential, we start with the totaldifferential of molar volume, for our current purposes expressed as (dln v)mur

N=

α(T ) dT − β(p) dp, also see Section 6 on page 112. Two commonly used func-tions for α and β are:

α(T ) = a1 + a2T + a3T 2 ,

β(p) = b11+b2 p .

1 F. D. Murnaghan. Proc. Natl. Acad. Sci. U. S. A., 30:244–247, 1944.


Let us start by integrating α(T ) at a given reference pressure p between thereference temperature T and an arbitrary system temperature T . This givesus: ln[v(T, p)/v(T, p)] = a1(T − T) + 1

2a2(T 2 − T 2 ) − a3(T -1 − T -1

) =(T − T)[a1 +

12a2(T + T) + a3(TT)-1]. Next we integrate β(p) at constant

T between the reference pressure p and an arbitrary system pressure p. Thisgives us: ln[v(T, p)/v(T, p)] = −b-1

2 b1 ln[(1 + b2 p)/(1 + b2 p)]. We can nowintegrate (dµ)T,N = v dp between the two states T, p and T, p. This producesµmur(T, p) = µ(T, p) +

∫ p

pv(T, p) dp which with a little bit of effort can be

rewritten as

µmur(T, p) = µ(T, p) + v(T, p)p∫

p

(1+b2 p1+b2 p

)−b1/b2 dp

= µ(T, p) + v(T, p)1+b2 pb2−b1

[(1+b2 p1+b2 p

)1−b1/b2 − 1]

= µ(T, p) +b1

b2−b1

vβ

[ββ

vv− 1

],

where β = β(p) and v = v(T, p). It is now easy to produce a completeequation of state µmur(T, p), provided that we have an expression for the com-pound’s standard chemical potential µ(T, p). For solid phases at high pres-sures and polymorphous compounds, it is normal to express the standard molarGibbs energy as a (kind of) power series of temperature:

µRT = . . . +

aT 2 +

bT + c + d ln

(TT

)+ eT + f T 2 + . . .

A trivial integration of µmur using the Euler method gives us the Gibbs energyfunction

Gmur(T, p,N) = Nµmur(T, p) .

§ 49 Murnaghan’s equation is derived by integrating the expansivity withrespect to T at a given p, and then integrating the compressibility with respectto p at constant T , but this provides no guarantee that our starting point wasthermodynamically consistent. Use differentiation to show that α = f (T ) andβ = g(p) genuinely represent a consistent basis for a model.

The Murnaghan model. It is easy to make apparently independent assump-tions without checking whether they are mutually coherent. For the model tobe consistent, the third-order Maxwell equation

∂2v∂T∂p

= ∂2v∂p∂T


must hold. Substituting in (∂v/∂T )p = v f (T ) and (∂v/∂p)T = −vg(p) gives us

−(∂v∂T

)p

g(p) ?=

(∂v∂p

)T

f (T ) ,

where f (T ) and g(p) are functions of different state variables. This avoidscircular references. If we rewrite the last equation slightly we get

v f (T )g(p) ?= vg(p) f (T ) .

Here the right-hand side is identical to the left-hand side, so the test has beensatisfied.

§ 50 Assume that thermal expansivity is constant, i.e. that a2 = 0 K-2 anda3 = 0 K. Find an expression for (∂cp/∂p)T,N that is true for this simplifiedMurnaghan model. What is the sign of the derivative?

Murnaghan CP. The fact that cp is pressure-dependent can be derived asshown below. Note the Maxwell relationship used at the start of the equation:

(∂cmurp

∂p

)T,N= ∂∂p

[T

(∂s∂T

)p,N

]

T,N= T ∂

∂T

[(∂s∂p

)T,N

]

p,N

= − T ∂∂T

[(∂v∂T

)p,N

]

p,N

= − T(∂(vα)∂T

)p,N

= − Tvα2

≤ 0

§ 51 For a one-component system to be thermodynamically stable, it mustsatisfy the conditions β > 0 and cp > 0. Are these conditions (always) met forMurnaghan’s equation?

Thermal instability. The first condition is satisfied for all T and p if b1 > 0and b2 > 0. With respect to the second condition, in Paragraph 50 we showedthat (∂cmur

p/∂p)

T,N< 0 for all V > 0. Since cp < ∞, there will always be a

pressure p < ∞ at which cmurp= 0, unless

∞∫

p

vmur dp?< ∞ .


It can be demonstrated that this is not the case, and that Murnaghan’s modeltherefore does not satisfy the requirement for thermal stability at all T, p.

93

CHAPTER 7

State changes at constant composition

163 Workouts in Thermodynamics, © 2001, 2011 Tore Haug-Warberg

The most important quantities in process engineering include energy, en-thalpy and entropy, generally expressed as functions of temperature and pres-sure (or density), and composition. These are particularly important to theenergy balances for media flowing through valves, turbines, compressors andpipelines. It is therefore of great practical relevance to have a fundamentalunderstanding of the thermodynamic properties of a fluid (of constant compo-sition).

In many cases it is sufficient to know U,H, S as functions of just one (com-mon) set of variables, but from a theoretical point of view there is great valuein being able to choose the variables that are most suited to the particular ap-plication. In this context, T, p, n and T,V, n are particularly relevant, as are thefunctions’ canonical variables.

§ 52 Find the total differentials of U(S ,V, n), H(S , p, n) and S (U,V, n) inthe specified sets of variables. Assume that the composition is in each caseconstant. How easy do you think it would be to integrate these differentials?

Revision. As a warm-up exercise, we will start by revising the canonicalexpressions for dU, dH and dS :

(dU)n = T dS − p dV ,(7.1)(dH)n = T dS + V dp ,(7.2)

(dS )n =1T

dU +pT

dV .(7.3)

The structure of these differentials is clearly simple, but this makes them allthe more difficult to integrate, as the equations of state are rarely or neverknown for T (S ,V, n) and p(S ,V, n), or alternatively T (S , p, n) and V(S , p, n),or 1

T(U,V, n) and p

T(U,V, n).

Canonical means conforming to a pattern or rule, and in thermodynamicsthis refers to the structure of the differentials. For example, S ,V, n are canonicalvariables for U, whereas T,V, n and T, p, n are not. This becomes clear ifwe compare the above equations with 7.11 and 7.22, which show that (dU)nbecomes more complex if we stray away from the canonical description. Other

97

98 7. STATE CHANGES AT CONSTANT COMPOSITION

important functions (stated with their associated canonical variables) include:H(S , p, n), A(T,V, n), G(T, p, n) and S (U,V, n). The relationship between thesefunctions can be derived from U, which in this case is a fundamental energyfunction:

(7.4) U(S ,V, n) = TS − pV +n∑

i=1µiNi .

Alternatively we can view S as a new fundamental relationship that gives usthe same information as U, in which case it is called a Massieu1 function:

(7.5) S (U,V, n) = 1T U +

pT V −

n∑i=1

µi

T Ni .

The remaining functions in the list can be obtained by performing Legendretransformations on the relevant fundamental relationship1. For instance, trans-forming U with respect to S gives rise to a new function, which is calledHelmholtz energy, which is defined by the equation A = U − TS . Thanks toour study of the theory behind Legendre transforms in Chapter 3, we know thatthe new function is particularly useful if we replace S with T as a free variable:

−S =(∂A∂T

)V,n,(7.6)

−p =(∂A∂V

)T,n,(7.7)

µi =(∂A∂Ni

)T,V,Nj!i

.(7.8)

This result is the reason behind the term canonical variables. Also note thatthe differential of A with respect to T is −S , whereas the differential of U withrespect to S is T . This duality means that S and T are referred to as conjugatevariables. The differentials of A with respect to V and n are the same as thoseof U, but remember that T has to replace S as one of the variables that is beingheld constant.

1. Non-canonical variables

We are now going to investigate to what extent we can (freely) choose be-tween the coordinates S ,V, n, T,V, n, T, p, n, U,V, n, etc. These are the canon-ical variables of U, A, G and S respectively, but could e.g. T,V, n serve as acommon set of variables for all of the functions? In order to get to the bottom ofthe matter, you need a thorough understanding of function transformations ingeneral and Legendre transforms in particular. That discussion is not based on

1 François Jacques Dominique Massieu, 1832–1896. French mathematician and physicist. 1 Ifyou start out from U(S ,V,n), the coordinates will always be positive. If on the other hand youuse S (U,V,n), the coordinates make more sense in relation to the physical system, in the sensethat it is easier to accept energy as a free variable than entropy.

1. NON-CANONICAL VARIABLES 99

thermodynamics, and so it may be helpful to leave the thermodynamic equa-tions out of it. We will therefore study an anonymous function f (x) with anassociated Legendre transform φ(ξ).

What we choose to call a state in thermodynamics is inextricably linked tothe information contained in the graph (x, f (x)). From Chapter 3 we know thatthe Legendre transform φ = f − ξx preserves all of the information about thethermodynamic state, assuming that we take φ to mean that φ(ξ) and ξ(x) =(∂ f /∂x). Hence the graph (ξ, φ(ξ)) is equivalent to (x, f (x)). At the same timewe found out that the function φ(x) does not possess that property. Let us lookat the possibility of using f (ξ) as an alternative to f (x) and φ(ξ). The followingdefinition can serve us as an example:

f (x) = ln(x) + c .

The transformed variable is defined by

ξ(x) =(∂ f∂x

)= x-1 ,

giving us x = ξ-1 and

f (ξ) = f (x(ξ)) = c − ln(ξ)

The range of f (x(ξ)) is exactly the same as the range of f (x). This assumesthat ξ(x) is an invertible function, but provided that this assumption holds, thetwo functions are identical until x is substituted by x(ξ). After substitution, thetwo functions are different. The function f (ξ) should therefore really have adifferent symbol from f (x), but in thermodynamic applications it is usual touse the argument to distinguish between the functions. This is not an idealsituation, but the same practice is used in multiple dispatch methods. The linkbetween the three functions is illustrated below, where x ∈ [1, e] has beenchosen as a suitable domain of definition for f (x). The figure shows the twographs f (x) : [1, e] → [c, c + 1] and ξ(x) : [1, e] → [1, e-1] along with f (x(ξ)) :[1, e-1]→ [c, c + 1].

e-1 1 e c c + 1

ξ(x) f (x)

f (ξ)

x(ξ)


It is particularly important to note that the partial derivative of f (x) is iden-tical to ξ(x) = (∂ f /∂x). Since ξ(x) is invertible, we can also determine theinverse function x(ξ) = ξ-1(x). After substituting x(ξ) into f (ξ) = f (x(ξ)), itis no longer possible to reverse the transformation. Information about x(ξ) hasbeen lost, and even if f (x) = f (ξ) throughout the domain of definition, it isimpossible to recreate x from f (ξ).

The complete state is inextricably linked to the graph (x, f (x)), for whichthe set of values of f only represent half of the information. Hence it is impossi-ble to recreate the state from f (ξ). In order to retain the information in (x, f (x)),we need both f (ξ) and x(ξ). Alternatively we can rely on the Legendre trans-form φ = f − ξx, which has the special property that (∂φ/∂ξ) = −x. An equiv-alent description of the state (x, f (x)) can then be stated in terms of (ξ, φ(ξ)).In other words, in a thermodynamic context we can recreate the informationcontained in U(S ,V,N) from the two functions U(T,V,N) and S (T,V,N), orfrom the Legendre transform A(T,V,N).

2. Using volume as a free variable

Chemical process simulation software often uses p(T,V, n) equations ofstate to predict the behaviour of fluids in phase transitions. Based on what wehave learnt so far, it seems wise to describe the system in terms of its Helmholtzenergy, and to derive the internal energy, enthalpy and entropy from A(T,V, n),but below we will look at a more direct approach to deriving U, H and S —partly because it is useful practice and partly because it is a key topic in mosttextbooks.

§ 53 Set out the total differentials of U(T,V), H(T,V) and S (T,V) in aform that can be integrated if C

ıgP

(T ) and p(T,V) are known state functions.

Differentials in T,V . One general approach to problems of this type is tostart out from one of the canonical differentials 7.1–7.3 and to transform itby systematically eliminating unwanted variables (in this case S ). The totaldifferential that we are after is

(7.9) (dU)n =(∂U∂T

)V,n

dT +(∂U∂V

)T,n

dV .

If we compare this differential with the equivalent differential in Eq. 7.1 wecan see that S needs to be replaced with T in order for us to make any progress.One possible approach is to convert dS into a differential expressed in termsof T and V:

(dS )n =(∂S∂T

)V,n

dT +(∂S∂V

)T,n

dV

=CV

T dT +(∂p∂T

)V,n

dV .(7.10)

2. USING VOLUME AS A FREE VARIABLE 101

This eliminates the problem of U being a free variable, as it is in Eq. 7.3, but thedefinition (∂S/∂T )V,n = CV/T does not bring us much closer to our goal — westill have some work to do. Also note that the Maxwell relation2 (∂S/∂V)T,n =

(∂p/∂T )V,n has been used in the last line. Let us combine Eqs. 7.1 and 7.10:1:

(dU)n = CV dT +[T

(∂p∂T

)V,n− p

]dV

= CV dT −(∂(p/T )∂(1/T )

)V,n

dV(7.11)

Eq. 7.11 can be integrated if we know CV (T,V) and p(T,V), but sinceC

ıgV= C

ıgP− NR and (∂(p/T )/∂(1/T ))ıg

V,n = 0 it is possible to replace CV (T,V)with C

ıgP

(T ) by going via an ideal gas reference state. This makes it much easierto integrate the equation. Note that the canonical differential of U in Eq. 7.1requires knowledge of altogether different and more complex equations of state.Incidentally, the differential in Eq. 7.11 is in the same form as the differentialin Eq. 7.9, from which it follows that

(∂U∂V

)T,n= −

(∂(p/T )∂(1/T )

)V,n,(7.12)

(∂U∂T

)V,n= CV ,(7.13)

where 7.13 represents an alternative definition of CV . Our next task is to de-termine the total differential of H(T,V). The easiest way to do that is to startfrom H = U + pV , but on this occasion we will derive dH from first principlesin order to demonstrate how it is done.

Differentials in T,V . With reference to Eq. 7.2 we will here express dp asa differential with respect to T and V:

(7.14) (dp)n =(∂p∂T

)V,n

dT +(∂p∂V

)T,n

dV .

If we substitute the above equation, along with Eq. 7.10, into Eq. 7.2, we getthe following result:

(7.15) (dH)n =

[CV + V

(∂p∂T

)V,n

]dT +

[V

(∂p∂V

)T,n+ T

(∂p∂T

)V,n

]dV .

Relationships equivalent to the ones that we found in Eqs. 7.12–7.13 can bederived for H, but they are of little practical value. Eqs. 7.10, 7.11 and 7.15

2 For the derivation of Maxwell relations, see Section 27 on page 38. 1 In orderto derive the second line from the first line we need to use the mathematical identity∂(y/x)∂(1/x)

≡ y + 1x

∂y

∂(1/x)≡ y − x

∂y

∂x, also known as the Gibbs–Helmholtz’ equation, from

Section 28.


summarise dS , dU and dH as differentials expressed in T,V coordinates. Allof them can be integrated, provided that you know the functions C

ıgP

(T )2 andp(T,V).

§ 54 Find the integral expressions for U,H and S by starting out from thedifferentials that were derived in Section 53. Use ∆fH

ıg(T) and S ıg(T, p) asthe constants of integration.

A selection of useful differentials were derived in Section 53 on page 100.Since U, H and S are state functions, we can largely choose to integrate thedifferentials in whatever way suits our purpose. In the case of internal energy,the most sensible way to integrate is as shown in Figure 7.1: first you integratedU from T to T assuming ideal gas behaviour, and then you integrate fromV = ∞ to V substituting in the relevant equation of state. It may appear we areswitching between an ideal gas and a real fluid as if by magic, but rememberthat

limV→∞

U = Uıg .

In other words, all real fluids approach ideal behaviour at the limit V → ∞ ⇔p → 0. This is precisely where the switch from ideal to real takes place, sothere is no discontinuity in the method.

Figure 7.1 Calculation ofthe internal energy of afluid at constant composi-tion. The most commonstandard state for fluidsis an ideal gas at T =

298.15 K and V = RT/p where p = 1 bar, butbecause (dU ıg)T,n " f (V)the standard state pres-sure is not important inthis context.

0 V -1 V -1

T

T

U ıg(T,V)

∞∫

V

dU ıg = 0

T∫

T

dU ıg

V∫

∞dU

U(T,V)

Integation of U(T,V). Assuming that there is an equation of state that canbe expressed in the form p = p(T,V), and that the heat capacity is a knownfunction of temperature, it is possible to integrate dU(T,V) in Eq. 7.11 to get

(7.16) U(T,V) = U ıg(T) +T∫

T

CıgV

dT −V∫

∞

(∂(p/T )∂(1/T )

)V,n

dV .

2 Normally Cıg

Pis included in reference works on thermodynamics — rather than C

ıg

V(T ).

2. USING VOLUME AS A FREE VARIABLE 103

Note that the integral from V to V = ∞ evaluates to zero in the case of an idealgas, because the state is independent of the volume1. This is evident in Eq. 7.11,where if you substitute in pıg/T = NR/V , you get (∂(p/T )/∂(1/T ))ıg

V,n = 0 and(dU)ıg

n = CıgV

dT .From the definition H = U + pV , the enthalpy can be determined using

Eq. 7.16, but here it is worth commenting on the standard state. Normally thestandard state is chosen to ensure that the elements have zero enthalpy in thestable aggregate state at 298.15 K and 1 bar (ideal gas, liquid or crystal).

Integration of H(T,V). The internal energy must therefore be expressed interms of the standard enthalpy, and not vice-versa:

U ıg(T) = Hıg(T) − (pV)ıg

= ∆fHıg(T) − NRT .(7.17)

The definition is trivial, but once the zero point has been chosen, there arecertain consequences if it is changed. One of the consequences is that1

Hıg(T) = ∆fHıg(T) ,(7.18)

U ıg(T) ! ∆fUıg(T)(7.19)

In principle, S is determined in the same way as U, but here we have tobear in mind that dS diverges at the limit V → ∞2. In order to get around thisproblem, dS is integrated in three steps, as shown in Figure 7.2 on the followingpage. First dS is integrated between V and V , and then it is integrated betweenT and T assuming ideal gas behaviour. Finally, dS −dS ıg is integrated betweenV = ∞ and V in accordance with the selected equation of state. Once again,switching between an ideal gas and a real fluid does not imply discontinuitybecause

limV→∞

S = S ıg ⇔ limV→∞

(∂p∂T

)V,n= NR

V.

1 This means that the internal energy of the fluid can be given without specifying its pressure, ifyou assume ideal gas behaviour. 1 With reference to Eq. 7.19, the molar heat of formation of acompound AaBb . . . Zz in the ideal gas state is defined as being: ∆fu

ıg

AaBb...Zz(T) = u

ıg

AaBb...Zz(T) −

auıgA (T) − bu

ıgB (T) − . . . − zu

ıgZ (T) = ∆fh

ıgAaBb...Zz

(T) − RT + aRT + bRT + . . . + zRT, wherethe coefficients a, b, . . . , z are the number of formula units of the elements A, B, . . . Z in the com-pound. For example, water has the chemical formula (H2)1 (O2)1/2 because H and O naturallyoccur as diatomic molecules. Only if a + b + . . . + z = 1 does ∆fu

ıg = ∆fhıg. 2 Integrating dS

in Eq. 7.10, assuming ideal gas behaviour, confirms that the expression diverges.


Integration of S (T,V). If we substitute (∂p/∂T )ıgV,n = NR/V into Eq. 7.10,

the integral of S is as shown below, but the limit V → ∞ remains undefined,and requires closer analysis:

S (T,V) = S ıg(T,V) +∞∫

V

NRV

dV +T∫

T

CıgV

TdT +

V∫

∞

(∂p∂T

)V,n

dV

= S ıg(T,V) + NR ln(

VV

)+

T∫

T

CıgV

TdT +

V∫

∞

[(∂p∂T

)V,n− NR

V

]dV .(7.20)

Note the term∫ V

∞ (NR/V) dV , which we have explicitly added and taken awayfrom the right-hand side in order to avoid divergence at the limit V → ∞.All real gases approach ideal gas behaviour when p → 0, and the volumeintegral in Eq. 7.20 is well-defined at this limit, since the integrand tends to zerothere. In thermodynamics, an integral that expresses the difference between athermodynamic property of a real gas and an ideal gas in the same state is oftencalled a residual. Chapter 12 covers this topic in greater detail.

Figure 7.2 Calculatingthe entropy of a fluidat constant composition.The hatched lines illus-trate the nominal stepsinvolved in integration.The steps used in practisediffer from the nominalones because the integraldiverges at the limitV → ∞. It is possible toswitch between the tworoutes because (dS ıg)T,n

" f (T ) and (dS ıg)V,n "

f (V). The most commonstandard state for fluidsis an ideal gas at T =

298.15 K and V = RT/p where p = 1 bar.

0 V -1 V -1

T

T

S ıg(T,V)

V∫

V

dS ıgT∫

T

dS ıg

V∫

∞dS −

V∫

∞dS ıg

S (T,V)

cancels

Unlike in the case of U, the standard state of S is a function of volume.However, the ideal gas entropy is generally tabulated at T = 298.15 K andp = 1 bar, and in spite of the fact that the volume is a free variable in thefunction, the standard volume V = NRT/p will have a numerical value. Wewill therefore rewrite Eq. 7.20 in a slightly different form, making use of the

3. USING PRESSURE AS A FREE VARIABLE 105

fact that CıgP= C

ıgV+ NR:

S (T,V) = S ıg(T, p) + NR ln(

pVNRT

)

+T∫

T

CıgP

T dT +V∫

∞

[(∂p∂T

)V,n− NR

V

]dV(7.21)

3. Using pressure as a free variable

Temperature and pressure occupy a unique place amongst the thermody-namic state variables. This is because we generally live under constant temper-ature and pressure conditions, and consequently a large number of experimentshave been performed at room temperature and atmospheric pressure. It is there-fore a useful exercise to derive the formulae discussed in the previous sectionagain, but this time using p as a free variable. Without going into details, the to-tal differentials of U(T, p), H(T, p) and S (T, p) can be derived in an analogousway to the equivalent differentials of internal energy, enthalpy and entropy ex-pressed in T,V coordinates:

(dU)n =

[CP − p

(∂V∂T

)p,n

]dT −

[p(∂V∂p

)T,n+ T

(∂V∂T

)p,n

]dp ,(7.22)

(dH)n = CP dT +(∂V/T∂1/T

)p,n

dp ,(7.23)

(dS )n =CP

TdT −

(∂V∂T

)p,n

dp .(7.24)

It is entirely possible to integrate these expressions if we know the heat capacityC

ıgP

(T ) and the equation of state V(T, p). In that case, we get

U(T, p) = H(T, p) − pV ,(7.25)

H(T, p) = ∆fHıg(T) +

T∫

T

CıgP

dT +p∫

0

(∂V/T∂1/T

)p,n

dp ,(7.26)

S (T, p) = S ıg(T, p) − NR ln(

pp

)+

T∫

T

CıgP

T dT −p∫

0

[(∂V∂T

)p,n− NR

p

]dp ,(7.27)

but in practice only the ideal gas law, the 2nd virial equation expressed aspV/NRT = 1 + Bp, and certain equations of state for solids are explicit func-tions of V(T, p, n). This means that Eqs. 7.26 and 7.27 are used less than theequivalent Eqs. 7.16 and 7.21.


4. The little Bridgman table

In an arbitrarily chosen system of constant composition, there are two pos-sible degrees of freedom. For practical (and historic) reasons, these degrees offreedom are expressed in terms of 9 experimentally determined properties thatare all related to physical chemistry: T , p, s, v, u, h, α, β and cp. The latterthree quantities in this list are defined as

α = v-1(∂v∂T

)p, β = −v-1

(∂v∂p

)T, cp = T

(∂s∂T

)p,

cf. Eq. 7.35 on page 112 and Eq. 7.24 on the preceding page. In view of therequirement for thermodynamic consistency, there must also be 9 − 3 − 2 = 4independent differentials that tell us what we need to know about the behaviourof the system. It is possible to imagine several alternative sets of differentials.Here we will select the set that describes the change in the system’s energy inthe simplest manner. This is a practical choice, in keeping with the philosophythat underlies the rest of this book:

(du)n = T ds − p dv ,

(dh)n = T ds + v dp ,

T (ds)n = cp dT − Tαv dp ,

p(dv)n = pvα dT − pβv dp .

The differentials are linearized functions (here as elsewhere), and the informa-tion that they contain can (here as elsewhere) be summarised in matrix form.Doing so will, in this case, make the next stages of our task significantly easier.We will choose to express the coefficient matrix in dimensionless form, withthe right-hand side equal to 0 J mol-1:

(7.28)

−1 0 1 −1 0 0

0 −1 1 0 0 1

0 0 −1 0 1 −Tα

0 0 0 −1 pvαc-1p−pβ

du

dh

T ds

p dv

cp dT

v dp

=

0

0

0

0

.

By performing a straightforward Gaussian elimination of the rows in the coeffi-cient matrix, taking the pivot elements from columns 1, 2, 3 and 6, it is possible

4. THE LITTLE BRIDGMAN TABLE 107

to derive the following set of coefficients:

−1 0 0 (−pβ + Tα)/(pβ) (cpβ − Tvα2)/(cpβ) 0

0 −1 0 (−1 + Tα)/(pβ) (cpβ − Tvα2 + vα)/(cpβ) 0

0 0 −1 Tα/(pβ) (cpβ − Tvα2)/(cpβ) 0

0 0 0 −1/(pβ) vα/(cpβ) −1

Back substitution assuming p dv and cp dT to be the free variables gives us:

(du)n = (cpβ − Tvα2)β-1 dT − (pβ − Tα)β-1 dv ,

(dh)n = (cpβ − Tvα2 + vα)β-1 dT − (1 − Tα)β-1 dv ,

(ds)n = (cpβ − Tvα2)T -1β-1 dT + αβ-1 dv ,

(dp)n = αβ-1 dT − v-1β-1 dv .

If we now bear in mind that

(∂p/∂T )v,n = αβ-1 ,

v (∂p/∂v)T,n = −β-1 ,

cv = cp − Tvα2β-1 ,

then we see that the result of the matrix operation is identical to the calcu-lations performed in Eqs. 7.11, 7.15, 7.10 and 7.14. This means that we haveuncovered a general method for calculating an arbitrarily chosen differential ex-pressed in the six primary variables: T , p, s, v, u og h. By systematically select-ing the relevant pivot elements, we can now calculate all of the 6× 5× 4 = 120differentials that describe the system. A Ruby program that calculates all ofthese differentials is described in Chapter H:3.1

Table 7.1 A condensed version of Bridgman’s table for calculating all of the6 × 5 × 4 = 120 partial derivatives of u, h, s, v, T and p, in a homogeneousphase at constant composition, i.e. for either one kilogram or one mol of the

substance (or alternatively the sum of all substances in the mixture).

(δT )p = −(δp)T = 1

(δv )p = −(δp)v = (∂v/∂T )p

(δ s )p = −(δp)s = cpT -1

(δu)p = −(δp)u = cp − p (∂v/∂T )p

(δh)p = −(δp)h = cp


Table 7.1 Bridgman table cont. . .

(δv )T = −(δT )v = − (∂v/∂p)T

(δ s )T = −(δT )s = (∂v/∂T )p

(δu)T = −(δT )u = T (∂v/∂T )p + p (∂v/∂p)T

(δh)T = −(δT )h = −v + T (∂v/∂T )p

(δ s )v = −(δv )s = cp (∂v/∂p)T T -1 + (∂v/∂T )2p

(δu)v = −(δv )u = cp (∂v/∂p)T + T (∂v/∂T )2p

(δh)v = −(δv )h = cp (∂v/∂p)T + T (∂v/∂T )2p − v (∂v/∂T )p

(δu)s = −(δ s )u = pcp (∂v/∂p)T T -1 + p (∂v/∂T )2p

(δh)s = −(δ s )h = −vcpT -1

(δh)u = −(δu)h = −pcp (∂v/∂p)T − pT (∂v/∂T )2p − vcp + pv (∂v/∂T )p

Table 7.2 A complete listing of all of the 6 × 5 × 4/2 = 60 total differentialsof u, h, s, v, T and p in a homogeneous phase at constant composition, i.e.either one kilogram or one mol of the substance (or alternatively the sum ofall substances in the mixture). The differentials are calculated by performingGaussian elimination on the system of Eqs. 7.28 on page 106 followed by

algebraic reduction and simplification in Maple®.

dh (u , s ) = cp

p(cpβ−Tvα2) du +(pβcp−pTvα2−cp+pvα)T

p(cpβ−Tvα2) ds

dh (u , v ) = cpβ−Tvα2+vα

cpβ−Tvα2 du +pβcp−pTvα2−cp+pvα

cpβ−Tvα2 dv

dh (u , T ) = 1−Tαpβ−Tα du +

pβcp−pTvα2−cp+pvαpβ−Tα dT

dh (u , p) = cp

cp−pvα du − (pβcp−pTvα2−cp+pvα)vcp−pvα dp

dh ( s , v ) = (cpβ−Tvα2+vα)Tcpβ−Tvα2 ds − cp

cpβ−Tvα2 dv

dh ( s , T ) = Tα−1α ds +

cp

Tα dT

dh ( s , p) = T ds + v dp

dh (v , T ) = Tα−1β dv +

cpβ−Tvα2+vαβ dT

dh (v , p) = cp

vα dv +cpβ−Tvα2+vα

α dp

dh (T , p) = cp dT + (1 − Tα)v dp

d p (u , h) = cp

(pβcp−pTvα2−cp+pvα)v du − cp−pvα

(pβcp−pTvα2−cp+pvα)v dh

d p (u , s ) = cp

p(cpβ−Tvα2)v du − (cp−pvα)Tp(cpβ−Tvα2)v ds

4. THE LITTLE BRIDGMAN TABLE 109

Table 7.2 Total differentials cont. . .

d p (u , v ) = αcpβ−Tvα2 du − cp−pvα

(cpβ−Tvα2)v dv

d p (u , T ) = 1(pβ−Tα)v du − cp−pvα

(pβ−Tα)v dT

d p (h , s ) = 1v dh − T

v ds

d p (h , v ) = αcpβ−Tvα2+vα

dh − cp

(cpβ−Tvα2+vα)v dv

d p (h , T ) = 1(1−Tα)v dh +

cp

(Tα−1)v dT

d p ( s , v ) = Tαcpβ−Tvα2 ds − cp

(cpβ−Tvα2)v dv

d p ( s , T ) = −1vα ds +

cp

Tvα dT

d p (v , T ) = −1vβ dv + αβ dT

d s (u , h) = −cp

(pβcp−pTvα2−cp+pvα)T du +p(cpβ−Tvα2)

(pβcp−pTvα2−cp+pvα)T dh

d s (u , v ) = 1T du +

pT dv

d s (u , T ) = −αpβ−Tα du +

p(cpβ−Tvα2)(pβ−Tα)T dT

d s (u , p) = cp

(cp−pvα)T du − p(cpβ−Tvα2)v(cp−pvα)T dp

d s (h , v ) = cpβ−Tvα2

(cpβ−Tvα2+vα)T dh +cp

(cpβ−Tvα2+vα)T dv

d s (h , T ) = αTα−1 dh +

cp

(1−Tα)T dT

d s (h , p) = 1T dh − v

T dp

d s (v , T ) = αβ dv +cpβ−Tvα2

βT dT

d s (v , p) = cp

Tvα dv +cpβ−Tvα2

Tα dp

d s (T , p) = cp

TdT − vα dp

dT (u , h) = Tα−1pβcp−pTvα2−cp+pvα

du +pβ−Tα

pβcp−pTvα2−cp+pvαdh

dT (u , s ) = Tαp(cpβ−Tvα2) du +

(pβ−Tα)Tp(cpβ−Tvα2) ds

dT (u , v ) = βcpβ−Tvα2 du +

pβ−Tαcpβ−Tvα2 dv

dT (u , p) = 1cp−pvα du − (pβ−Tα)v

cp−pvα dp

dT (h , s ) = Tαcp

dh + (1−Tα)Tcp

ds

dT (h , v ) = βcpβ−Tvα2+vα

dh + 1−Tαcpβ−Tvα2+vα

dv

dT (h , p) = 1cp

dh + (Tα−1)vcp

dp

dT ( s , v ) = βTcpβ−Tvα2 ds − Tα

cpβ−Tvα2 dv

dT ( s , p) = Tcp

ds + Tvαcp

dp


Table 7.2 Total differentials cont. . .

dT (v , p) = 1vα dv +

βα dp

du (h , s ) = p(cpβ−Tvα2)cp

dh − (pβcp−pTvα2−cp+pvα)Tcp

ds

du (h , v ) = cpβ−Tvα2

cpβ−Tvα2+vαdh − pβcp−pTvα2−cp+pvα

cpβ−Tvα2+vαdv

du (h , T ) = pβ−Tα1−Tα dh +

pβcp−pTvα2−cp+pvαTα−1 dT

du (h , p) = cp−pvαcp

dh +(pβcp−pTvα2−cp+pvα)v

cpdp

du ( s , v ) = T ds − p dv

du ( s , T ) = −(pβ−Tα)α ds +

p(cpβ−Tvα2)Tα dT

du ( s , p) = (cp−pvα)Tcp

ds +p(cpβ−Tvα2)v

cpdp

du (v , T ) = −(pβ−Tα)β dv +

cpβ−Tvα2

β dT

du (v , p) = cp−pvαvα dv +

cpβ−Tvα2

α dp

du (T , p) = (cp − pvα) dT + (pβ − Tα)v dp

dv (u , h) = −(cpβ−Tvα2+vα)pβcp−pTvα2−cp+pvα

du +cpβ−Tvα2

pβcp−pTvα2−cp+pvαdh

dv (u , s ) = −1p du + T

p ds

dv (u , T ) = −βpβ−Tα du +

cpβ−Tvα2

pβ−Tα dT

dv (u , p) = vαcp−pvα du − (cpβ−Tvα2)v

cp−pvα dp

dv (h , s ) = −(cpβ−Tvα2)cp

dh +(cpβ−Tvα2+vα)T

cpds

dv (h , T ) = βTα−1 dh +

cpβ−Tvα2+vα1−Tα dT

dv (h , p) = vαcp

dh − (cpβ−Tvα2+vα)vcp

dp

dv ( s , T ) = βα ds − cpβ−Tvα2

Tα dT

dv ( s , p) = Tvαcp

ds − (cpβ−Tvα2)vcp

dp

dv (T , p) = vα dT − vβ dp

5. Heat capacity

Heat capacity is a measure of the thermal inertia of the system. If thisinertia is very high, a lot of energy is needed to raise the temperature by onedegree. For instance, if you were to measure the heat capacity of water at 1 atmwithout being aware of the vapour–liquid transition, you would conclude thatcp → ∞ at 100 C. But the integral of cp dT over a small temperature rangedT ∈ 〈100 − ε, 100 + ε〉 will be finite, and equal to what is defined as the heatof vaporization. Analogously, substances with a low heat capacity, e.g. carbon,

5. HEAT CAPACITY 111

for which cp ≈ 3.5 J mol-1 K-1 in the temperature range T ∈ 〈0, 300〉K, requireonly a small amount of energy for a large temperature change. Hence heatingone gramme of carbon from 0→ 300 K will require about the same amount ofenergy as is needed to vaporize just 0.04 g of water.

For practical reasons it is normal to define the heat capacity of a closedsystem in two different ways: you measure how energy varies with temperatureat either constant pressure, or at constant volume. A third option, which iswidely used in solid and fluid mechanics, is to assume adiabatic conditionsand to look at how energy varies with temperature along an isentrope. Therelationship between these heat capacities can be derived from the volumetricproperties of the fluid. In the next section we will look at the relationshipbetween the first two of the abovementioned heat capacities.

§ 55 Entropy and heat capacity are closely linked, and by comparing thedifferentials dS (T,V)n and dS (T, p)n from 7.10 on page 100 and 7.24 onpage 105, it is possible to obtain two alternative relationships between CP andCV . Show that these relationships can be expressed as:

i) CP −CV = −T(∂V∂T

)2

p,n

(∂V∂p

)-1

T,n,(7.29)

ii) CP −CV = −T(∂p∂T

)2

V,n

(∂p∂V

)-1

T,n.(7.30)

Heat capacity. A total differential contains the same information whetherit is expressed in T,V or T, p coordinates. Based on Eqs. 7.10 and 7.24, for(dS )n we can write

(7.31) CV

TdT +

(∂p∂T

)V,n

dV =CP

TdT −

(∂V∂T

)p,n

dp .

If we assume constant volume (or pressure) we get

(7.32) CP −CV = T(∂p∂T

)V,n

(∂V∂T

)p,n.

We can now choose to get rid of either p or V as a dependent variable. If wesubstitute dp = 0 into the total differential of p, we can show3 that

(7.33)(∂V∂T

)p,n= −

(∂p∂T

)V,n

(∂p∂V

)-1

T,n.

Similarly, by substituting dV = 0 into the total differential of V , we can showthat

(7.34)(∂p∂T

)V,n= −

(∂V∂T

)p,n

(∂V∂p

)-1

T,n.

3 The total differential of p at constant composition is: (dp)n = (∂p/∂T )V,n dT + (∂p/∂V)T,n dV.Assume constant pressure i.e. dp = 0 and divide by dT . Then isolate the derivative: (dV /dT )p,n = (∂V/∂T )p,n = . . .


Substitution of the last two equations into Eq. 7.32 produces the two expres-sions that we set out to find1.

6. Compressibility and expansivity

Heat capacity at constant pressure was one of the first thermodynamic prop-erties to be investigated systematically. Other historically important propertiesinclude isobaric expansivity and isothermal compressibility:

(7.35) α = v-1(∂v∂T

)p, β = −v-1

(∂v∂p

)T.

For some applications it is useful to use these quantities as they give a morecompact set of formulae. In the case of molar volume, for instance, it gives us

(7.36) (dv)n = vα dT − vβ dp ,

while for molar entropy we get

(ds)n =cp

T dT − αv dp ,(7.37)

(ds)n =cv

T dT + αβ dv .(7.38)

These expressions are obtained by substituting 7.35 into the total differentialsof volume and entropy; see Section 55 on the previous page for the necessarydetails. Incidentally, Eqs. 7.29–7.30 in Section 55 are expressed in terms of ex-tensive variables. In the literature on thermodynamics, intensive variables aregenerally used, which makes the size of the system irrelevant to the problem,and one widely used alternative to Eq. 7.29 is the molar form

(7.39) cp − cv =Tvα2

β .

7. Ideal gas

Ideal gases are an important concept in thermodynamics, and it is reas-suring to know that this simple model gives a good description of many realsystems, e.g. in process engineering and meteorology. This section is there-fore one of the most important ones in the whole book, at least in terms of itspractical applications.

§ 56 What are the expansivity α and compressibility β of an ideal gas mix-ture, expressed in terms of the variables T and p? And what is C

ıgP− C

ıgV

forthe mixture?1 With a pressure-explicit equation of state, it may be sensible to work in reduced variables. Foryour own benefit, show that (cp − cv)/R = zcTr (∂pr/∂Tr)2

vr ,x (∂pr/∂vr)-1Tr ,x for one mole of fluid.

7. IDEAL GAS 113

Ideal gas I. If we combine the ideal gas law pıg = NRT/V with the defini-tions in Eq. 7.35, it follows that

αıg = NRPV= T -1 and βıg = NRT

P2V= p-1 ,

which gives us the molar relationship cıgp− c

ıgv= pv/T = R when substituted

into Eq. 7.39. For an arbitrary system size, we can say that

CıgP−C

ıgV= NR .

Extra exercise: Use the Eqs. 7.29 and 7.30 to verify this result.

§ 57 Rewrite the equations in Sections 55 and 54 on page 102 assumingideal gas behaviour.

Ideal gas II. In general the results speak for themselves, but is worth notingthat U ıg and Hıg are only functions of T (and not of p or V):

U ıg(T ) = ∆fHıg(T) − NRT +

T∫

T

CıgV

dT

= ∆fHıg(T) − NRT +

T∫

T

CıgP

dT ,(7.40)

Hıg(T ) = U ıg(T,V) + (pV)ıg

= ∆fHıg(T) − NRT +

T∫

T

CıgV

dT + NRT

= ∆fHıg(T) +

T∫

T

CıgP

dT ,(7.41)

S ıg(T,V) = S ıg(T, p) +T∫

T

CıgV

TdT + NR ln

(pV

NRT

)

= S ıg(T, p) +T∫

T

CıgP

T dT − NR ln(

NRTpV

),(7.42)

S ıg(T, p) = S ıg(T, p) +T∫

T

CıgP

T dT − NR ln(

pp

)(7.43)


§ 58 Use cıgp

(298.15 K) to obtain an approximation for the enthalpy Hıg(T )−Hıg(298.15 K) of argon, nitrogen and methane in the temperature range 0–1000 K. Draw the function in Matlab and compare it with the values fromJANAF1. Comment on the discrepancies.

The enthalpy function. The expression for enthalpy becomes ∆H(T ) =c

ıgp

(T − T), where T = 298.15 K. The predictive accuracy of the functioncan be seen on the left-hand side of Figure 7.3 on the next page. The discrep-ancy between the calculated and measured values is barely noticeable for ar-gon, slightly bigger for nitrogen and biggest for methane (see Matlab-programH:1.3). This difference between the elements is due to the internal degrees offreedom of the molecules. Argon is a monatomic gas without any rotationaland vibrational energy, nitrogen has one vibrational and two rotational degreesof freedom, while methane has nine vibrational and three rotational degreesof freedom. At 298.15 K it is essentially only the rotational energy that isrelevant, whereas the vibrational energy only comes into play at higher temper-atures. This is why the enthalpy curves turn upwards, and why the deviationincreases with the size of the molecule.

§ 59 Use cıgp

(298.15 K) to obtain an approximation for the entropy S ıg(T,p)−S ıg(298.15 K, p) of argon, nitrogen and methane in the temperature range0−1000 K. Draw the function in Matlab and compare the curves with the valuesfrom JANAF1. Comment on the discrepancies.

The entropy function. The expression for entropy becomes ∆S (T, p) =c

ıgp

ln(T/T), where T = 298.15 K. The predictive accuracy of the functioncan be seen on the right-hand side of Figure 7.3 on the facing page. Unsurpris-ingly, the distance between the calculated values and the experimental onesincreases as T moves away from T. An explanation of the behaviour at hightemperatures follows from Section 58, but what happens at low temperatures?You should particularly note that S (0, p) < 0 (see inset figure). Negative en-tropy is theoretically possible and errors of prediction of this kind can havemajor consequences in conjunction with simulating real processes. The Mat-lab program H:1.3 gives all of the details.

§ 60 Show that the following relations are true for an ideal gas mixture if

1 JANAF thermochemical tables. 3rd edition. Part I,II. J. Phys. Chem. Ref. Data, Suppl., 14(1),1985.

7. IDEAL GAS 115

0 500 1000

-10

0

10

20

30

0 500 10000

50

100

150

200

250

T [K]T [K]

Enthalpy

0

Entropy

H[k

J]

S[J/K

mol

]Figure 7.3 Approximated (linearised) ideal gas enthalpy and entropy calculatedfor Ar, N2 and CH4, compared to precise values taken from the JANAF tables.

the heat capacity is constant over the interval of integration:

(T2T1

)ıg

S ,n=

(p2p1

) γ−1γ

S ,n,(7.44)

(p2p1

)ıg

S ,n=

(V2V1

)−γS ,n,(7.45)

(T2T1

)ıg

S ,n=

(V2V1

)1−γ

S ,n.(7.46)

What will the exponent in Eq. 7.44 be for an isochoric process? And what is itfor an isothermal process in Eq. 7.45? And for an isobaric process in Eq. 7.46?

Isentropic relations. Use the differential in Eq. 7.37 as a starting point andthen assume ideal gas behaviour for α. For a isentropic process we also knowthat ds = 0, so we can set up the following integral:

T2∫

T1

cıgp

(dln T )s,n =

p2∫

p1

αıgv dp = Rp2∫

p1

dln p .

Integrate between states 1 and 2, assuming that the heat capacity is constantover this range. This assumption holds perfectly for monatomic gases, whereasfor other gases it is necessary to use a mean value:

ln(T2T1

)ıg

s,n= R

cıgp

ln(

p2p1

)s,n,

(T2T1

)ıg

s,n=

(p2p1

) γ−1γ

s,n.


Here the heat capacity ratio is defined as being γ = cp/cv. Note that R = cıgp−c

ıgv

,which has been used to obtain the final line. The other relations referred to inthe question can be derived by analogous methods starting from a combinationof Eqs. 7.37 and 7.38. For an isothermal, isobaric or isochoric change of state,the ideal gas law immediately tells us that:

(T2T1

)ıg

V,n=

(p2p1

)V,n,

(p2p1

)ıg

T,n=

(V2V1

)-1

T,n,

(T2T1

)ıg

p,n=

(V2V1

)p,n

These are evidently related to 7.44–7.46, but there is no value of γ capable ofconverting that system of equations into this one.

8. Van der Waals equation of state

The van der Waals equation represents an important milestone in our un-derstanding of fluids. By itself, the equation is too inaccurate to be of anypractical use, but it has laid the foundations for a whole series of cubic2 equa-tions of state that do have practical applications. A number of these equations(RK, SRK, PR) are very widely used in relation to liquid–gas separation andgas treatment equipment.

§ 61 Substitute the van der Waals equation of state (6.29) into Section 54on page 102. Go on to show that the internal energy can be expressed

UVdW(T,V) = U ıg(T)Reference

+ ∆U ıg(T )CV integral

− aN2

VResidual

.

Van der Waals U. Let us first state the van der Waals equation in the formp(T,V, n); cf. Eq. 6.35. The form of the equation of state is important, becausesubsequently it is integrated with respect to the total (rather than molar1) vol-ume V . Then we differentiate the equation of state with respect to T , whichgives us

(∂(p/T )∂(1/T )

)VdW

V,n= − aN2

V2 .

If we substitute this into the expression for U(T,V) in Section 54 on page 102(Eq. 7.16), we get

UVdW(T,V) = U ıg(T) +T∫

T

CıgV

dT +V∫

∞

aN2

V2 dV

2 Known as cubic because v(p) can be expressed as the solution to a cubic equation. 1 Mixingup total and molar values is a common mistake in thermodynamics!

8. VAN DER WAALS EQUATION OF STATE 117

= ∆fHıg(T) − NRT︸!!!!!!!!!!!!!!!!!︷︷!!!!!!!!!!!!!!!!!︸

Uıg(T)

+T∫

T

CıgV

dT − aN2

V.(7.47)

The physical model for the internal energy of a van der Waals fluid involves anarbitrarily chosen reference state, a non-linear temperature term and a compo-sition term that is proportional to the system’s molar density.

§ 62 Find an expression for cVdWv

based on Section 61. Determine thefunction of cVdW

p− cVdW

vand draw it on a suitable diagram. What happens as

the state approaches the mixture’s critical point? What is the upper temperaturelimit and what happens at extremely high pressure?

Van der Waals CP. Differentiating UVdW with respect to T at constant Vgives us

CVdWV =

(∂U∂T

)VdW

V,n= C

ıgV,

provided that a is a constant parameter in the equation. This shows, quiteunexpectedly, that heat capacity at constant volume is the same as for an idealgas. Experimentally this is not true. In order to determine the heat capacity atconstant pressure we need to derive the van der Waals Eq. 6.35 again:

(∂pVdW

∂T

)V,n= NR

V−Nb,(7.48)

(∂pVdW

∂V

)T,n= −NRT

(V−Nb)2 +2aN2

V3 .(7.49)

Now substitute the differentials into Eq. 7.30 on page 111 and express the resultin terms of molar variables:

(cp − cv)VdW =−T

(R

v−b

)2

2av3 − RT

(v−b)2

.

If we substitute in a = 9RTc/8vc and b = vc/3, the equation can be rewrittenusing reduced variables:

(7.50) (cp−cv)VdW

R=

1

1 − (3vr−1)2

4Trv3r

.

According to the van der Waals theory, the volume is defined in the range〈b,∞〉. This means that vr ∈ 〈1/3,∞〉, and moreover that cVdW

p− cVdW

v→ ∞ in


100

101

0

2

4

6

8

10

1

1.2

1.5

25

vr

c p−

c vR

Figure 7.4 Dimensionless heat capacities calculated fromthe van der Waals equation of state. The numbers in the

figure refer to Tr = 1.0, 1.2, 1.5, 2.0, 5.0.

Eq. 7.50 if both vr → 1 and Tr → 1. At temperatures Tr < 1, the gas and liquidstates can exist in equilibrium, but our method is not designed to deal with thatsituation. We will therefore stick to supercritical temperatures Tr > 1.

A few selected isotherms have been drawn on Figure 7.4 using the Matlabprogram H:1.6. Note that cVdW

p− cVdW

v→ R due to ideal gas behaviour at high

temperatures Tr >> 1 and low pressures vr >> 1. Funnily enough, the samelimit also applies at extremely high pressures i.e. vr → 1/3. In spite of the factthat at constant volume the heat capacity varies in the same way as that of anideal gas, at constant pressure it deviates greatly. It is particularly worth notingthat cVdW

p→ ∞ at the critical point, which is in line with experimental values.

§ 63 The inversion temperature of a fluid is defined as the temperaturewhere (∂T/∂p)H,n = 0. What are the practical implications of this? Determinethe function for the van der Waals equation of state and draw the graph in apr, Tr diagram.

Joule–Thomson coefficient. This exercise would become incredibly un-wieldly if we were to derive everything from first principles. For the sake ofclarity we will therefore employ a schematic solution with references to otherchapters (this is not generally an advisable approach, but every rule is made tobe broken). Our starting point is a relevant expression for the Joule–Thomson


coefficient which we derived in Section 92 on page 173:

κ =(∂T∂p

)H,n=

T

(∂p

∂T

)

V,n+ V

(∂p

∂V

)

T,n

T

(∂p

∂T

)2

V,n−CV

(∂p

∂V

)

T,n

.

Next we take the partial derivatives of the van der Waals equation from Sec-tion 62 on page 117

(∂p∂T

)VdW

V,n= NR

V−Nb ,

(∂p∂V

)VdW

T,n= −NRT

(V−Nb)2 +2aN2

V3 ,

and substitute this result into the expression for κ:

κVdW =−b + (v−b)2

RT2av2

R + cv − cv(v−b)2

RT2av3

.

In Section 62 we showed that the heat capacity cv of a van der Waals fluid isthe same as that of an ideal gas. In addition we need the relationship betweenthe parameters a, b and Tc, pc as discussed in Section 46 on page 89, and therelationship between c

ıgp

and cıgv

from Section 55 on page 111. In summary, wecan state that:

cVdWv = c

ıgv

vc pc

RTc= 3

8

cıgp= c

ıgv+ R b = 1

3 vc

cıgp= γcıg

va = 3pcv2

c =98 RTcvc

Substituted into the expression for κVdW this gives us

κVdWRvc=− 1

3 +(3vr−1)2

4Trv2r

γγ−1 −

1γ−1

(3vr−1)2

4Trv3r

The inversion state is defined at κVdW = 0. In this state the enthalpy doesnot change with pressure, and hence the temperature remains unchanged evenif there is a slight fall in pressure. Assuming that the denominator is finite


throughout the interval of solution, the criterion for inversion will be fulfilledwhen the numerator is zero. If we solve with respect to vr we get

(vVdWr )κ=0 =

1

3−√

43 Tr

.

As a final trial the answer needs to be converted into reduced coordinates. Weknow the Van der Waals equation in reduced coordinates from Section 48 onpage 90

pVdWr = 8Tr

3vr−1 −3v2

r,

and it is easy enough to substitute in (vVdWr )κ=0 from the equation above. That

gives us our final result:

(pVdWr )κ=0 = 24

√3Tr − 12Tr − 27 .

This expression can usefully be compared with the empirical correlation (pr)κ=0 =

24.21−18.54T -1r −0.825T 2

r , which gives a relatively accurate approximation ofthe experimental values for N2, CO2 and some simple hydrocarbons2. In spiteof the fact that the two equations are entirely different in form, the graphs thatthey produce are qualitatively similar; cf. Figure 7.51.

2 Donald G. Miller. Ind. Eng. Chem. Fundam., 9(4):585–589, 1970. 1 Theoretical expressionsand empirical correlations are in some ways mathematical half-brothers — they have the sameorigins and areas of application, but in terms of their content and form they are completelydifferent from each other.


0 5 100

1

2

3

4

5

6

7

Miller

Van der Waals

pr

Tr

Figure 7.5 Inversion temperature calculated from the vander Waals equation of state compared with an empirical

correlation obtained by D. G. Miller (1970).

equations of state - ntnu

Documents