7 Equations of State

The properties of fluids can be defined in two ways, either by the use of tabulated data (e.g. steam tables) or by state equations (e.g. perfect gas law). Both of these approaches have been developed by observation of the behaviour of fluids when they undergo simple processes. It has also been possible to model the behaviour of such fluids from ‘molecular’ models, e.g. the kinetic theory of gases. A number of models which describe the relationships between properties for single component fluids, or constant composition mixtures, will be developed here.

7.1 Ideal gas law

The ideal and perfect gas laws can be developed from a number of simple experiments, or a simple molecular model. First the experimental approach will be considered.

If a fixed mass of a single component fluid is contained in a closed system then two processes can be proposed:

(i) the volume of the gas can be changed by varying the pressure, while maintaining the temperature constant;

(ii) the volume of the system can be changed by varying the temperature, while maintaining the pressure constant.

The first process is an isothermal one, and is the experiment proposed by Boyle to define Boyle’s law (also known as Mariotte’s law in France). The second process is an isobaric one and is the one used to define Charles’ law (also known as Gay-Lussac’s law in France).

The process executed in (i) can be described mathematically as

v = v( P I T (7.1)

while the second one, process (ii), can be written

v = v(T) ,

Since these processes can be undergone independently then the relationship between the three properties is

(7.3)

122 Equations of state

Equation (7.3) is a functional form of the equation ofstare of a single component fluid. It can be seen to obey the two-property rule, which states that any property of a single component fluid or constant composition mixture can be defined as a function of two independent properties. The actual mathematical relationship has to be found from experiment (or a simulation of the molecular properties of the gas molecules), and this can be derived by knowing that, if the property, v, is a continuous function of the other two properties, p and T, as discussed in Chapter 6, then

dv = ( $)T dp + ( $)p dT (7.4)

Hence, if the partial derivatives (av/ap) , and (av/aT), can be evaluated then the gas law will be defined. It is possible to evaluate the first derivative by a Boyle’s law experiment, and the second one by a Charles’ law experiment. It is found from Boyle’s law that

pv =constant

giving

P =- - V

Similarly, it is found from Charles’ law that

V _ - - constant T

giving

V

($)p =r Substituting eqns (7.6) and (7.8) into eqn (7.4) gives

V V dv = - - dp +- - dT

P T

which may be integrated to give

= constant T

(7.5)

(7.7)

(7.8)

(7.9)

(7.10)

Equation (7.10) is known as the ideal gas law. This equation contains no information about the internal energy of the fluid, and does not define the specific heat capacities. If the specific heat capacities are not functions of temperature then the gas is said to obey the perfect gas law: if the specific heat capacities are functions of temperature (Le. the internal energy and enthalpy do not vary linearly with temperature) then the gas is called an ideal gas.

It is possible to define two coefficients from eqn (7.4), which are analogous to concepts used for describing the properties of materials. The first is called the isothermal compressibility, or isothermal bulk modulus, k. This is defined as the ‘volumetric strain’

Van der Waals’ equation of state 123

(7.1 1)

The negative sign is introduced because an increase in pressure produces a decrease in volume and hence (av/ap) < 0; it is more convenient to have a coefficient with a positive value and the negative sign achieves this. The isothermal compressibility of a fluid is analogous to the Young’s modulus of a solid.

The other coefficient that can be defined is the coeficient of expansion, B. This is defined as the ‘volumetric strain’ produced by a change in temperature, giving

(7.12)

The coefficient of expansion is analogous to the coefficient of thermal expansion of a solid material.

Hence, eqn (7.4) may be written

dv dV v v _ - - -kdp +- BdT (7.13)

The ideal gas law applies to gases in the superheat phase. Th~s is because when gases are superheated they obey the kinetic theory of gases in which the following assumptions are made:

0 molecules are solid spheres; 0 molecules occupy a negligible proportion of the total volume of the gas; 0 there are no forces of attraction between the molecules, but there are infinite forces of

repulsion on contact.

If a gas is not superheated the molecules are closer together and the assumptions are less valid. This has led to the development of other models which take into account the interactions between the molecules.

7.2 Van der Waals’ equation of state

In an attempt to overcome the limitations of the perfect gas equation, a number of modifications have been made to it. The two most obvious modifications are to assume:

e the diameters of the molecules are an appreciable fraction of the mean distances between them, and the mean free path between collisions - this basically means that the volume occupied by the molecules is not negligible;

0 the molecules exert forces of attraction which vary with the distance between them, while still exerting infinite forces of repulsion on contact.

First, if it is assumed that the molecules occupy a significant part of the volume occupied by the gas then eqn (7.10) can be modified to

%T P = -

v - b (7.14)

where b is the volume occupied by the molecules. This is called the Clausius equation of state.

124 Equations of state

Second, allowing for the forces of attraction between molecules gives the van der Waals’ equation of state, which is written

%T a p = - - -

v - b v 2 (7.15)

Figure 7.1 shows five isotherms for water calculated using van der Waals’ equation (the derivation of the constants in the eqn (7.15) is described below). It can be seen that eqn (7.15) is a significant improvement over the perfect gas law as the state of the water approaches the saturated liquid and vapour lines. The line at 374°C is the isotherm at the critical temperature. This line passes through the critical point, and follows closely the saturated vapour line (in fact, it lies just in the two-phase region, which indicates an inaccuracy in the method). At the critical temperature the isotherm exhibits a point of inflection at the critical point. At temperatures above the critical temperature the isotherms exhibit monotonic behaviour, and by 500°C the isotherm is close to a rectangular hyperbola, which would be predicted for a perfect gas. At temperatures below the critical isotherm, the isotherms are no longer monotonic, but exhibit the characteristics of a cubic equation. While this characteristic is not in agreement with empirical experience it does result in the correct form of function in the saturated liquid region - which could never be achieved by a perfect gas law. It is possible to resolve the problem of the correct pressure to use for an isotherm in the two-phase region by considering the Gibbs function, which must remain constant during the evaporation process (see Chapter 1). This results in a constant pressure (horizontal) line which obeys an equilibrium relationship described in Section 7.4. It can also be seen from Fig 7.1 that the behaviour of a substance obeying the van der Waals’ equation of state approaches that of a perfect gas when

0 the temperature is above the critical temperature 0 the pressure is low compared with the critical pressure.

400

350

300 c d

250 e - 2 200 $ v1

2 150

100

50

0

.oo 1 .o 1 .1 1

Specific volume / (m3/kg)

Fig. 7.1 Isotherms on a p - v diagram calculated using van der Waals’ equation

Law of corresponding states 125

Exercise

Evaluate the isothermal compressibility, k, and coefficient of expansion, j9, of a van der Waals’ gas.

7.3 Law of corresponding states

If it is assumed that all substances obey an equation of the form defined by the van der Waals’ equation of state then all state diagrams will be geometrically similar. This means that if the diagrams were normalised by dividing the properties defining a particular state point ( p, v , T ) by the properties at the critical point ( pc, vc, T,) then all state diagrams based on these reducedproperties will be identical. Thus there would be a general relationship

(7.16) vR = f ( P R , TR)

where vR = v /vc = reduced volume; pR = p / p c = reduced pressure; TR = T/Tc = reduced temperature.

To be able to define the general equation that might represent all substances it is necessary to evaluate the values at the critical point in relation to the other parameters. It has been Seen from Fig 7.1 that the critical point is the boundary between two different forms of behaviour of the van der Waals’ equation. At temperatures above the critical point the curves are monotonic, whereas below the critical point they exhibit maxima and minima. Hence, the critical isotherm has the following characteristics: @p/av>, and (a2p/av2), = 0.

Consider van der Waals’ equation

RT a p = - - -

v - b v 2

This can be differentiated to give the two differentials, giving

RT 2a + -

2RT 6a

and at the critical point

RT, 2a - + - = o 3 (Vc-bI2 vc

2RTc 6a

(vc-bI3 vC -0

4

giving

(7.17)

(7.18)

(7.19)

(7.20)

8a a 27bR’ 27b2

vC=3b; T C = - p c = - (7.21)

126 Equations of state

Hence

(7.22)

The equation of state for a gas obeying the law of corresponding states can be written

or

(7.23)

(7.24)

Equation (7.24) is the equation of state for a substance Obeying Van der Waals’ equation: it should be noted that it does not explicitly contain Values of a and b. It is possible to obtain a similar solution which omits a and b for any two-parameter state equation, but such a solution has not been found for state equations with more than two parameters. The law of corresponding states based on van der Waals’ equation does not give a very accurate prediction of the properties of substances over their whole range, but it does demonstrate some of the important differences between real substances and perfect gases. Figure 7.1 shows the state diagram for water evaluated using parameters in van der Waals’ equation based on the law of corresponding states. The parameters used for the calculation of Fig 7.1 will now be evaluated.

The values of the relevant parameters defining the critical point of water are

p c = 221.2 bar; v, = 0.00317 m3/kg; Tc = 374.15°C

which gives

v 0.00317

3 3 b=“= = 0.0010567 m3/kg

a = 3pCv,2 = 3 x 221.2 x 0.003172 = 0.0066684 bar m6/kg20.003172 0.0066684barm6/kg2

8 x 221.2 x 0.00317 3 x (374.15 + 273)

R = = 0.002889 bar m3/ kg K

Thus, the van der Waals’ equation for water is

0.0066684 - 0.002889T P =

(v- 0.0010567) V 2 (7.25)

Note that the value for ‘R ’ in this equation is not the same as the specific gas constant for the substance behaving as a perfect gas. This is because it has been evaluated using values of parameters at the critical point, which is not in the region where the substance performs as a perfect gas.

It is possible to manipulate the coefficients of van der Waals’ equation to use the correct value of R for the substance involved. This does not give a very good approximation for the critical isotherm, but is reasonable elsewhere. Considering eqn (7.22), the term for vc

2

c

v

3 3 T

can be eliminated to give

Law of corresponding states 127

(7.26)

Substituting the values of p c and Tc into these terms, using R = %/18, gives

27 x (8.3143 x 103/18)2 x (374.1 + 273)2 a =

64 x 221.2 x 10’

= 1703.4Pam6/kg2 = 0.017034 barm6/kg2 8.3143 x lo3 x (374.1 + 273)

18 x 8 x 221.2 x 10’ b = = 0.0016891 m3/kg

which results in the following van der Waals’ equation for water:

0.004619 T 0.0 17034 - P = (V - 0.0016891) V 2

(7.27)

It can be readily seen that eqn (7.27) does not accurately predict the critical isotherm at low specific volumes, because the value of b is too big. However, it gives a reasonable prediction of the saturated vapour region, as will be demonstrated for the isotherms at 200”C, 300°C and the critical isotherm at 374°C. These are shown in Fig 7.2, where the predictions are compared with those from eqn (7.25).

400

350

300 l d

. 8

200 2 ul g 150

100

50

0

- Equation 25

--c Equation 27

Saturated l iquid + vapour lines

.001 .o 1 .1 1

Specif ic volume / (m3/kg)

Fig. 7.2 Comparison of isotherms calculated by van der Waals’ equation based on eqns (7.25) and (7.27)

It is interesting to compare the values calculated using van der Waals’ equation with those in tables. This has been done for the isotherm at 200”C, and a pressure of 15 bar (see Fig 7.3), and the results are shown in Table 7.1.

128 Equations of state

Table 7.1 Specific volume at 200°C and 15 bar

Specific volume from Specific volume from Specific volume Region eqn (7.25)/ (m3/kg ) eqn (7.27) / (m 3/kg 1 from tables/(m3/kg)

Saturated liquid 0.00 16 0.0017 0.001 157

Saturated vapour 0.09 0.14 0.1324

2 5

20 h c m e 1 15

e, c

2 2 10 (II

PI

5

0

001

I , ' ( ' 1 I

.o 1 .1 1

Specific volume / (m3 /kg)

Fig. 7.3 Expanded diagram of 200°C isotherms, showing specific volumes at 15 bar

The above diagrams and tables show that van der Waals' equation does not give a good overall representation of the behaviour of a gas in the liquid and mixed state regions. However, it is a great improvement on the perfect, or ideal, gas equation in regions away from superheat. It will be shown later that van der Waals' equation is capable of demonstrat- ing certain characteristics of gases in the two-phase region, e.g. liquefaction. There are other more accurate equations for evaluating the properties of substances, and these include

0 the virial equation:

0 the Beattie-Bridgeman equation:

(7.28)

RT A p = 7 (1 - e ) (v + B ) - 7

V V

Isotherms or isobars in the two-phase region 129

where

A =Ao(i - f),

0 the Bertholet equation:

RT a p=---

v - b Tv2

0 the Dieterici equation:

RT -alRTv p"- e v - b

C B = B o l - - , e = - ( :) vT3

Returning to the van der Waals' gas, examination of eqns (7.22) shows that

z, = - pcvc = - = 0.375 RT, 8

(7.29)

(7.30)

(7.31)

(7.32)

This value z, (sometimes denoted p,) is the compression coefficient, or compressibility factor, of a substance at the critical point. In general, the compression coefficient, z, is defined as

P V RT

z = - (7.33)

It is obvious that, for a real gas, z is not constant, because z = 1 for a perfect gas (i.e. in the superheat region), but it is 0.375 (ideally) at the critical point. There are tables and graphs which show the variation of z with state point, and these can be used to calculate the properties of real gases.

7.4 Isotherms or isobars in the two-phase region

Equation (7.15) predicts that along an isotherm the pressure is related to the specific volume by a cubic equation, and this means that in some regions there are three values of volume which satisfy a particular pressure. Since the critical point has been defined as an inversion point then the multi-valued region lies below the critical point, and experience indicates that this is the two-phase (liquid + vapour) region. It is known that pressure and temperature are not independent variables in this region, and that fluids evaporate at constant temperature in a constant pressure chamber; hence these isotherms should be at constant pressure. This anomaly can be resolved by considering the Gibbs energy of the fluid in the two-phase region. Since the evaporation must be an equilibrium process then it must obey the conditions of equilibrium. Considering the isotherm shown in Fig 7.4, which has been calculated using eqn (7.27) for water at 300°C, it can be seen that in the regions from 3 to 1 and from 7 to 5, a decrease in pressure results in an increase in the specific volume. However, in the region between 5 and 3 an increase in pressure results in an increase in specific volume: this situation is obviously unstable (see also Chapter 6). It was shown in Chapter 1 that equilibrium was defined by dG I p , T L 0. Now

d g = v d p - s d T (7.34)

130 Equations of state

and hence along an isotherm the variation of Gibbs function from an initial point, say, 1 to another point is

(7.35)

This variation has been calculated for the region between points 1 and 7 in Fig 7.4, and is shown in Fig 7.5.

0 0.005 0.01 0.015 0.02 0.025 0.03 Specific volume / (m3/kg)

Fig. 7.4 Variation of pressure and specific volume for water along the 30O0C isotherm

It can be seen from Fig 7.5 that the Gibbs function increases along the isotherm from point 1 to point 3. Between point 3 and point 5 the Gibbs function decreases, but then begins to increase again as the fluid passes from 5 to 7. The regions between 1 and 3 and 5 and 7 can be seen to be stable in terms of variation of Gibbs function, whereas that between 3 and 5 is unstable: this supports the argument introduced earlier based on the variation of pressure and volume. Figure 7.5 also shows that the values of Gibbs energy and pressure are all equal at the points 2, 4 and 6, which is a line of constant pressure in Fig 7.4 This indicates that the Gibbs function of the liquid and vapour phases can be equal if the two ends of the evaporation process are at equal pressures for an isotherm. Now considering the region between the saturated liquid and saturated vapour lines: the Gibbs function along the ‘isotherm’ shown in Fig 7.4 will increase between points 2 and 3, and the substance would attempt to change spontaneously back from 3 to 2, or from 3 to 4, and hence point 3 is obviously unstable. In a similar manner the stability of point 5 can be considered. An isobar at 100 bar is shown in Fig 7.5, and it can be seen that it crosses the Gibbs function line at three points - the lowest Gibbs function is at state 1. Hence, state 1 is more stable than the other points. If that isobar were moved down further until it passed through point 5 then there would be other points of higher stability than point 5 and the system would tend to move to these points. This argument has still not defined the equilibrium line

Problems 131

200

175

1 so

100

7s

so

3

5 5

I

Variation in Gibbs function along isotherm

Fig. 7.5 Variation of Gibbs function for water along the 300°C isotherm

between the liquid and vapour lines, but this can be defined by ensuring that the Gibbs function remains constant in the two-phase region, and this means the equilibrium state must be defined by constant temperature and constant Gibbs function. Three points on that line are defined by states 2, 4 and 6. The pressure which enables constant Gibbs function to be achieved is such that

5,4” vdp = 0

and this means that

(7.36)

(7.37) -I

Region I Region 11

which means that the area of the region between the equilibrium isobar and line 2-3-4 (Region I) must be equal to that between the isobar and line 4 - 5 - 6 (Region II ). This is referred to as Marwell’s equal area rule, and the areas are labelled I and II in Fig 7.4.

7.5 Concluding remarks

A number of different equations of state have been introduced which can describe the behaviour of substances over a broader range than the common perfect gas equation. Van der Waal’s equation has been analysed quite extensively, and it has been shown to be capable of defining the behaviour of gases close to the saturated vapour line. The law of corresponding states has been developed, and enables a general equation of state to be considered for all substances. The region between the saturated liquid and vapour lines has been analysed, using Gibbs energy to define the equilibrium state.

132 Equations of state

PROBLEMS

The Dieterici equation for a pure substance is given by

v - b Determine

(a) the constants a and b in terms of the critical pressure and temperature; (b) the compressibility factor at the critical condition; (c) the law of corresponding states.

Derive expressions for (&,/dv), for substances obeying the following laws:

ST e-(a/%Tv) (a) P = -

v - b

R T a v - b Tv2

(b) p = - - -

U c + -. R T (c) p = - - ___

v - b V ( V - b ) v3

Discuss the physical implication of the results.

The difference of specific heats for an ideal gas, c ~ , ~ - cV,,, = 3. Evaluate the difference in specific heats for gases obeying (a) the van der Waals', and (b) the Dieterici equations of state. Comment on the results for the difference in specific heat for these gases compared with the ideal gas.

Derive an expression for following expression:

R T a p = - - -

v - b T v 2

the law of corresponding states for a gas represented by the

Problems 133

A, for a gas obeying the state equation

5 p v = ( l + a ) % T

where a is a function of temperature alone, that the specific heat at constant pressure is given by

d2 ( a T )

dT2 cp = -%T ~ h P + CP"

where cpo is the specific heat at unit pressure.

6 The virial equation of state is

Compare this equation with van der Waals' equation of state and determine the first two virial coefficients, b, and b,, as a function temperature and the van der Waals' constants.

Determine the critical temperature and volume (T,, v,) for the van der Waals' gas, and show that

27 T, b - _ 2 - v c 3 (

BV. ) [b, = 1; b2= ( b - u/%T)]

7 The equation of state for a certain gas is

where A is constant.

then the value of the specific heat at constant pressure at the state (T , p ) is given by Show that if the specific heat at constant pressure at some datum pressure p o is cpo

cp- cPo=%TA e-AT(2-AT)(p-po)

8 How can the equation of state in the form of a relationship between pressure, volume and temperature be used to extend limited data on the entropy of a substance.

A certain gas, A, has the equation of state

pv=%T(l+ a p )

where a is a function of temperature alone. Show that

Another gas B behaves as an ideal gas. If the molar entropy of gas A is equal to that of gas B when both are at pressure p o and the same temperature T, show that if the

5 Show

134 Equations of state

pressure is increased to p with the temperature maintained constant at T , tht, entropy of .gas B exceeds that of gas A by an amount

%(p -po) a + T - ( :;) 9 A gas has the equation of state

-- pv - a - b T RT

where u and b are constants. If the gas is compressed reversibly and isothermally at the temperature T' show that the compression will also be adiabatic if

a 2b

T' = -