Solution By definition, the Joule–Thomson coefficient is m = (∂T/∂P)H.

In Jacobian form,

6.6 FUGACITY
The concept of fugacity was introduced by G.N. Lewis (1901) and is widely used in solution
thermodynamics to represent the behaviour of real gases. The name fugacity is derived from the Latin
for ‘fleetness’ or the ‘escaping tendency’. It has been used extensively in the study of phase and
chemical reaction equilibria involving gases at high pressures. Though the ‘fugacity’ is mainly
applied to mixtures, the present discussion is limited to pure gases.
For an infinitesimal reversible change occurring in the system under isothermal conditions, Eq. (6.18)
reduces to
dG = V dP
For one mole of an ideal gas V in the above equation may be replaced by RT/P, so that

………(6.117)
Equation (6.117) is applicable only to ideal gases. If, however, we represent the influence of
pressure on Gibbs free energy of real gases by a similar relationship, then the true pressure in the
above equation should be replaced by an ‘effective’ pressure, which we call fugacity f of the gas. The
following equation, thus, provides the partial definition of fugacity.
dG = RT d(ln f)………(6.118)
Equation (6.118) is satisfied by all gases whether ideal or real. Integration of this equation gives
G = RT ln f + C………(6.119)
where C is a constant of integration that depends upon the temperature and nature of the gas. Fugacity
has the same dimension as pressure, usually atmosphere or bar.
6.6.1 Standard State for Fugacity
Consider the molar free energies of a gas in two states both at the same temperature. Let G1 and G2
be the free energies and f1 and f2 be the corresponding fugacities in these states. By Eq. (6.119), the
change in free energy is

………(6.120)
The free energy change can be experimentally measured and by the above equation the measured free
energy change gives the ratio of fugacities f2/f1. The fugacity in any state can be evaluated if the
fugacity is assigned a specific value in a particular reference state.
For an ideal gas integration of Eq. (6.117) gives the free energy change as

………(6.121)
Whereas Eq. (6.121) is applicable only to ideal gases, Eq. (6.120) is valid for all fluids, ideal or
real. It follows that in the case of ideal gases, f2/f1 = P2/P1, or fugacity is directly proportional to
pressure. The proportionality constant is chosen to be unity for convenience. That is, f/P = 1 or f = P,
for ideal gases. The fugacity is always equal to the pressure for an ideal gas. However, for real
gases, fugacity and pressure are not proportional to one another, and f/P is not constant. As the
pressure of the gas is reduced, the behaviour of the real gas approaches that of an ideal gas. That is, at
very low pressures, the fugacity of a real gas should be the same as its pressure. So the gas at a very
low pressure P0 is chosen as the reference state and it is postulated that the ratio of fugacity to
pressure at this state is unity. Thus the definition of fugacity is completed by stating that

Thus, the standard state of a real gas is a hypothetical state in which the gas is at a pressure P0 where
it behaves perfectly. By this choice, the standard state has the simple properties of an ideal gas. If the
standard state were chosen as the one for which f is equal to say, 1 bar, the standard state of different
gases would have different and complex properties. If the standard state chosen were the gas at zero
pressure, the free energy would have become – at the standard state. The choice of the
hypothetical standard state standardises the interaction between the particles by setting them to zero.
Since all intermolecular forces are absent in the standard state chosen, the differences in the standard
free energies of different gases arise solely from the internal structure and properties of the
molecules, and not from the way they interact with each other.
Equation (6.122), which sets the fugacity of the real gas equal to its pressure at low pressures,
permits the evaluation of absolute values for fugacities at various pressures. It is this property that
makes fugacity a widely accepted thermodynamic property in practical calculations.

6.6.2 Fugacity Coefficient

The ratio of fugacity to pressure is referred to as fugacity coefficient and is denoted by f. It is
dimensionless and depends on nature of the gas, the pressure, and the temperature. Integrating
Eq. (6.118) between pressures P and P0,

For ideal gases, by Eq. (6.120), G = G0 + RT ln P/P0. Combining this result with Eq. (6.124) we see
that the free energy of a real gas = free energy of an ideal gas + RT ln f. The quantity RT ln f,
therefore, expresses the entire effect of intermolecular interaction.
Since all gases becomes ideal as pressure approaches zero, we can say that
f P…as…P 0
f 1…as…P 0
6.6.3 Effect of Temperature and Pressure on Fugacity
In Eq. (6.123), G0 and f0 refer to the molar free energy and fugacity respectively at a very low
pressure where the gas behaves ideally. This equation can be rearranged as

H is the molar enthalpy of the gas at the given pressure and H0 is the enthalpy at a very low pressure.
H0 – H can be treated as the increase of enthalpy accompanying the expansion of the gas from
pressure P to zero pressure at constant temperature. Equation (6.125) indicates the effect of
temperature on the fugacity.
The effect of pressure on fugacity is evident from the defining equation for fugacity [Eq. (6.118)].
dG = V dP = RT d(ln f)………(6.118)
which on rearrangement gives:

6.6.4 Determination of Fugacity of Pure Gases

Using compressibility factor, Z. The compressibility factor Z of a real gas is the ratio of its
volume to the volume of an ideal gas at the same temperature and pressure.

As (Z – 1)/P is finite as pressure approaches zero, there is no difficulty in using Eq. (6.127) for the
evaluation of f. The values of the compressibility factor, Z, from zero pressure to pressure P are
calculated from the volume of the gas at the corresponding pressures. The integral in
Eq. (6.127) is found out graphically by plotting (Z – 1)/P against P.
EXAMPLE 6.22 Derive an expression for the fugacity coefficient of a gas obeying the equation of
state P(V – b) = RT and estimate the fugacity of ammonia at 10 bar and 298 K, given that
b = 3.707 10–5 m3/mol.
Solution Since, P(V – b) = RT, we have,

Equation (6.127) becomes

Therefore, fugacity f = 10.151 bar.

Using generalised charts. Using Eq. (6.127) in reduced form, a generalised chart similar to the
generalised compressibility chart can be prepared for predicting the fugacity of gases.

The integral in Eq. (6.128) is evaluated graphically at constant temperature by taking compressibility
factors from the isotherms on the generalised compressibility chart. The fugacity coefficient is then
plotted against reduced pressure (Pr) for various constant reduced temperature (Tr) values. This
provides a generalised chart for fugacity of all gases as shown in Fig. 6.7.
If experimental volumetric data are not available, this chart can be used for approximate calculation
of fugacity, provided the critical temperature and pressure are known. The accuracy of the results
depends upon how closely the generalised compressibility charts predicts the actual P-V-T behaviour
of gases.
Using residual volumes. Equation (6.118) relates the fugacity of the gas to the molar
volume V at temperature T and pressure P
dG = V dP = RT d(ln f)………(6.118)
The residual volume a is defined as the difference between V and the volume occupied by one mole
of an ideal gas at the same temperature and pressure.

This result is integrated between a very low pressure and the given pressure P. At low pressures, f/P
= 1 and the required integral is

To find f, the residual volume a derived from experimentally determined values of molar volumes at
different pressures are plotted against P. Refer Fig. 6.8. The area under the curve between pressures
0 and P is equal to the integral in Eq. (6.130).
EXAMPLE 6.23 From the P-V-T data for a gas it is found that = – 556.61 J/mol. Find the
fugacity of the gas at 50 bar and 300 K.
Solution Using Eq. (6.130), we obtain

Using equations of state. We have seen that Eq. (6.118) defines fugacity as
dG = V dP = RT d (ln f)
On integrating this between pressure P0 where fugacity is f0 and pressure P where fugacity is f , we
get the following result.

If an analytical equation of state is available, and if it can be put in a form in which V is expressed
explicitly as a function of P, the integral in Eq. (6.131) can be readily evaluated. On the other hand, if
P is expressed as a function of V, the integral is determined by integration by parts. We can use the
following identity for this purpose:
where V0 is volume of the gas at pressure P0. Since the gas behaves ideally under this condition,
P0V0 = RT and Eq. (6.133) becomes

Equation (6.134) can be used for evaluating the integral in Eq. (6.131).
EXAMPLE 6.24 Find the fugacity coefficient at 1 bar, 5 bar, and 10 bar for a gas that follows the
equation of state PV = RT(1 – 0.00513 P), where P is pressure in bar.
Solution According to Eq. (6.118),

EXAMPLE 6.25 Show that the fugacity of a gas obeying the van der Waals equation of state is given
by

where a and b are van der Waals constants.

Solution The van der Waals equation [Eq. (3.29)] can be written in the following form:
Since V0 is very large compared to b, V0 – b V0. Further, V0 can be replaced by RT/P0 as V0 is
the volume of a gas at a very low pressure P0 at which ideal gas equations are obeyed by the gas.
Also, as V0 is very large, a/V0 can be neglected. With these simplifications, the above equation
becomes

EXAMPLE 6.26 Calculate the fugacity of pure ethylene at 100 bar and 373 K. The van der Waals
constants are a = 0.453 J m3/mol2, b = 0.571 10–4 m3/mol, molar volume at 100 bar and 373 K =
2.072 10–4 m3/mol.
Solution Substitute a = 0.453, b = 0.571 10–4, V = 2.072 10–4, R = 8.314, and T = 373 into
Eq. (6.139). Note that RT/(V – b) be multiplied by 10–5 for dimensional consistency.
Thus we get ln f = 4.3 and f = 73.7 bar.
Using values of enthalpy and entropy. Equation (6.123) indicates the free energy change
between the given state where free energy and fugacity are G and f respectively, and a standard state
where the free energy and fugacity are G0 and f0 respectively. By the definition of free energy, G = H
– TS, and G0 = H0– TS0, where H0 and S0 are the enthalpy and entropy values at the standard state.
Using these, Eq. (6.123) becomes

Assuming that the gas behaves ideally at the reference state, f0 = P0, the pressure at the standard
state. The fugacity can be calculated using the values of H, H0, S and S0 in Eq. (6.140)
EXAMPLE 6.27 Determine the fugacity and fugacity coefficient of steam at 623 K and 1000 kPa
using enthalpy and entropy values from steam tables. Assume that steam behaves ideally at 101.3 kPa.
Data from steam tables: At 1000 kPa and 623 K, H = 3159 kJ/kg; S = 7.3 kJ/kg K. At
101.3 kPa and 623 K, H = 3176 kJ/kg; S = 8.38 kJ/kg K.
Solution Since steam behaves ideally at 101.3 kPa, fugacity at this pressure = 101.3 kPa. Using Eq.
(6.140),

Approximate method for estimation. Experimental evidences suggest that at moderate

pressures, the value of PV for any gas is a linear function of its pressure at constant temperature. The
functional relationship between PV and P may be written as PV = RT + kP, where k is a constant. The
residual volume a, by definition is a = V – RT/P = k. It means that the residual volume a is constant
and is independent of pressure. Substituting this result in Eq. (6.130),

At moderate pressures, f/P is close to unity and therefore, ln (f/P) (f/P) – 1. (Note: When x
approaches unity, ln x is approximately equal to x – 1). Equation (6.141) can be modified as
Equation (6.142) can be used to determine the approximate value of the fugacity of a gas from its
pressure and molar volume.
EXAMPLE 6.28 The density of gaseous ammonia at 473 K and 50 bar is 24.3 kg/m3. Estimate its
fugacity.
Solution The molar volume of ammonia under the given conditions is
V = 17/(24.3 1000) m3/kmol
Pressure, P = 50 105 N/m2
Using Eq. (6.142), we get

6.6.5 Fugacities of Solids and Liquids

Every solid or liquid has a definite vapour pressure although it may be immeasurably small, in some
cases. At this pressure, the solid (or the liquid) will be in equilibrium with its vapour. When two
phases of a substance are in thermodynamic equilibrium, the molar free energies in both phases
should be equal. This follows from the criterion of phase equilibrium, which will be discussed in
detail in Chapter 8. By this criterion the molar free energy of the liquid (or the solid) in equilibrium
with its vapour is equal to the molar free energy of the vapour. That is, GL = GV and GS = GV, where
the superscripts L, S and V refer to liquid, solid and gas respectively. Since the molar free energy is
related to the fugacity as G = RT ln f + C, where C is constant that depends only on temperature, it
follows that
fL = fV, fS = fV………(6.143)
Equation (6.143) means that the fugacity of solid (or liquid) is equal to the fugacity of the vapour with
which it is in equilibrium, provided that the reference state is taken to be the same in each case. If the
vapour pressure is not very high, the fugacity of the vapour would be equal to the vapour pressure;
hence, the fugacity of a liquid (or a solid) is approximately equal to its vapour pressure.
If the vapour pressure is very high and the vapour cannot be treated as ideal gas its fugacity is related
to the saturation pressure as in Eq. (6.142)
 ………(6.142)
PS is the saturation pressure of the gas and fsat is the saturation fugacity. The latter should in turn be
equal to the fugacity of solid or liquid at the desired temperature and the saturation pressure, by Eq.
(6.143). Since, RT d(ln f) = V dP and the liquid can be assumed to be incompressible, the fugacity of
the liquid at any other pressure P is readily obtained as

………(6.144)
where V is the molar volume of the liquid.
EXAMPLE 6.29 Calculate the fugacity of liquid water at 303 K and 10 bar if the saturation pressure
at 303 K is 4.241 kPa and the specific volume of liquid water at 303 K is 1.004 10–3 m3/kg.
Solution The molar volume is

EXAMPLE 6.30 Calculate the fugacity of n-butane in the liquid state at 350 K and 60 bar. The
vapour pressure of n-butane at 350 K is 9.35 bar. The molar volume of saturated liquid at 350 K is
0.1072 10–3 m3/mol. The fugacity coefficient for the saturated vapour at 350 K is 0.834.
Solution The fugacity of saturated vapour at 350 K = 0.834 9.35 = 7.798 bar. Therefore, fugacity
of saturated liquid at 350 K = 7.798 bar = fsat. Using Eq. (6.144),

Thus the fugacity of the liquid at 60 bar and 350 K, f = 9.4 bar.

6.7 ACTIVITY
The vapour pressures of relatively non-volatile solids and liquids may be extremely low, so, an
experimental determination of their fugacity is impractical. When dealing with such substances, it
would be convenient to work with another function called activity rather than with fugacity itself.
‘Activity’ is, in fact, relative fugacity and is defined as the ratio of fugacity to fugacity in the standard
state. It finds wide application in the study of homogeneous chemical reaction equilibria involving
solids and liquids. Activity is denoted by the letter a, where

The standard state at which fugacity is f0 is chosen arbitrarily, but the temperature in the standard
state is the same as the temperature in the given conditions. For gases, the standard state fugacity is
chosen by convenience to be unity, and therefore, fugacity and activity are numerically equal.
The change in the free energy accompanying the process in which the substance is undergoing a
change of state from the standard state to the given conditions is related to the activity of the substance
as

The assumption of constant V is a good approximation and will not introduce much error for solids
and liquids up to very high pressures, provided the temperature is well below the critical value.
Comparison of Eqs. (6.146) and (6.148) shows that

The concept of activity is particularly useful in the study of solutions. The commonly used standard
states and their properties are discussed in detail in Chapter 7.
EXAMPLE 6.31 Determine the activity of solid magnesium (MW = 24.32) at 300 K and 10 bar if the
reference state is 300 K and 1 bar. The density of magnesium at 300 K is 1.745 103 kg/m3 and is
assumed constant over this pressure range.
Solution Using Eq. (6.149), we obtain

6.7.1 Effect of Pressure and Temperature on Activity

From Eq. (6.146) we see that,
Equation (6.154) predicts the effect of pressure on activity.

6.8 DEPARTURE FUNCTIONS AND GENERALISED CHARTS

The methods for the evaluation of thermodynamic properties from experimental P-V-T data or
analytical equations of state were discussed earlier. If these data are not available or if very accurate
values of the properties are not required, rough estimate of the thermodynamic properties can be
made through the use of departure functions or residual properties . The residual properties are
defined as the difference between the thermodynamic property at the specified temperature and
pressure and the property that the substance would have exhibited at the same temperature and
pressure, had it been an ideal gas. Representing the properties in the ideal gas state with the
superscript id, the residual enthalpy (HR) and residual entropy (SR) are defined as
HR = H – Hid………(6.155)
SR = S – Sid………(6.156)
HR and SR are also known as enthalpy departure and entropy departure , respectively. These
represent hypothetical property changes because a gas cannot be both real and ideal at a given P and
T.
Equations (6.155) and (6.156) are differentiated with respect to pressure to get the following results.
The values of Z and (∂Z/∂T)P are calculated directly from the experimental P-V-T data and the
integrals in the preceding two equations are evaluated by numerical or graphical methods. Analytical
integration is possible when Z is expressed as an equation of state. Thus, HR and SR and all other
residual properties are readily evaluated. We see that there exists a direct connection between the
experimental data and the residual properties, which makes the latter a very valuable tool for
evaluation of thermodynamic properties.
Once the residual enthalpy and entropy are known, the enthalpy and entropy of the gas can be
calculated using Eqs. (6.155) and (6.156). The enthalpy and entropy of the ideal gas at pressure P and
temperature T for use in these equations can be determined by an arbitrary choice of the reference
state at pressure P0 and temperature T0 where the enthalpy and entropy values are H0 and S0
respectively.
Z and (∂Z/∂Tr)P required for the evaluation of the integral in the above equations can be obtained
from the generalised compressibility charts. The HR/RTC or the SR/R values thus calculated are
plotted against reduced pressure with reduced temperature as parameter to give the generalised
enthalpy departure chart and the entropy departure chart, respectively. [See Hougen O.A., K.M.
Watson and R.A. Ragatz, Chemical Process Principles, Part II.]
It is observed that the uncertainty in Z taken from the generalised compressibility chart is only around
2.5 per cent. However, a greater uncertainty is expected in the reduced residual properties as their
calculation involves the derivatives of Z.

6.9 THERMODYNAMIC DIAGRAMS

Thermodynamic properties of pure fluids or mixtures of constant composition are determined when
any two properties are specified. Thermodynamic diagrams provide these properties in terms of two
independent variables. Various combinations of these independent variables give rise to different
type of thermodynamic diagrams. These diagrams find wide applications in computation of
thermodynamic properties and thermodynamic analysis of processes. We have already used the T-S
diagrams in the analysis of refrigeration cycles and the vapour-power cycles in Chapter 5.
6.9.1 Types of Diagrams
P-H diagram. In a P-H diagram the pressure P (or, ln P) is the Y-axis and enthalpy H is the
X-axis. The general appearance of a P-H diagram is shown in Fig. 6.9. This chart is particularly used
for calculating the heat loads and temperature changes in refrigeration systems. The refrigeration
cycles have both constant pressure (evaporators and condensers) and constant enthalpy (throttling
valve) processes and the P-H diagrams are very useful in analysing them. The dome-shaped envelope
encloses the two-phase region. Within the envelope, horizontal lines of constant pressure and constant
temperature are shown. Temperature lines are nearly vertical in the liquid-phase region; are
horizontal in the two-phase region and drop steeply in the vapour-phase region. The constant entropy
lines are shown as continuous lines in the vapour as well as in the two-phase region. Constant volume
lines show a point of inflection as they cross from the vapour-phase region to the two-phase region.
Figure 6.9 also shows lines of constant quality (x).

H-T diagram. Enthalpy H forms the Y-axis and temperature T is the X-axis. A typical H-T diagram
is shown in Fig. 6.10. These are useful in the calculations in throttling processes as well as in the
constant pressure flow processes.
The vertical distance between the saturated vapour curve and the saturated liquid curve gives the heat
of vaporisation at a particular temperature and pressure.
T-S diagram. A typical T-S diagram is shown in Fig. 6.11. These are useful in following the
temperature changes in isentropic processes. A reversible adiabatic process is essentially isentropic
and would be represented by a vertical line on the T-S diagram. In the case of turbines and
compressors, a vertical line drawn from the initial pressure to the final pressure shows the
path followed by the fluid undergoing reversible adiabatic operation. Horizontal lines of constant
pressure, lines of constant enthalpy, and constant quality are drawn within the two-phase region. The
former two lines are drawn in the liquid-phase and vapour-phase regions as well. The horizontal
distance between the saturation curves is l/T, where l is the heat of vaporisation at temperature T.
H-S diagram (the Mollier diagram). Figure 6.12 shows the enthalpy-entropy chart, popularly
known as the Mollier diagram.
The energy requirements in flow processes, in general, and the temperature changes involved in
isentropic and isenthalpic processes are easily determined using these charts. The lines of constant
temperature and pressure shown within the two-phase region, separate in the vapour region into
pressure lines that rise continuously and temperature lines that drop and eventually become
horizontal. Lines of constant quality are also shown in the two-phase region.

6.9.2 Construction of Thermodynamic Diagrams

Thermodynamic charts are prepared by first calculating the ideal gas properties by the standard
methods, followed by measuring the enthalpy and entropy departures to correct for pressure. The
departure functions are obtained from the generalised charts for departure functions. The critical
pressure, temperature and an expression for ideal gas heat capacity as a function of temperature are
the additional required data. The construction of T-H and T-S diagrams is discussed below.
Construction of T-H diagram. Here we discuss how the enthalpy is calculated as function of
pressure at three temperatures. The temperatures chosen are T0, T and TC. T0 is a reference or base
temperature; T is any temperature below the critical temperature and TC is the critical temperature.
Properties below T0 are not generally required for practical calculations. For the present calculations
we assume that the substance exist as saturated liquid at the base temperature. The saturation pressure
is denoted by P0. In Fig. 6.13, the point A represents the system at the reference state.
The various steps in the calculations are given below:
1. Enthalpy HA at point A is assumed to be zero.
2. Enthalpy at point B, HB = HA + l0, where the heat of vaporisation at temperature T0 and pressure
P0 is determined from experimental data or using the enthalpy departure charts. If the latter is
used, , where and are the residual enthalpies of saturated vapour and
saturated liquid respectively at Pr = P0/PC and Tr = T0/TC.
3. The enthalpy HC at point C = HB – where = HB – HC, the residual enthalpy at Pr = P0/PC
and Tr = T0/TC.
4. The enthalpy at D is calculated using ideal gas heat capacity C P, which is independent of
pressure.

5. The point E represents the condition of the gas at any pressure, say P1 and temperature T above the
dew point. HE = HD + (HE – HD) = HD + , where is the residual enthalpy at Pr = P1/PC
and Tr = T/TC and is evaluated from the generalised charts.
6. Point F represents condition of the gas at saturation pressure P2 and temperature T. The residual
 enthalpy determined at Pr = P2/PC and Tr = T/TC gives . The enthalpy at F, HF = + HD.
7. Point G has the same reduced properties as point F. HG – HD is obtained from the liquid line of
the two-phase envelope of the generalised enthalpy departure chart. Let this be . The enthalpy
at G, HG = – HD.
8. For any pressure above the bubble point pressure, say P3, the enthalpy departure function, =
HH – HD, where is the residual enthalpy corresponding to Pr = P3/PC and
Tr = T/TC. Therefore, HH = + HD.

9. Point I represents the ideal gas state at temperature TC. Enthalpy at I, HI = .

10. Critical conditions are indicated by point J. The enthalpy departure function
is obtained at Pr
= Tr = 1. The enthalpy at J, HJ = HI + . The enthalpy at any temperature and pressure can be
calculated in this way and the T-H diagram can be constructed.
Construction of T-S diagram. The T-S diagram (refer Fig. 6.14.) is constructed in a similar way
as a T-H diagram, with additional steps for calculating the effect of pressure on entropy of an ideal
gas.

1. Point A represents the entropy of the saturated liquid at the reference temperature, T0.