0stprocess fugacity

The concept of “fugacity” was introduced by G.N. Lewis (1901) and is widely used in solution thermodynamics to
represent the behavior of real gas.
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 gas at high pressures

Although the fugacity is mainly applied to mixtures, the present discussion is limited to pure gases.

By definition of G we have

dG = Vdp – SdT

For an infinitesimal reversible change occurring in the system under isothermal conditions (dT=0), we have equation:

dG = V dP

For one mole of an ideal gas (V=RT/P), V in the above equation may be replaced by RT/P so that

dP
dG =RT =RT d ¿
P
(6.117)
[HTD: Note that (ln(x))’=1/x or d(ln(x)) = dx/x that means d(lnP) = dP/P]
This equation 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. (Actually, utilizing ideal gas condition,
where P = f, we replace P by f).

dG=RT d ¿
(6.118)
This equation is satisfied by all gases whether ideal or real. (With taking into account that the proposed parameter
“fugacity f” must be equal to P at ideal gas condition).
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 gas.
Fugacity has the same dimension as pressure, usually atmosphere or bar.

Standard State for Fugacity

Consider the molar free energies of a gas in two states both at the same temperature.
Let G₁ and G ₂ be the free energies and f₁ and f₂ be the corresponding fugacities in these states.
By equation (6.119), the change in free energy is
f2
∆ G=G 2−G 1=RT ln
f1
(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 f₁/f₂.
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, (f = P) integration of Eq. (6.117) gives the free energy change as

Page 1 of 36
 P2
∆ G=G2−G1=RT ln
P1
(6.121)
Whereas Eq. (6.121) is applicable only to ideal gas (due to f = P condition), Eq. (6.120) is valid for all fluids, ideal or
real (as variable is (variable in “general case”) f instead of P (value of the variable in “specific case” where f = P).
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 (f₂/P₂ = f₁/P₁ = 1).
That is, f/P = 1 or f = P, for ideal gas.
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 reduced, the behavior of the real gas approaches that of the ideal gas.
That is, at very low pressure, 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

f f
lim =1∨ →1 as P → 0
P →0 P P
(6.122)
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 G⁰ at standard state.
The choice of the hypothetical standard state standardizes 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.
Eq. (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.

Fugacity coefficient

The ratio of fugacity to pressure is referred to as “fugacity coefficient” and is denoted by φ=f/P.
It is dimensionless and depends on nature of the gas, the pressure, and the temperature.
Integrating Eq. (6.118) between pressure P and P⁰,
0 f
G−G =RT ln
f0
(6.123)
Since f⁰ = P⁰ and let f = φP, we can write the above equation as
0 Pφ 0
G=G + RT ln 0
=G + RT ¿
P

or
P
G=G 0+ RT ln 0
+ RT ln φ
P
(6.124)
0 P
For ideal gases, by Eq. (6.121), we could obtain G=G + RT ln 0 . Combining this result with Eq. (6124) we see that
P
the free energy of a real gas = free energy of an ideal gas + RT ln φ.

Page 2 of 36
The quantity RT ln f, therefore, expresses the entire effect of intermolecular interaction.

Since all gases become ideal gas as pressure approaches zero, we can say that

f → P as P→0
f/P → 1 as P→0

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

f G G0
R ln 0
= −
f T T
or
G G0
R ( ln f −ln f 0 ) ¿ −
T T

Differentiate this with respect to temperature at constant pressure.

( ( )) ( ( ))
0
G G
∂ ∂
[( ) ( )] T T
0
∂ ln f ∂ ln f
R − = −
∂T P ∂T P ∂T P ∂T P

Substituting the Gibbs – Helmholtz equation

( ) ∂ ( GT )
∂T P
=
−H
T2

Into the above result and observing that f⁰ is equal to the pressure and is independent of temperature (served as
constant), we get

[( ) ]
0
∂ ln f −H H
R −0 = +
∂T P T 2 T2
Or

( ∂ ln f
∂T )
P
=
H 0 −H
RT2
(6.125)
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 ¿
which on rearrangement gives:

( ∂∂lnPf ) = RTV T

(6.126)

Page 3 of 36
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.

V V
Z= =
V ideal RT
P
Or
V Z
=
RT P

Introducing this in Eq. (6.126) and rearranging, the following result is obtained.

( ∂∂lnPf ) = RTV T

( ∂∂lnPf ) = RTV = ZP
T

Or
d¿

The above result, as such, is of not much use for the determination of fugacity, because as P → 0, Z/P → ∞. This
difficulty can be overcome if we add and subtract dP/P on the right hand side of the preceding equation.

d¿
Note that dP/P=d(ln P) as (ln P)’=1/P, then we have
d¿

Rearranging it we have
d¿

or

( Pf )=( Z−1) dPP

d ln

when this is integrated between 0 and P we get

P
ln
f
=∫
P 0
Z −1
P
dP ( )
(6.127)
or
P
lnφ=∫
0
( Z−1
P )
dP

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 value of the compressibility factor, Z, from zero pressure to pressure P are calculated from the volume of the gas
at the corresponding pressure.
The integral in Eq. (6.127) is found out graphically by plotting (Z-1)/P against P.

Example:

Page 4 of 36
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 E-5 m3/mol.
Solution:
Since P(V-b)=RT, we have
PV =RT + Pb

As z=PV/RT, then we have

PV Pb
Z= =1+
RT RT
P
Applying (Eq. 6.127) ln
f
=∫
P 0
Z −1
P ( )
dP we have
P
f b Pb
ln =ln φ=∫ dP=
P 0 RT RT

Where φ is the fugacity coefficient. For ammonia at 10 bara and 298 K,

−5 5
f (10 ×10 )(3.707 ×10 )
ln =ln =0.015
10 8.314 × 298

Therefore, fugacity f = 19.151 bara.

Using generalized charts.

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

Pr
f Z−1
ln =∫ d Pr
P 0 Pr
(6.128)
The integral in Eq. (6.128) is evaluated graphically at constant temperature by taking compressibility factors from
isotherms on generalized compressibility chart. The fugacity coefficient is then plotted against reduced pressure (Pr)
for various constant reduced temperature (Tr) values. This provides a generalized chart for all fugacity of all gases as
shown in Fig.

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
generalized compressibility charts predicts the actual P-V-T behavior of gases.

Using residual volumes

Eq. (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)

The residual volume α is defined as the difference between V and the volume occupied by one mole of an ideal gas
at the same temperature and pressure.
RT
α =V −
P
Substitution of this in Eq. (6.118) gives

d¿

Note that d(ln(P))=dP/P as (ln(P))’=1/P, then rearranging it we obtain:

Page 5 of 36
 d¿

d¿

Eventually, we obtained
d ln( Pf )= RTα dP
(6.129)
This result is integrated between a very low pressure and the given pressure P. At low pressure, f/P=1 and the
required integral is
P
ln ( )
f
=
1
P RT 0
∫ α dP
(6.130)
To find f, the residual volume α derived from experimentally determined values of molar volumes at different
pressures are plotted against P. Refer the Fig. The area under the curve between pressures 0:and P is equal to the
integral in Eq. (6.130).

Example:
50
From the P-V-T data for a gas it is found that ∫ α dP=−556.61 J /mol . Find the fugacity of the gas at 50 bara and
0
300 K.
Solution:
Using Eq. (6130), we obtain
f −556.61
ln = =−0.2232 , therefore, f = 40 bara
P 8.314 ×300

USING EQUATION OF STATE

We have seen that Eq. (6.118) defines fugacity as

dG=VdP=RT d ¿

Or
d¿
(6.131)
In isothermal conditions (T is constant), on integrating this between pressure P⁰ where fugacity is f⁰ and pressure P
where fugacity is f, we get the following result.

P
f 1
ln 0 =
f
∫ V dP
RT P 0

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.
∫ V dP=PV −∫ P dV
(6.132)
Then the integral in Eq. (6.131) becomes
P V

∫ V dP=PV −P0 V 0−∫ P dV

0 0
P V
(6.133)

Page 6 of 36
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
P V

∫ V dP=PV −RT −∫ P dV
0 0
P V
(6.134)
Equation (6.134) can be used for evaluating the integral in Eq. (6.131).

Example:
Find the fugacity coefficient at 1 bara, 5 bara, and 10 bara for a gas that follows the equation of state PV=RT(1-
0.00513 P), where P is pressure in bara.
Solution:
According to Eq. (6.118)
RT d ¿
Therefore,
P
f
d ln =−0.00513 dP∨ln φ=−0.00513∫ dP=−0.00513 P
P 0

This gives the fugacity coefficient as 0.995 at 1 bara, 0.975 at 5 bara, and 0.950 at 10 bara.

Example:
Show that the fugacity of a gas obeying the Van de Waals equation of state is given by
b 2a RT
ln f = − + ln
V −b RTV V −b
Where a and b are van de Waals constants.
Solution:
The Van de Waals equation can be written in the following form

RT a
P= − 2
V −b V
(6.135)
Substituting Eq. (6.134) in to Eq. (6.131), we get

( )
V
f 1
ln 0 = PV −RT −∫ P dV
f RT V
0

(6.136)
Rearranging Eq. (6.135), we get
RT b a
PV −RT = −
V −b V
(6.137)
Integrating Eq. (6.137), we get (noting that d(ln(x))=dx/x and d(1/x)= – dx/x²)

|
V V V V
RT a a
∫ P dV =∫ dV −∫ 2 dV =RT ln (V −b )|V +
V
0

V
0
V
0 V −b V V
0 V V
0

V V V

∫ P dV =∫ VRT
−b
a V −b a a
dV −∫ 2 dV =RT ln 0 + − 0
V −b V V
V V
0 0 0
V V

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

Page 7 of 36
 V
(V −b) P0 a
∫ P dV =RT ln RT + V =RT ln P0 + RT ln VRT
0
−b a
+
V
V
(6.138)
Substituting results (6.137) and (6.138) into Eq. (6.136)
ln
f
f
0
= (
1 RT b a
− −RT ln P0 −RT ln
RT V −b V
V −b a
RT
−
V )
f b 2a 0 V −b
ln = − −ln P −ln
f 0
V −b V RT

Rearranging and noting that –ln(x/y)=ln(y/x,we have

f 0 b 2a RT
ln +ln P = − + ln
f 0
V −b V V −b
As P⁰ = f⁰, eventually, we have

b 2a RT
lnf = − + ln
V −b RTV V −b
(6.139)

Example:
Calculate the fugacity of pure ethylene at 100 bara and 373 K. The van de Waals constants are a = 0.453 J m³/mol²,
b= 0.571 x 10E-4 m³/mol, molar volume at 100 bara and 373 K = 2.072 x E-4 m³/mol.
Solution:
Substitute a = 0.453, b = 0.571xE-4, V= 2.072xE-4, R = 8.314, and T = 373 into Eq. (6.139). Note that RT/(V-b) be
multiplied by 10E-5 for dimensional consistency.
Thus we get ln f = 4.3 and f = 73.7 bara.

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

( H −H 0 )−T (S−S 0)=RT ln f 0

f

f 1
ln 0
=
RT
[ ( H−H 0 ) −T ( S−S0 )]
f
(6.140)
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:
Determine the fugacity and fugacity coefficient of steam at 623 K and 1000 kPa(a) using enthalpy and entropy values
from steam tables. Assume that steam behaves ideally at 101.325 kPa.
Data from steam tables: At 1000 kPa and 623 K, H = 3159 kJ/kg; S = 7.3 kJ/kg K. At 101.325 kPa and 623 K, H = 3176
kJ/kg, S = 8.38 kJ/ kg K.
Solution:
Since steam behaves ideally at 101.325 kPa, fugacity at this pressure = 010.325 kPa. Using Eq. (6.140),

Page 8 of 36
 f 1
ln =
101.325 (8.314 /18) ×623
[ (3159−3176 ) −623 (7.3−8.38 ) ]=2.279
φ = exp(2.279) = 9.76691
f = 9.76691 x 101.325 = 989.4 kPa = 9.89 bar

Approximate method for estimation:

Experimental evidences suggest that at moderate pressure, 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 α, by definition is α = V – RT/P = k. It means that the residual volume α is
a constant and is independent of pressure. Substituting this result in Eq. (6.130),

P
ln( )
f
=
1
P RT 0
∫ α dP
(6.130)
Integrating it and noting that α is constant we have
f αP
ln =
P RT
(6.141)
At moderate pressure, f/P is close to unity and therefore, ln(f/P) → (f/P) – 1. (Note that When x approaches 1, ln x is
approximately equal to x – 1). Equation (6.141) can be modified as

f αP
−1=
P RT

As α = V-RT/P, having

( ) ( )
2 2
αP RT P VP
f = 1+ P=P+ V − =P+ −P
RT P RT RT
2
VP
f=
RT
(6.142)

Equation (6.142) cab be used to determine the approximate value of the fugacity of a gas from its pressure and
molar volume. (with moderate pressure conditions)

Example:
The density of gaseous ammonia at 473 K and 50 bara is 24.3 kb/m3. Estimate its fugacity.
Solution:
The molar volume of ammonia under the given conditions is
V = 17/ (24.3 x 1000) m3/kmol
Pressure, P = 50 x 10E5 N/m2
Using Eq. (6.142), we get
f =17 ׿ ¿

Fugacities of Solids and Liquids

Every solid or liquid has a definite vapor pressure although it may be immeasurably small, in some cases. At this
pressure, the solid (or the liquid) will be in equilibrium with its vapor. 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 later. By this criterion the molar free energy of the liquid (or
the solid) in equilibrium with its vapor is equal to the molar free energy of the vapor. That is, G L= 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

Page 9 of 36
the fugacity as G = RT ln f + C, where C is constant that depends only temperature (but independent with respect to
pressure), it follows that
L V S V
f =f ∧f =f
(6.143)
Equation (6.143) means that the fugacity of solid (or liquid) is equal to the fugacity of the vapor with which it is in
equilibrium, provided that the reference state is taken to be the same in each case. If the vapor pressure is not very
high, the fugacity of the vapor would be equal to the vapor pressure; hence, the fugacity of a liquid (or a solid) is
approximately equal to its vapor pressure.
If the vapor pressure is very high and the vapor pressure cannot be treated as ideal gas its fugacity is related to the
saturation pressure as in Eq. (6.142)
2
V ( PS )
sat
f =
RT
(6.142)
PS is the saturation pressure of the gas and f sat is the saturation fugacity. The latter should in turn be equal to the
fugacity of solid or liquid at the desired temperature and 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 readily
obtained as
f V
ln = ( P−P S )
f sat
RT
(6.144)
Where V is the molar volume of the liquid.

Example:

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 x E-3 m3/kg.
Solution:
The molar volume is
V = 1.004 x 10E-6 x 18 = 18.072 x 10E-6 m3/mol
Assuming that vapor behaves as an ideal gas, we have

PS =4.241 kPa=0.0424 ¯¿ f sat

Using Eq. (6.144)

f 18.072 ×10−6
ln = × ( 10−0.0424 ) ×105=7.1435× 10−3
f
sat
8.314 × 303

Therefore, f = 0.0427 bar

Example:
Calculate the fugacity of n-butane in the liquid state at 350 K and 60 bar. The vapor pressure of n-butane at 350 K is
9.35 bar. The molar volume of saturated liquid at 350 K is 0.1072x10E-3 m3/mol. The fugacity coefficient for the
saturated vapor at 350 K is 0.834.
Solution:
The fugacity of saturated vapor at 350 K = 0.834 x 9.35 = 7.798 bar. Therefore, fugacity of saturated liquid at 350 K =
7.798 bara = fsat. Using Eq. (6.144),

f 0.1072 ×10−3
ln sat = × ( 60−9.35 ) ×105=0.18659
f 8.314 × 350
Thus the fugacity of the liquid at 60 bara and 350 K, f =9.4 bara

ACTIVITY

Page 10 of 36
The vapor pressure 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
f
a= 0
f
(6.145)
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 of state from the standard state to the given conditions is related to the activity of the substance as

f
∆ G=RT ln =RT ln a
f0
(6.146)
Since, dG = V dP – S dT, the change in the free energy when the substance is compressed isothermally (dT=0) is given
by
∆ G=∫ V dP
(6.147)
Assuming that the substance is incompressible (V=constant) between the standard state pressure P⁰ and the given
pressure P, Eq. (6.147) can be integrated as
0
∆ G=V ( P−P )
(6.148)
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
V 0
ln a= ( P−P )
RT
(6.149)
The concept of activity is particularly useful in the study of solutions. The commonly used standard states are their
properties are discussed in details later.

Example:
Determine the activity of solid magnesium (MW=24.32) at 300 K and 10 bara if the reference state is 300 K and 1
bara. The density of magnesium at 300 K is 1.745 E-3 kg/m3 and is assumed constant over this pressure range.
Solution:
Using Eq. (6.149), we obtain
5
24.32×(10−1) ×10 −3
ln a= 6
=5.029× 10
1.745× 10 x 8.314 ×300
Therefore, a = 1.00504

Effect of Pressure and Temperature on Activity

From Eq. (6.146) we see that

0
∆ G=G−G =RT ln a

G G0
R ln a= −
T T
(6.150)
Differentiating with respect to T at constant pressure,

Page 11 of 36
 ( )( ) ( ) ( )
0
G G
∂ ∂
R (
∂ ln a
∂T P
= )
T
∂T P
−
T
∂T P
(6.151)
Using Gibbs-Helmholtz equation in the above equation, we see that

R ( ∂ ln a
∂T P
= )
H 0 −H
RT2
(6.152)
Equation (6.152) predicts the effect of temperature on activity. Combining Eq2 (6.146) and (6.147) we get, for
constant temperature,

RT d ln a = V dP
(6.153)

R ( ∂∂lnPa ) = RTV
T
(6.154)
Equation (6.154) predicts the effect of pressure on activity

Departure functions and generalized charts

The methods for 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 use of departure functions or
residual properties. The residual properties are defined as the difference between the thermodynamic property at
the specific temperature and pressure and the property that 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

H R=H −H id
(6.155)
S R=S−S id
(6.156)
HR and SR are also known as enthalpy departure and entropy departure, respectively. These represent hypothetic
property changes because a gas cannot be both real and ideal gas at a given P and T.
Equations (6.155) and (6.156) are differentiated with respect to pressure to get the following results

( ) ( ) ( )
∂ HR
=
∂H
∂P T ∂ P T
−
∂ H id
∂P T
(6.157)

( ) ( ) ( )
∂ SR
=
∂S
∂ P T ∂P T
−
∂ S id
∂P T
(6.158)
We have already seen that the entropy and enthalpy vary with pressure as

( ∂∂ SP ) =−( ∂∂ VT )
T P
(6.24)

( ∂∂ HP ) =V −T ( ∂∂TV )
T P
(6.49)
For ideal gas V=RT/P; therefore,

Page 12 of 36
 ( ) ( ) =−RP
id id
∂H ∂S
=0∧
∂P T ∂P T
(6.159)
The molar volume V=ZRT/P, where Z is the compressibility factor. Then

( ∂∂TV ) = ZRP + RTP ( ∂∂ TZ ) = RP [ Z+T ( ∂∂ ZT ) ]

P P P
(6.160)
Substituting Eq. (6.160) into Eqs (6.49) and (6.24), the following equations are obtained:

( )∂H
∂P T
=
−R T 2 ∂ Z
P ∂T ( ) P
(6.161)

( ∂∂ SP ) =−RP [ Z+T ( ∂∂ ZT ) ]
T P
(6.162)
Equations (6.157) and (6.158) can now be written as

( ) ( ∂∂ ZT )
R 2
∂H −R T
=
∂P T P P
(6.163)

( ) = −RP ( Z−1) − RTP ( ∂∂ ZT )

R
∂S
∂P T P
(6.164)
Integrating these equations for an isothermal change from pressure P = 0, where HR = 0 and SR = 0, to pressure P,
P
R
H =−R T
2
∫( ∂∂TZ ) dp
0 P P
(6.165)

[ ( ) dPP ]
P
dP ∂Z
H =−R ∫ ( Z−1 )
R
+T
0 P ∂T P
(6.166)
The values of Z and (∂Z/∂T)P are calculated directly from 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, H R 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.
T
H =H +∫ C P dT
id 0 id

0
T
(6.167)
T
dT P
Sid =S 0+∫ CidP −R ln 0
T
0 T P
(6.168)
Equations (6.165) and (6.166) can be put into a reduced form nothing that P =PcPr, T = TcTr, dP=PcdPr and dT=TcdTr
Pr

( )
HR ∂Z d Pr
=−T 2r ∫
RTc 0
∂Tr Pr Pr
(6.169)

Page 13 of 36
 [ ( )]
Pr
SR ∂Z d Pr
=−∫ ( Z−1 )+T r
R 0
∂Tr Pr Pr
(6.170)
Z and (∂Z/∂Tr)P required for the evaluation of the integral in the above equations can be obtained from the
generalized 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 generalized enthalpy departure chart and the entropy
departure chart, respectively.
It is observed that the uncertainty in Z taken from the generalized compressibility chart is only around 2.5 percent.
However, a greater uncertainty is expected in the reduced residual properties as their calculation involves the
derivatives of Z.

THERMODYNAMIC DIAGRAMS

Types of diagrams

P-H diagram

H-T diagram
T-S diagram
H-S diagram (the Mollier diagram)

Construction of thermodynamic diagrams

Construction of T-H diagram

Construction of T-S diagram

PROPERTIES OF SOLUTIONS

We have seen in Chapter 6 that the thermodynamic properties of homogeneous pure substances depend only on
the state of the system.

The relationships developed for pure fluids are not applicable to solutions and need modification. The
thermodynamic properties of solutions and heterogeneous systems consisting of more than one phase are
influenced by the addition and removal of matter. The term solution includes homogeneous mixtures of two or more
components in the gas, liquid or solid phase. The pressure, temperature and the amount of various constituent
present determine the extensive state of solution; and pressure, temperature and composition determine the
intensive state.

Partial molar properties

The properties of a solution are not additive properties of its components. For example, the volume of a solution is
not the sum of the volume of the pure components constituting the solution. It means that when a substance
becomes part of a solution it loses its identity. But it still contributes to the property of the solution as is evident
from the fact that by changing the amount of substance, the solution property also changes. Thus we need a new set
of concepts that enable us to apply thermodynamics to solutions of variable composition. In this connection, the
concept of partial molar properties is of great use. The term partial molar property is used to designate the property
of a component when it is in admixture with one or many components. To be more precise, the partial molar
property of a particular component in a mixture measures the contribution of that component to the mixture
property.
If Mt is the total value of any extensive thermodynamic property of a solution, the partial molar property M i of the
component i in the solution is defined as

Page 14 of 36
 ( ) ( )
t
∂nM ∂M
M i= =
∂ ni T , P ,n j≠ i ∂ ni T , P , n j≠ i
(7.1)
In the Eq. (7.1), n is the total number of moles and M is the molar property of the solution. n i denotes the number of
moles of component I in solution, so that n = Σ n i.

In general, any partial molar property M i is the increase, in the property M t of the solution resulting from the
addition at constant temperature and pressure, of one mole of that substance to such a large quantity of the system
that its composition remains virtually unchanged.
It is an intensive property and its value depends only on the composition at the given temperature and pressure. The
subscript n j ≠i j indicates that the number of moles of all components in the solution other than the number of
moles of i are kept constant.

Physical meaning of partial molar properties

To understand the physical meaning of partial molar properties let us consider the partial molar volume, the simplest
partial molar property to visualize. It is the contribution that a component in the solution makes to total volume.
Consider an open beaker containing a huge volume of water.
Assume that one mole of water is added to it. The volume increases by 18 x 10E-6 m3, which is the molar volume of
pure water. If the same amount of water is added to a large amount of pure ethanol taken in the beaker, the
increase in volume will be approximately 14 x 10E-6 m3, which is the partial molar volume of water in pure ethanol.
The difference in the increase in volumes can be explained thus: the volume occupied by a given number of water
molecules depends on the molecules surrounding them. When water is mixed with a large volume of alcohol, there
is so much alcohol present that each water molecule is surrounded by pure ethanol. Consequently, the packing of
the molecules would be different from that in pure water, and the molecules occupy lesser volume.
If one mole water is added to an equimolar mixture of alcohol and water, the increase in volume of the solution
would be different from that resulted when the same quantity was added to pure alcohol.
The partial molar properties of the components of a mixture vary with composition because the environment of
each type of molecule changes as the composition changes.
The intermolecular forces also get changed resulting in the changes in the thermodynamic properties of solutions
with compositions. The variation of partial molar volumes with concentration is show in Fig 7.1 for ethanol €- water
(W) system.

We have seen that the effective molar volume of water added to the ethanol-water solution, i.e. the partial molar
volume V W in the solution is less than the molar volume VW of pure water at the same temperature and pressure.
To be specific, when pure water is added to an ethanol water solution of volume V t and allowed sufficient time for
heat exchange so that temperature remains the same as that before addition, the increase in volume of the solution

t
∆ V =∆ nW V W

where ∆nw is the moles of water added. The increase in volume is given by
(7.2)
Equation (72.) can be written
t
∆V
VW=
∆ nW
(7.3)
In this process, a finite drop of water was added which may cause a finite change in composition. If V W were to
represent a property of the solution, it must be based on data for the solution at this composition. For an
infinitesimal amount of water added, Eq. (7.3) becomes

( )
t t
∆V ∂V
V W = lim =
∆ n → 0 ∆ nW
W
∂ nW

Page 15 of 36
Since temperature, pressure and number of moles of alcohol are kept constant during addition of water,

( )
t
∂V
VW=
∂ nW T , P ,n E
(7.4)
Where nE represents the number of moles of alcohol present in the solution.
In general, the partial molar volume V i of component i is defined as

( )
t
∂V
V i=
∂ ni T , P ,n j≠ i
(7.5)
And it denotes the incremental change in mixture volume which occurs when a small quantity of component I is
added at constant pressure and temperature. The amount of I added is so small that no detectable change in
composition occurs. While the molar volume is always positive, the partial molar volume may even be negative. The
partial molar volume of MgSO4 in water at infinite dilution (i.e. in the limit of zero concentration) is – 1.4 x E-6
m3/mol which means that the addition of one mole of MgSO4 to a large volume of water results in a decrease in
volume of 1.4 x E-6 m3. The contraction may be due to breaking up and subsequently collapse of the open structure
of water as the ions becomes hydrated.
Though different from molar properties of the pure components, to get a physical picture of the concept of partial
molar properties, we can treat them as the molar properties of the components in solution. However, it is to be
borne in mind that the components of a solution are intimately intermixed and cannot have individual properties of
their own. The partial molar properties in fact, represent the contribution of individual components constituting the
solution to the total solution property as described in the following section.

Partial molar properties and properties of solution

Consider any thermodynamic extensive property (such as volume, free energy, heat capacity, etc.), its value for a
homogeneous system being completely determined by the temperature, pressure and the amounts of various
constituents present. Let M be the molar property of a solution and Mt is total property. Then, Mt = nM, where n is
n1 + n2 + n3+ … . Here, n1, n2, n3,… are the number of moles of the respective components 1,2, 3… of the system.
The solution property is a function represented by
Mt = f(P,T,n1,n2,n3,…,ni,…) (7.6)
If there is a small change in the pressure, temperature and amounts of various constituents, then

( ) ( ) ( ) ( )
t t t t
t ∂M ∂M ∂M ∂M
dM= dP+ dT + d n1 +…+ d ni +…
∂P T,N ∂T P, N ∂ n1 P ,T , n 2 ,n 3 … ∂ ni P ,T , n j≠ i
(7.7)
The subscript N in the first two partial derivatives indicate that the number of moles is kept constant, and the
subscript n j ≠i indicates that the number of moles of all components other than that of component I is kept constant.
At constant temperature and pressure, dP and dT are zero, so that Eq. (7.7) reduces to

( )
n t
∂M
d M =∑
t
d ni
i =1 ∂ ni P ,T ,n j≠ i
(7.8)
The partial derivatives appearing on the right-hand side are, by Eq. (7.1), the partial molar properties M i . That is, at
constant temperature and pressure,
n
d M t=∑ M i d ni
i =1
(7.9)
It is evident that the partial molar properties M i are not extensive properties, but are intensive properties of the
solution. They depend, therefore, not upon the total amount of each constituent, but only upon the composition, or
the relative amounts of the constituents. If we add several constituents simultaneously to a given solution at

Page 16 of 36
constant temperature and pressure, keeping the ratio of the various constituents constant, the partial molar
properties are not changed. Then, the change in property.

t
d M =M 1 d n1+ M 2 d n2+ …=( M 1 x 1 + M 2 x 2+ …)dn
(7.10)
where xi represents mole fraction of component i in the solution. Integration of Eq. (7.10) yields

t
M =( M ¿ ¿1 x1 + M 2 x 2 +…)n=M 1 n1 + M 2 n2 +… ¿ ¿
(7.11)
Therefore, Mt, the total property of the solution, is the sum of partial molar properties of the constituents each
weighted according to its number of moles. That is
M =∑ ni M i
t

(7.12)
This equation along with Eq. (7.9), which can be written in the following form serves as the relationship between
partial molar properties and total solution properties.

dM =∑ M i d ni
t

(7.13)
For one mole of the solution it can be easily shown (see Example) that
M =∑ x i M i
(7.14)
dM =∑ M i d xi
(7.15)
Thus the molar volume V of a solution made up of components 1, 2, … can be written as

V =x 1 V 1+ x2 V 2 …

We see that the partial molar property M i of any constituent may be regarded as the contribution of one mole of
that constituent to the total value of the property under the specified conditions. In other words, the partial molar
properties may be treated exactly as though they represent the molar properties of the components in the solution

Example:
Give an alternative derivation for Eq. (7.12) and (7.14) starting from Eq. (7.9)
Solution:
Equation (7.9) gives
dM =∑ M i d ni
t

Using xin = ni, d(xi,n) = ndxi + xidn and dMt = d(nM) = ndM + M dn, where M, as pointed out earlier, is the molar
property. Equation (7.9) becomes
n dM + M dn=∑ M i ( x i dn+n d xi )
On rearranging the above result, we get
( M −∑ M i x i ) dn=( ∑ M i d x i−dM ) n
N represents the total amount of various constituents and dn the changes in the total number of moles.
One is free to choose any value for n as well as dn. In short, n and dn can be independently changed.
For all possible values of n and dn, the above equation is to be satisfied. This is possible only if the terms in brackets
reduce to zero.
M −∑ M i xi =0 or M =∑ M i xi
(7.14)
∑ M i d x i−dM =0 or dM =∑ M i d xi
(7.15)
Multiplying Eq. (7.14) by n, we get
nM =M t =∑ n i M i

Page 17 of 36
(7.12)

Example:
Will it be possible to prepare 0.1 m3 of alcohol-water solution by mixing 0.03 m3 alcohol with 0.07 m3 pure water?
If possible, what volume should have been mixed in order to prepare a mixture of the same strength and of the
required volume? Density of ethanol and water are 789 and 997 kg/m3 respectively. The partial molar volumes of
ethanol and water at the desired compositions are: Ethanol = 53.6x 10E-6 m3/mol; water = 18x10E-6 m3/mol.
Solution:
Let us first find out the number of moles of ethanol and water mixed and their mole fractions in the resultant
mixture.
Moles of ethanol in the solution = (0.03 x 789x10E3)/46 = 514.57 mol
Moles of water in solution = (0.07 x 997 x 10E3)/18 = 3877.22 mol
Mole fraction of ethanol desired = 514.57/(514.57+3877.22) = 0.1172
Mole fraction of water = 1 – 0.1172 = 0.8828
Actual volume of solution is
514.57 x 53.6E-6 + 3877.22 x 18 x 10E-6 = (0.027558 + 0.06979) = 0.09737 m3
That is, by mixing 0.03 m3 alcohol with 0.07 m3 water, we would get only 0.09737 m3 of solution.
To prepare 0.1 m3 of solution the volumes to be mixed are:
Ethanol= (0.03/0.09737) x 0.1 = 0.03081 m3 and Water = (0.07/0.09737) x 0.1 = 0.07189 m3

Determination of partial molar properties

Method 1 (Analytical)
If the volume of a solution is known as a function of its composition, the partial molar volume of a constituent may
be found by partial differentiation with respect to the amount of that constituent.

( )
t
∂V
V i=
∂ ni T , P ,n j≠ i

Method 2 (Graphical)
Let Vt, the volume of the solution containing a fixed amount of one of the constituents (say, n1) is known for several
values of the amount of other constituents (say, n2). We may plot Vt against n2. See Fig. 7.2 The slop of the tangent
to the curve is (∂Vt/∂n2)P,T,n1 which, by definition is V̅ 2, the partial molar volume of component 2. The volume of
solution is assumed so large that no significant change in composition occurs when n2 is changed. This method has
limitation of not yielding values of V̅ 1 directly. Also, it is not advisable to use this method for determination of V̅ 2
when n2 is large compared to n1. The method of tangent intercepts is free from such limitations and is therefore
preferred for the determination of partial molar properties.

Method 3 (The tangent-intercept method)

This is also a graphical method widely used for the determination of partial molar properties of both components in
a binary solution. The molar volume V is plotted against mole fraction of one of the components (say, x2, the mole
fraction of component 2). To determine the partial molar volumes, draw the tangent to the curve at the desired
mole fraction. The intercept that this tangent makes with the vertical axis at x2 = 1 gives V̅ 2 and the intercept on the
vertical axis at x2 = 0 (or x1 = 1) gives V̅ 1. In Fig. 7.3 BD = V̅ 2 and AC = V̅ 1.

To prove this result, consider a binary solution containing n₁ moles of component 1 and n₂ moles of component 2.
Let the total volume be Vt and let V be molar volume. Then
Vt = nV = (n₁ + n₂) V
(7.16)
Differentiating Eq. (7.16) with respect to n₁, keeping n₂, T, and P constant (Note (uv)’=uv’+u’v and (u+v)’ = u’+v’)

( )
∂Vt
∂ n1 T , P ,n2
=V 1=V +( n1+ n2)
( ∂∂ nV )
1 T , P ,n2

Page 18 of 36
(7.17)
The mole fraction x₂ is given by
n2
x 2=
n1 +n2
(7.18)
Differentiating Eq. (7.18) with respect to n1 keeping n2 constant, we get (note (1/x)’ = – 1/x²)

( )
∂ x2
∂ n1 n2
=
−n2
(n1 +n2 ) 2
=
−x 2
n 1 + n2
Which can be rearranged as
n1 +n2 −x2
=
d n1 d x2
(7.19)
Equation (7.19) can be substituted in Eq. (7.17) to yield the following:

∂V
V 1=V −x 2
∂ x2
(7.20)
Similarly, it can be shown that
∂V
V 2=V −x 1
∂ x1
(7.21)
Since, for a binary solution, x₁ = 1 – x₂ and dx₁ = - dx₂, Eq. (7.21) can be put in another form:
∂V
V 2=V +(1−x 2 )
∂ x2
(7.22)
In Fig 7.3 the length BD = BE + ED, where BE is the slope of the tangent at P times the length PE. That is,
BE = (1 – x₂)(∂V/∂x₂)
And ED = V, the molar volume at the mole fraction x₂. Thus
BD = V + (1- x₂)(∂V/∂x₂)
Which, by Eq. (7.22) is V̅ ₂. Similarly, the length
AC = FC – FA = V – x₂ (∂V/∂x₂) = V̅ ₁
The above methods are applicable for the determination of various other partial molar properties also. Of the
various mixture properties, only the volume can be determined absolutely. For the determination of other
properties like G̅ i, H̅ i, etc., it becomes necessary to work with property changes on mixing (discussed later) like ∆G,
∆H, etc., The method of tangent intercept for the determination of, say G̅ ₁ and G̅ ₂, requires the plot of ∆G per mole
versus x₂.

Chemical potential

The chemical potential, denoted by the symbol μ, is a widely used thermodynamic property. It is used as an index of
chemical equilibrium in the same manner as temperature and pressure are used as indices of thermal and
mechanical equilibrium. The chemical potential μᵢof component I in a solution is the same as its partial molar free
energy in the solution, G̅ ᵢ. That is, chemical potential of a component i in a solution can be defined as

( )
t
∂G
μi=G i =
∂ ni T , P ,n j
(7.30)
The total free energy Gᵗ of a solution is a function of pressure, temperature and number of various components.
t
G =f (P ,T ,n 1 , … , ni , …)
(7.31)
The total differential dGᵗ is

Page 19 of 36
 ( ) ( ) ( )
t t t
∂G ∂G ∂G
dT + ∑
t
dG= dP+ d ni
∂P T ,N ∂T P, N ∂ ni T , P , n j≠ i
(7.32)
Then using Eq.(7.30),

( ) ( )
t t
∂G ∂G
dT + ∑ μi d ni
t
dG= dP+
∂P T ,N ∂T P, N
(7.33)
We have shown that for a closed system, when there is no change in the amount of various contituents,
dG =VdP−SdT (6.18)
Considering the total properties of the system,
t t t
d G =V dP−S dT
From which, it follows that

( )
∂ Gt
∂T P, N
=−S ,
t
( )
∂ Gt
∂P T ,N
=V
t

(7.34)
Eq. (7.33) can be written as
d G =V dP−S dT + ∑ μ i d ni
t t t

(7.35)
This is fundamental relationship for changes in the free energy of a solution. At constant temperature and pressure,
the change in the free energy is due to entirely to the changes in the number of moles and is given by
d G T ,P =∑ μi d ni
t

(7.36)
By reasoning analogous to that used in the derivation of Eq. (7.12), we have, at constant temperature and pressure,
G = ∑ μ i ni
t

For binary solution, he molar free energy of the solution is

G = x₁ μ₁ + x₂ μ₂
The chemical potential of a component is thus seen to be the contribution of that component to the Gibbs free
energy of the solution. The chemical potential is an important property of solution extensively used in the study of
phase and chemical equilibria.

Effect of Temperature and Pressure on Chemical Potential

Effect of temperature

Consider Eqs. (7.30) and (7.34). Differentiate Eq. (7.30) with respect to temperature. Then

[ ]
∂ μi
∂T P ,N
=
∂2 G t
∂ T ∂ ni
(7.37)
Differentiating Eq. (7.34) with respect to ni, we get

[ ]
t 2 t
∂S ∂ G
− =
∂ ni P ,T ,n j ∂ ni ∂T
(7.38)
Equations (7.37) and (7.38) imply that

[ ]
∂ μi
[ ]
t
∂S
=− =−S i
∂T P ,N ∂ ni P ,T ,n j
(7.39)
Where Si is the partial molar entropy of the component I in the solution. This result, though gives the variation of
chemical potential with temperature, can be put in a more useful form [compared with Eq. (6.73)] as follows: Since
G = H – TS, G̅ ᵢ = H̅ ᵢ – TS̅ᵢ, μᵢ = H̅ ᵢ – TS̅ᵢ
Page 20 of 36
We can write
μi−H i
−Si=
T
(7.40)
We know that (u/v)’=(u.(1/v))’=u’.(1/v)+u.(-1/v²)=(u’-u)/u²

[ ]
∂ ( μT )
∂T
i

P,N
Substituting Eqs. (7.39) and (7.40) into the above equation, we get,
=
T ( ∂∂Tμ )−μ
T2
i
i

(7.41)
[ ]
∂ ( μT )
∂T
i

P,N
=
−H i
T
2

Equation (7.41) predicts the effect of temperature on chemical potential.

Effect of pressure

Equation (7.30) and (7.34) are further differentiated to develop equations that predict the effect of pressure on
chemical potential. Differentiating Eq. (7.30) with respect to pressure, we obtain

[ ]
∂ μi 2 t
∂G
=
∂P T,N ∂ P ∂ ni
(7.42)
Differentiating Eq. (7.34) with respect to ni,
∂2 G t
∂ ni ∂ P
=
∂Vt
∂n i [ ] P ,T ,n j≠ i
=V i

(7.43)
Compare Eq. (7.42) with Eq. (7.43)

( ∂∂ μP )i

T,N
=V i
(7.44)
The rate of change of chemical potential with pressure is thus equal to the partial molar volume of the constituent.

Fugacity in solutions

The concept of fugacity was discussed in Chapter 6 with reference to pure substances. It was pointed out that
fugacity is a useful concept in dealing with mixtures. For pure fluids, te definition of fugacity is provided by Eq.
(6.118) and (6.122):
dG=RT d ¿

f
lim =1
P →0 P

The fugacity of a component I in a solution (gaseous, liquid or solid) is defined analogously by

d μi=RT d ¿
(7.51)

Page 21 of 36
 fi
lim =1
P →0 Pi
(7.52)
Here μᵢ is the chemical potential, fᵢ the fugacity and Pᵢ is the partial pressure of component i in the solution. For ideal
gas mixture, the fugacity of a component is equal to its partial pressure.
All gaseous mixtures behave ideally on approaching zero pressure. The partial pressure is defined as product of total
pressure and mole fraction of i (yᵢ) in the mixture.
Pi= y i P
(7.53)

Fugacity in gaseous solutions

The fugacity of a component I in a gaseous solution is given by Eq. (7.51). Equation (7.44) gives the effect of pressure
on chemical potential.

( )
∂ μi
∂P T,N
=V i
Where Vi is the partial molar volume of the component I in the solution. Rearranging this equation

d μi=V i dP
(7.54)
Compare Eq. (7.54) with Eq. (7.51). We get

RT d ¿
(7.55)
d¿

Subtracting d(ln P̅ i), where P̅ i is partial pressure of component i in the gas mixture, from both sides,

( )
d ln
fi
=
Pi RT
1
¿

(7.56)
Since P̅ ᵢ = yᵢ P, where yᵢ is the mole fraction,
d¿
At constant composition, d(ln yi) = 0, so that the above equation reduces to

d¿
Substituting this in Eq. (7.56) we obtain

( ) (
d ln
fi
=
Pi RT
1
V i−
RT
P
dP )
As P  0, f̅i = P̅ i, and the above equation can be readily integrated to give

( )
P

( )
fi 1
ln
Pi
=ln φi= ∫ V i− RT
RT 0 P
dP

(7.57)
Where φ̅ᵢ denotes the fugacity coefficient of a component in solution.
fi fi
φ i= =
Pi y i P
(7.58)
For a mixture of ideal gases, we have the following simple equation of state:

P V t= ( n1 +n2 +n3 + … ) RT
Page 22 of 36
 ( )
t
∂V RT
V i= =
∂ ni T , P ,n j P
Substituting this into Eq. (7.57), it follows that (for ideal gas mixture)
f i ¿ Pi = y i P
(7.59)
Which states that the fugacity of a component in a mixture of ideal gases is equal to the partial pressure of that
component in the mixture.
However, this is not true for real gases.

( )
P

( )
fi 1 RT
The equation (7.57) ln
Pi
=ln φi= ∫
RT 0
V i−
P
dP

To provides the means for computing fugacities in real gaseous solution. But this requires the evaluation of Vi as a
function of pressure, which in turn requires the knowledge of how the solution volume varies with composition at
each pressure. These types of data are rarely available, and hence rigorous calculation of fugacity in gaseous mixture
using Eq. 7.57) is rarely done.

Lewis-Randall Rule

As the calculation of fugacity in a mixture of gases through general equation Eq. (7.57) is very difficult, we devise a
model for mixtures known as the “ideal solution model” the fugacity of which can be easily evaluated. The fugacity in
actual solution is then determined by taking into account the deviation of the actual solution from this ideal model
behavior. As an ideal gaseous solution we can consider a gas mixture formed without any volume change on mixing
components. A gas mixture that follows the Amagat’s law is an ideal gaseous solution. For such solution, the volume
of the mixture is a linear function of the mole numbers at a fixed temperature and pressure. That is,

V = ∑ ni V i
t

(7.60)
Where Vi is the molar volume of pure I at the same temperature and pressure. For such ideal solutions,

( )
t
∂V
V i= =V i
∂ ni T , P ,n j
(7.61)
Note that the right-hand side of Eq. (7.57) reduces to the same result as that given by Eq. (6.128) where the residual
volume for the pure component is given by α = Vᵢ – RT/P. That is, for pure component at a temperature T and
pressure P.
fi 1 P
ln = ∫
P RT 0
V i−
RT
P
dP ( )
(7.62)
And for component i in a gas mixture at the same temperature and pressure,
P

( )
fi 1 RT
ln = ∫
Pi RT 0
V i−
P
dP
(7.63)
Subtracting Eq. (7.62) from (7.63),
P
fi P 1
ln = ∫ ( V −V i ) dP
f i Pi RT 0 i
(7.64)
Since P̅ i = yi P, we can simplify the above equation to the following form:

Page 23 of 36
 P
fi 1
ln = ∫ ( V −V i ) dP
y i f i RT 0 i
On substitution of Eq. (7.61), in the preceding equation, we get,
fi
ln =0∨f i= y i f i
yi f i
(7.65)
Which is commonly known as Lewis-Randall rule or Lewis fugacity rule. Its states that fugacity of a component in an
ideal solution is directly proportional to the mole fraction of the component in the solution. In Eq. (7.65), f i̅ is the
fugacity of the species I in an ideal gaseous solution, and fi is the fugacity of pure I evaluated at the temperature and
pressure of the mixture. Thus, we have now, f i̅ = yi fi for ideal gaseous solution and, fi̅ = P̅ i = yi P for ideal (perfect)
gases.
For a gas mixture to behave as an ideal solution, it requires only that the molar volume in the pure state and partial
molar volume in the solution be the same, or V̅ i = Vi. For the mixture to be an ideal gas it requires that V̅ i = Vi = RT/P,
which means that the molar volumes of all the components are same whether in the mixture or in the pure state.
For ideal solutions, the volumes of components may different from one another. In short, the concept of an ideal
gaseous solution is less restrictive than that of a mixture of ideal gases.

The Lewis-Randall rule is a simple equation and is therefore widely used for evaluating fugacities of components in
gas mixtures. It allows the fugacity of a component in the mixture to be calculated without any information about
the solution except its composition. However, it is not reliable because of severe simplification inherent in Amagat’s
law of additive volumes. But at high pressures it is often a very good assumption, because, at liquid like densities,
fluids tend to mix little or no change in volume. Lewis fugacity rule is valid for systems where the intermolecular
forces in the mixture are similar to those in the pure state. Thus, it can be said that this rule is valid

1.At low pressure when the gas phase behaves ideally.

2.At any pressure if the component is present in excess
3.If the physical properties of the components are nearly the same
4.At moderate and high pressure, the Lewis-Randall rule will give incorrect result if the molecular properties of the
components are widely different and the component under consideration is not present in excess.

Fugacities in Liquid solutions

Calculation of fugacity of a component in a liquid solution using (7.57) is not practical because the volumetric data at
constant temperature and pressure and composition are rarely available. These data are required for integration
over the entire range of pressure from the ideal gas state to the pressure of the solution including the two-phase
region. For calculation of fugacities in liquid solutions, another approach is used. We define an ideal solution whose
fugacity can be easily calculated knowing its composition and measure the departure from ideal behavior for the real
solution. A quantitative measure of the deviation from ideality is provided by the function known as activity
coefficient which will be discussed Section 7.6

Ideal Solutions and Raoult’s Law

A solution in which the partial molar volumes of the components are the same as their molar volumes in the pure
state is called ideal solution. There is no volume change when the components are mixed together to form an ideal
solution.
That is, for an ideal solution V = ΣxiV̅ i = Σxi Vi, where V is the molar volume of the solution, Vi and V̅ i are the molar
volume and partial molar volume respectively of the component i, and xi is the mole fraction of component i in the
solution. If a mixture of two liquids is to behave ideally, theoretical considerations reveal that the two types of
molecules must be similar. The environment of any molecule and hence the force acting on it is then not appreciably
different from that existing in the pure state. We have shown that for ideal gaseous solutions, the Lewis-Randall rule
is applicable which states that fugacity of each constituent is directly proportional to the number of moles of the
constituent in the solution. The Lewis-Randall rule is applicable to ideal liquid solution also. It can be written as

f i= y i f i
Page 24 of 36
(7.66)
Where f̅i is fugacity of component i in the solution, fi is the fugacity of i in the pure state, and xi is the mole fraction
of component I in the solution.
While the ideal solution model is adequate for many gas mixtures for reasonable temperature and pressure, the
same is not true for the case of liquid solutions. Very few solutions follow Eq. (7.66) over the entire composition
range. Ideal liquid solution behavior is often approximated by solutions comprised of molecules not too different in
size and chemical nature. Thus a mixture of isomers (e.g. ortho-, meta-, and para-xylene), adjacent members of
homologous series of organic compounds (e.g. n-hexane and n-heptane, ethanol and propanol, benzene and
toluene, ethyl bromide and ethyl iodide) etc., are expected to form ideal solutions.

Raoult’s Law

The criterion of phase equilibria permits us to replace the liquid phase fugacities fi and fi with fugacities in the gas
L V
phase with which the liquid is in equilibrium. Thus, f i =f i under equilibrium. Here superscripts V and L refer to the
vapor phase and liquid phase respectively. Thus, fugacity f i̅ in Eq. (7.66) is equal to the fugacity of constituent I in the
vapor phase. If the vapor phase is assumed to be ideal gas, which is true if the pressure is not too high, the vapor
V
phase fugacity f i is the same as partial pressure P̅ i of component in the vapor. If the liquid phase is pure I, the
S
fugacity of pure I in the vapor phase can be replaced with the vapor pressure Pi . Under these conditions the Lewis-
Randall rule, Eq. (7.66), becomes
S
Pi=x i P i
(7.67)
This expression is known as Raoult’s Law. This is a simplified form of the Lewis-Randall rule. Whereas the Lewis-
Randall rule is obeyed by all ideal solutions, the Raoult’s Law is applicable to ideal solution if the vapor phase with
which it is in equilibrium is an ideal gas.
Raoult’s law provides a very simple expression for calculating the fugacity of a component in the liquid mixture
which is the same as the partial pressure of the component in the vapor. It says that the partial pressure is directly
proportional to the mole fraction in the liquid solution. Ideal solutions which conform to Raoult’s law over the entire
range of concentrations are rare. A frequently cited example for ideal solutions is mixture of optical isomers of
organic compounds. Raoult’s law applies only over a limited concentration range.

Henry’s Law and Dilute Solutions

Solutions conforming to Raoult’s Law over the entire concentration range are rare as pointed out earlier. A solution,
any of whose components does not obey Raoult’s law is designated as non-ideal solution. Even non-ideal solutions
exhibit a common form of ideal behavior over a limited concentration range where the fugacity f i̅ (or, the partial
pressure P̅ i) is directly proportional to the concentration in the liquid. This behavior is exhibited by the constituent as
its mole fraction approaches zero, and is generalized by Henry’s law.

f i=x i K i
(7.68)
Pi=x i K i
(7.69)
Often, the solute portion of non-ideal liquid solution can be assumed to follow Henry’s law. P̅ i is the partial pressure
of the solute over the solution, xi is its mole fraction in the solution and Ki is a proportionality constant known as
S
Henry’s law constant. Ki may be greater or less than Pi , the vapor pressure of the solute at the temperature and
total pressure in question. When Ki and Psi are equal, Henry’s law and Raoult’s law are identical. Henry’s law may be
thought of as a general rule of which Raoult’s law is special case. Henry’s law is obeyed in all solutions by the solute
at extremely low concentrations. Essentially all liquids will obey Henry’s law close to mole fraction zero, but many
will deviate from the law above 0.01-0.02 mole fraction. And almost all liquids deviate above 0.1 mole fraction. But
in some exceptional cases, Henry’s law is found to be obeyed quite well up to xi = 0.5.
For ideal solutions, the partial fugacity (or partial pressure) of a component is proportional to its mole fraction. For a
real solution it has been found experimentally that as the mole fraction of the component approaches unity, its

Page 25 of 36
fugacity approximates to the value for an ideal solution, though at lower mole fractions, the behavior departs
markedly from ideal behavior.
In Fig 74. The fugacity curve becomes asymptotic to the straight line showing ideal behavior as mole fraction
approaches unity. In a dilute solution, the component present in larger proportions designated as solvent, obeys
Raoult’s law even though it may depart from ideal solution behavior in a more concentrated solution. As the mole
fraction of the solute – the component present in smaller proportions – approaches zero, it will conform to the ideal
behavior predicted by Henry’s law. Thus, we can generalize by saying that the solute in a dilute solution obeys
Henry’s law and the solvent obeys Raoult’s law. It can be show that over the range of compositions where the
solvent obeys Raoult’s law. It can be shown that over the range of compositions where the solvent obeys Raoult’s
law, the solute obeys Henry’s law

Ideal behavior of Real Solutions

The ideal behavior exhibited by non-ideal solutions can be summarized by the following mathematical statements

fi
lim =f [ Lewis−Randall rule]
xi →1 xi i
(7.70)
fi '
lim =f [Henr y s law ]
xi → 0 xi i
(7.71)

Henry’s Law and Gas Solubility

Since the solubility of the gases in liquids is usually very low, the mole fraction of a gas in a saturated liquid solution
is very small. The solute gas obeys Henry’s law and therefore its fugacity (or the partial pressure) would be directly
proportional to its mole fraction, the proportionality constant being the Henry’s law constant (Eq. 7.69). In other
words, the mole fraction or the solubility of the gas in the liquid is proportional to the partial pressure of the gas
over the liquid as given by
Pi
x i=
Ki
(7.72)
Where Ki is the Henry’s law constant.

Activity in solutions

The activity with reference to pure substance was defined (see Eq. (6.145)) and the concept was discussed in Chapter
6.
The activity of a component in a solution can be defined in a similar way. It is the ratio of fugacity of a component in
the solution in a given condition to the fugacity of that component in the standard state. It is denoted by ai:
fi
a i= 0
fi❑
(7.73)
Since the fugacities are related to the chemical potential as

μi=RT ln f i +C

μ0i =RT ln f 0i +C
It follows that
fi
∆ μi =RT ln =RT ln ai
f 0i

Page 26 of 36
(7.74)
∆ μi = μi –μ0i is the increase in the chemical potential of species I when it is brought into solution from its standard
state.
The concept of activity plays an important role in solution thermodynamics because activity can be related to
compositions directly. For example, let the standard state of a substance be the pure component at the temperature
and pressure of the solution. Then the activity of that component becomes equal to its mole fraction in the case of
ideal solutions and is a strong function of mole fraction in the case of real solutions
fi fi
a i= 0
=
f i
fi
For ideal solutions as f i=x i f i , the activity ai = xi. For real solution, the activity can be shown to be equal to the
product of activity coefficient and mole fraction. The activity coefficient is discussed later
The term activity is a ratio without dimensions. It is a widely used function in solution thermodynamics, particularly
in dealing with property changes of mixing. The relationship between property change of mixing and activity is
discussed later.

Selection of Standard States

The numerical values of activity depend upon the choice of the standard state, this choice being based largely on
experimental convenience and reproducibility. For all such standard states, the temperature is the same as the
temperature of the solution under study and it is not a fixed value. Following are the commonly accepted standard
states:

Gases:

Two standard states are common:

1.The pure component gas in its ideal state at 1 bara. At this state, the fugacity is unity if expressed in bara. The
activity becomes
fi fi
a i= = =f i
f0
i
1
That is, the activity of a component in a mixture of gases is equal to its fugacity, numerically. If the mixture behaves
as an ideal gas at the given conditions the activity and partial pressure are the same. This standard state is used in
the study of chemical reaction equilibrium.

2.The pure component gas at the pressure of the system. With this choice the activity of each component in ideal
gas solution becomes equal to its mole fraction.

fi f i xi f i
a i= = = =xi
f0
i
fi fi

This standard state becomes hypothetical at temperatures where the total pressure exceeds the saturation pressure
of the component gas in the pure state. Vapor-liquid equilibrium studies conventionally use this standard state.

Liquids:

Two standard states are common for liquids also.

1.The pure component liquid at a pressure of 1 bara. This state is hypothetical if the vapor pressure of the pure
liquid exceeds 1 bara.

2.The pure liquid at the pressure of the system. This state becomes hypothetical at temperatures above the critical
or saturation temperature of the pure liquid. This standard state is used in vapor-liquid equilibrium studies.

Solids
Page 27 of 36
The standard state chosen for solid is usually the pure component in the solid state at a pressure of 1 bara.

ACTIVITY COEFFICIENTS

We have already seen that the concept of ideal solution enables us to calculate the fugacity of a component in the
liquid solution from the knowledge of its concentration in the solution and its fugacity in the pure state. The
calculation of fugacity of a component in a real solution should take into account the degree of departure from ideal
behavior.
Activity coefficients measure the extent to which the real solution departs from ideality. Activity coefficient of the
component I in solution is denoted by γᵢ and is defined by the following relationship

f i=γ i x i f i0
(7.75)
0
Where f i is the fugacity in the standard state. For ideal solutions γi = 1, and we have

f i=x i f 0i
(7.76)
Which is the same as the Lewis-Randall rule (Eq.(7.66)) with the pure liquid at the system pressure as the standard
state.
Two types of ideal behavior are observed; the first conforms to Lewis-Randall rule (or Raoult’s law) in which case
0
f i =f i , the fugacity of the pure species at the system pressure and the second type conforms to an ideal dilute
0
solution behavior (the Henry’s law), in which case f i =K i , the Henry’s law constant. Depending upon the standard
state on which they are based, the activity coefficients can take different numerical values. For standard state in the
sense of Lewis-Randall rule or Raoult’s law,
f i=γ i x i f i
(7.77)
fi a
γ i= = i
xi f i xi
(7.78)
a i=γ i x i
(7.79)
Where ai is the activity of I in the solution. Eq. (7.77) is, in fact, Lewis fugacity rule modified by the factor γi to correct
for deviation from ideality. This equation should reduce to Raoult’s law as x approaches unity and to Henry’s law as x
approaches to zero. For this to be possible, γ must equal unity as mole fraction approaches unity (Raoult’s law
region) and Ki/fi , as mole fraction, approaches zero (Henry’s law region). In terms of partial pressure Eq. (7.77) may
well be written as
S
Pi=γ i x i Pi
(7.80)
If the hypothetical state, where the pure component fugacity = Henry’s law constant, is chosen as the standard state,
we get,
'
f i=γ i x i K i
(7.81)
Pi=γ 'i x i K i
(7.82)
Then the activity coefficient approaches unity as x approaches zero. In Eq. (7.81) and (7.82), γ’I is the activity
coefficient referred to infinite dilution.
When activity coefficients are defined with reference to an ideal solution in the sense of Raoult’s law, then for each
component I,
γᵢ → 1 as xᵢ → 1
On other hand, if activity coefficients are defined with reference to an ideal dilute solution, then
γᵢ → 1 as xᵢ → 1 (solvent)

Page 28 of 36
 γ’₂ → 1 as x₂ → 1 (solute)
Activity coefficients with reference to ideal dilute solution would be useful when dealing with liquid mixtures that
cannot exist over the entire composition range as happens, for example, in a liquid mixture containing gaseous
solute. If the critical temperature of the solute is lower than the temperature of the mixture, then a liquid phase
cannot exist as x2 → 1, and the relations based on an ideal mixture in the sense of Raoult’s law can be used only by
introducing a hypothetical standard state for component 2. However, relations based on an ideal dilute solution
eliminate this difficulty.
Activity coefficients are very strong functions of concentration of solution. The variation of γ with x over the entire
range of composition is usually complex, but can often be roughly approximated in binary solutions by the empirical
equations such as the one proposed by Porter:

2 2
ln γ 1=b x 2 , ln γ 2=b x 1

Where b is an empirical constant. These relationships apply best when the components are not too dissimilar in
structure and polarity.

Effect of Pressure on Activity Coefficients

The effect of pressure on fugacity was derived in Chapter 6. (Eq. 6.126)

( ∂∂lnPf ) = RTV
i

T
i

Here fi is the fugacity of pure I and Vi is its molar volume. In a similar way it can be shown that, f i̅ , the fugacity of I in
solution varies with pressure according to

( ∂∂lnPf ) = RTV
i

T
i

(7.83)
Combining Eq. (6.126) with Eq. (7.83), we get

( )
fi
∂ ln
f i V i−V i
=
∂P T RT
(7.84)
According to Eq. (7.77) fi̅ /fi = γi xi so that Eq. (7.84) can be written as

( ∂ ln γ i x i
∂P T
=)V i−V i
RT
(7.85)
As the mole fraction xi is independent of pressure (∂ln xi/∂P)=0, and hence

( ∂P T )
∂ ln γ i
=
V i −V i
RT
(7.86)
The molar volume Vi and Vi correspond to the particular phase under consideration. For liquid solutions, the effect
of pressure on activity coefficients is negligible at pressure below atmospheric. For gaseous mixtures, activity
coefficients are nearly unity at reduced pressured below 0.8.

Effect of Temperature on Activity Coefficients

The effect of temperature on fugacity of a pure substance was given by Eq. (6.125) as

( ∂ ln f i
∂T P )
=
H i0−H i
RT2
Similarly, for the substance in the solution

Page 29 of 36
 ( )
0
∂ ln f i H i −H i
= 2
∂T P RT
(7.87)
Combining the above two equations, and noting that f i̅ /fi = γi xi where xi is independent of temperature,

( ∂∂Tln γ ) = HR−H
i

T P
i
2
i

(7.88)
Equation (7.88) gives the effect of temperature on activity coefficients. The term (Hi – H̅ i) is the partial heat of mixing
of component I from its pure state to the solution of given composition both in the same state of aggregation and
pressure. For gaseous mixtures, this term is negligible at low pressure.

Example:
The partial pressure of acetone (A) and chloroform (B) were measured at 298 K and are reported below:
xA 0 0.2 0.4 0.6 0.8 1.0
,bara 0 0.049 0.134 0.243 0.355 0.457
,bara 0.386 0.288 0.187 0.108 0.046 0

Calculate the activity and activity coefficient of chloroform in acetone at 298 K.

(a) Base on the standard state as per Lewis-Randall rule
(b) Based on Henry’s law constant
Solution:
The Henry’s law constant was determined in Example 7.11. KB = 0.217 bar. The vapor pressure of pure chloroform,
PSB = 0.386 bara. The activity was defined by Eq. (7.73) and activity coefficient by Eq. (7.75). Combining these two
we get,
ai = γi xi=
Based on the Lewis-Randall rule, the activity,
Pi
a i= S
Pi
(7.89)
Based on the Henry’s law, the activity,

Pi
a ' i=
Ki
(7.90)
The activity coefficient based on the Lewis-Randall rule is
fi ai
γ i= 0
=
f i xi xi
(7.91)
The activity coefficient based on the Henry’s law is
fi a 'i
γ ' i= =
K i x i xi
(7.92)
The above equations are used to calculated the activity and activity coefficients for different concentrations. A
sample calculation is provided below for the second set where
xA = 0.2, xB = 0.8, P̅A= 0.049 bara, P̅ B = 0.288 bara, KB = 0.211 bara, PSB = 0.386 bara.

(a) For the standard state referred to the Lewis-Randall rule:

PB 0.288
a B= S
= =0.75
P B
0.386

Page 30 of 36
 a B 0. 75
γB = = =0. 94
x B 0. 8

(b) For the standard state referred to the Henry’s law:

P B 1.288
a ' B= = =1.33
K B 0. 217

a ' B 1.33
γ 'B = = =1.66
x B 0.8
The above calculations are repeated for other concentrations and the results are given below:

xB 0 0.2 0.4 0.6 0.8 1.0

a 0 0.12 0.28 0.48 0.75 1.0
a’ 0 0.21 0.50 0.86 1.33 1.78
γ 0.60 0.70 0.80 0.94 1.0
γ’ 1.0 1.05 1.25 1.43 1.66 1.78

Page 31 of 36
PHASE EQUILIBRIA

A system is said to be in a state of equilibrium if it shows no tendency to depart from that state either by energy
transfer through the mechanism of heat and work or by mass transfer across the phase boundary.
Since a change of state is caused by a driving force, we can describe a system at equilibrium as one in which there
are no driving forces for energy or mass transfer. That is, for a system in a state of equilibrium, all forces are in exact
balance. It may be noted here that the state of equilibrium is different from a steady state condition.
Under steady state there exist net fluxes for material or energy transfer across a plan surface placed anywhere in the
system.
Under equilibrium the net flux is zero.
Transfer of material or energy across phase boundaries occurs till equilibrium is established between the phases. In
our daily experience, we come across a number of processes in which materials are transferred from one phase to
another. During breathing we take oxygen from air through the lungs and dissolve it in the blood. During the
preparation of tea or coffee we extract the soluble components in the powder into boiling water. Dilute aqueous
solution of alcohol is concentrated by distillation in which vapor rich in alcohol is produced from boiling solution. The
phase equilibrium thermodynamics is of fundamental importance in many branches of science, whether physical or
biological. It is particularly important in chemical engineering, because majority of manufacturing processes involve
transfer of mass between phases either during the preparation of the raw materials or during the purification of the
finished products. Gas-liquid absorption, distillation, liquid-liquid extraction, leaching, adsorption, etc., are some of
important separation techniques employing mass transfer between phases.
In addition to these, many industrial chemical reactions are carried out under conditions where more than one
phase exist. A good foundation in phase equilibrium thermodynamics is essential for the analysis and design of these
processes.
In this chapter due emphasis is given to the development of the relationship between the various properties of the
system such as pressure, temperature and composition when a state of equilibrium is attained between the various
phases constituting the system. The temperature-pressure-composition relationships in multiphase system at
equilibrium form the basis for the quantitative treatment of all separation processes.

The two types of phase equilibrium problems that are frequently encountered are:

1.The determination of composition of phases which exist in equilibrium at a known temperature and pressure
2.The determination of conditions of temperature and pressure required to obtained equilibrium between phases of
specified compositions.

Criteria of phase equilibrium

Consider a homogeneous closed system in a state of internal equilibrium. The criteria of internal thermal and
mechanical equilibrium are that the temperature and pressure be uniform throughout the system. For a system to
be in thermodynamic equilibrium, additional criteria are to be satisfied.

Page 32 of 36
Consider a closed system consisting of two phases of a binary solution, for example, the vapor and liquid phases of
an alcohol-water solution. The requirement of uniformity of temperature and pressure does not preclude the
possibility of transfer of mass between the phases.
If the system is in thermodynamic equilibrium mass transfer also should not occur. It means that additional criteria
are necessary for establishing the state of thermodynamic equilibrium.
A system can interact with the surrounding reversibly or irreversibly.
In the reversible process, a state of equilibrium is maintained throughout the process. So it can be treated as a
process connecting a series of equilibrium states. The driving forces are only infinitesimal in magnitude and the
process can be reversed by infinitesimal changes in the external conditions.
However, all irreversible processes tend towards a state of equilibrium. We have shown in above Chapter un
“Clausius inequality”
dQ
dS ≥
T
(4.44)
In this equation, the equality sign refers to a reversible process which can be treated as a succession of equilibrium
states and the inequality refers to the entropy change for a spontaneous process whose ultimate result would be an
equilibrium state.
The first law of thermodynamics expressed mathematically by Eq. (2.5) can be written as
dQ = dU + dW (8.1)
Substituting Eq. (8.1) into Eq. (4.44), we get
T dS ≥ dU + dW
dU ≤ T dS – dW
(8.2)
dW in Eq. (8.2) may be replaced by P dV so that
dU ≤ T dS – P dV (8.3)
Eq. (8.3) is valid for cases where external pressure is the only force and the work is, therefore, the work of expansion
only. By this, we exclude other effects like those due to gravitational and electromagnetic fields and surface and
tensile forces. Equation (8.3) can be treated as the combined statement of the first and second law of
thermodynamics applied to a closed system which interact with its surroundings through heat transfer and work of
volume displacement.
This equation is utilized for deriving the criteria of equilibrium under various sets of constraints, each set
corresponding to a physically realistic or commonly encountered situation. These different criteria are discussed
now.

Constant U and V.

An isolated system does not exchange mass, heat or work with the surroundings. In Eq. (8.1), dQ = , dW = 0 and
hence dU = 0.
A well-insulated vessel of constant volume would closely approximate this behavior. Thus in Eq. (8.3) dU = 0 and dW
= 0 so that
d S U ,V ≥ 0
(8.4)
The entropy is constant in a reversible process and increases in a spontaneous process occurring in a system of
constant U and V. Since an irreversible process leads the system to an equilibrium state, the entropy is maximum at
equilibrium when no further spontaneous processes are possible.

Constant T and V:

Helmholtz free energy is defined by Eq. (6.1)

A = U – TS
Rearranging Eq. (6.1), we get
U = A + TS
dU = dA + T dS + S dT
Substitute this result in Eq. (8.3)( dU ≤ T dS – P dV) and rearrange the resulting expression to the following form

Page 33 of 36
 dA + T dS + S dT ≤ T dS – P dV
dA ≤ – P dV – S dT
(8.5)
Under the restriction of constant temperature and volume, the latter implying no work, the equation simplifies to

d A T ,V ≤ 0
(8.6)
Equation (8.6) means that the spontaneous process occurring at constant temperature and volume is accompanied
by a decrease in the work function and consequently, in a state of thermodynamic equilibrium under these
conditions the Helmholtz free energy or the work function is a minimum.

Constant P and T

Equation (6.6) defines Gibbs free energy as

G = H – TS
Since H = U + PV we can write Eq. (6.6) as
G = U + PV – TS
Taking the differentials
dG = dU + P dV + V dP – T dS – S dT
Rearranging these as
dU = dG – P dV – V dP + T dS + S dT
and combining this result with Eq. (8.3) (dU ≤ T dS – P dV), we obtain
dG – P dV – V dP + T dS + S dT ≤ T dS – P dV
dG ≤ V dP – S dT
(8.7)
At constant temperature and pressure, Eq. (8.7) reduces to

d GT , P ≤ 0
(8.8)
Equation (8.8) means that the free energy either decreases or remains unaltered depending upon whether the
process is spontaneous or reversible. It implies that for a system in equilibrium at a given temperature and pressure
the Gibbs free energy must be minimum.

Since most chemical reactions and many physical changes are carried out under conditions of constant temperature
and pressure, Eq. (8.8) is the commonly used criterion of thermodynamic equilibrium. It also provides a very
convenient and simple test for the feasibility of a proposed process. No process is possible which results in an
increase in Gibbs free energy of the system, because according to Eq. (8.8) the Gibbs free energy always decreases in
a spontaneous process and in the limit of the reversible process, the free energy doesn’t change at all.

In the equilibrium state, differential variations can occur in the system at constant temperature and pressure
without producing any change in the Gibbs function. Thus, the equality in Eq. (8.8) can be used as the general
criterion of equilibrium or as a thermodynamic statement that characterizes the equilibrium state.

dG = 0 (at constant T and P) (8.9)

To apply this criterion for phase equilibrium problems we need formulate an expression for dG as function of the
number of moles of the components in various phases and set it equal to zero.
This equation along with the mass conservation equations provides the solutions to phase equilibrium problems.

Criterion of stability

It can be shown that the criterion of equilibrium Eq. (8.8) can be used to formulate the criterion of stability for a
binary mixture. When two pure liquids at a given temperature T and pressure P are mixed together, the resulting
mixture should have a lower free energy at the same temperature and pressure. This is because the mixed state is an

Page 34 of 36
equilibrium state or stable compared to the unmixed state. The molar free energy of the mixture is thus less than the
sum of the molar free energies of the constituents for all possible concentrations. That is,
G−∑ xi G i< 0
(8.10)
The left-hand side in the above equation is the free energy change on mixing ∆G. Therefore,
∆G < 0 (8.11)
When the free energy change on mixing ∆G is plotted against x1 – the mole fraction of constituent 1 in the binary
mixture – the resulting curve is one of the two types shown in Fig. 8.1. The upper curve is for binary mixture, which is
miscible for the entire concentration range. Assume that the points A and B represent two binary mixtures of
composition xA and xB respectively. Points on the dotted line AB represent the composition as well as ∆G of the
mixture of two phases obtained when solutions represented by the points A and B are mixed together. Since the line
AB is above the solid curve that represents the free energy of the miscible solution, the free energy of the mixture in
the miscible state is the minimum and the mixture exists as a single homogeneous phase. However, this argument is
not valid for the lower curve in Fig . The dotted line MN represents the free energy of the two-phase mixture
obtained when two binary mixtures of composition xM and xN, respectively, are mixed together. It lies below the ∆G
curve of the homogeneous solution. Any point on the line MN represents the ∆G that would result for systems
consisting of two phases of mole fraction xM and xN. Thus, when the system moves from solid curve to the dotted
line MN, there is a decrease in the free energy. That is, the system attains stability when it moves from a
homogeneous to s heterogeneous state. Therefore, for mixtures of composition between points M and N, the
equilibrium or stable state consists of two immiscible phases. We see that the second derivative of ∆G with
respective t x1 is always positive for stable liquid phase and if it becomes zero or negative, phase separation occurs.
This criterion of stability is that at constant temperature and pressure the free energy change on mixing ∆G, its first
and second derivatives are all continuous functions of the concentration x and

2
d ∆G
>0
d x2
at constant T and P

Example:

Page 35 of 36
Page 36 of 36