4
1 Nucleation Theory
The Clausius–Clapeyron equation can be used to find the relationship between
pressure and temperature along phase boundaries [2, 3]. Feynman had a unique way
of explaining key thermodynamic concepts in his lecture series (44–46), including the
Clausius–Clapeyron equation in series (45) [2]. His illustrations of the Carnot cycle,
the efficiency of an ideal engine, and the ratchet model are among the most elegant
explanations I have come across on these topics, and the readers are recommended
to read the Feynman lecture series for an alternative derivation. Here, we follow the
approach adopted by Reif [3].
The general expression for the Gibbs free energy is [3, 4]
G ≡ U − T S + PV
(1.1.5)
where G is the Gibbs free energy, U is the internal energy, T is the system temperature,
S is the system entropy, P is the system pressure, and V is the system volume. The
Gibbs free energy of a system is also equal to the total of the chemical potential of
all the molecules in the system.
G = μN
(1.1.6)
where μ is the chemical potential and N is the number of the molecules in the system.
Now we consider two phases of a closed single-component system, (1) and (2), that
have the chemical potential of μ 1 and μ 2 , respectively. If we denote the number of
the molecules in each phase as N 1 and N 2 , then the total Gibbs free energy of the
system is
G = μ 1 N 1 + μ 2 N 2
(1.1.7)
Since the system is closed, the total number of molecules, N 1 + N 2 , is a constant.
Then,
dG = μ 1 dN 1 + μ 2 dN 2 = μ 1 dN 1 + μ 2 d(constant − N 1 )
= μ 1 dN 1 − μ 2 dN 1 = (μ 1 − μ 2 )dN 1
(1.1.8)
For the two phases to be in equilibrium, dG = 0, and hence, (μ 1 − μ 2 ) = 0 or
μ 1 = μ 2
(1.1.9)
In other words, the transfer of a molecule from one of the two phases to the other
does not change the system free energy when μ 1 = μ 2 .
The physical meaning of a phase boundary is almost clear now. Since G = G(P,
T ) [3, 4], if P and T were such that μ 1 > μ 2 then the minimum value of G in
Eq. (1.1.7) would be achieved if all the molecules transformed from Phase (1) to
Phase (2). Likewise, if P and T were such that μ 1 < μ 2 then the minimum value
of G in Eq. (1.1.7) would be achieved if all the molecules transformed from Phase
(2) to Phase (1). Thus, the two phases could only coexist in equilibrium when μ 1 =
μ 2 because a transformation of a number of molecules from one phase to the other
1 Nucleation Theory
The Clausius–Clapeyron equation can be used to find the relationship between
pressure and temperature along phase boundaries [2, 3]. Feynman had a unique way
of explaining key thermodynamic concepts in his lecture series (44–46), including the
Clausius–Clapeyron equation in series (45) [2]. His illustrations of the Carnot cycle,
the efficiency of an ideal engine, and the ratchet model are among the most elegant
explanations I have come across on these topics, and the readers are recommended
to read the Feynman lecture series for an alternative derivation. Here, we follow the
approach adopted by Reif [3].
The general expression for the Gibbs free energy is [3, 4]
G ≡ U − T S + PV
(1.1.5)
where G is the Gibbs free energy, U is the internal energy, T is the system temperature,
S is the system entropy, P is the system pressure, and V is the system volume. The
Gibbs free energy of a system is also equal to the total of the chemical potential of
all the molecules in the system.
G = μN
(1.1.6)
where μ is the chemical potential and N is the number of the molecules in the system.
Now we consider two phases of a closed single-component system, (1) and (2), that
have the chemical potential of μ 1 and μ 2 , respectively. If we denote the number of
the molecules in each phase as N 1 and N 2 , then the total Gibbs free energy of the
system is
G = μ 1 N 1 + μ 2 N 2
(1.1.7)
Since the system is closed, the total number of molecules, N 1 + N 2 , is a constant.
Then,
dG = μ 1 dN 1 + μ 2 dN 2 = μ 1 dN 1 + μ 2 d(constant − N 1 )
= μ 1 dN 1 − μ 2 dN 1 = (μ 1 − μ 2 )dN 1
(1.1.8)
For the two phases to be in equilibrium, dG = 0, and hence, (μ 1 − μ 2 ) = 0 or
μ 1 = μ 2
(1.1.9)
In other words, the transfer of a molecule from one of the two phases to the other
does not change the system free energy when μ 1 = μ 2 .
The physical meaning of a phase boundary is almost clear now. Since G = G(P,
T ) [3, 4], if P and T were such that μ 1 > μ 2 then the minimum value of G in
Eq. (1.1.7) would be achieved if all the molecules transformed from Phase (1) to
Phase (2). Likewise, if P and T were such that μ 1 < μ 2 then the minimum value
of G in Eq. (1.1.7) would be achieved if all the molecules transformed from Phase
(2) to Phase (1). Thus, the two phases could only coexist in equilibrium when μ 1 =
μ 2 because a transformation of a number of molecules from one phase to the other