2.8 Thermodynamics of Real Gases
111
with μ ≡ G/N being the chemical potential for a pure substance in a specified
phase. In the event, for example, that we have both liquid and gaseous phases of
the substance present at the same point (v, P ) of the P v-diagram, the total Gibbs
energy is given by G = G + G g and the total number of atoms/molecules is given
by N = N + N g , in which the subscripts , g denote the liquid and gas phases,
respectively. The total differential of G at the specified point (v, P ) in the phase
diagram can then be expressed as
dG =
∂G
∂N
N g
dN +
∂G g
∂N g
N
dN g .
(2.8.25a)
Now, as the total number of atoms/molecules of the substance is fixed, dN ≡ 0,
and dN = −dN g . Given that the partial derivatives in expression (2.8.25a) for dG
define the chemical potentials μ and μ g for the two phases of the substance under
consideration, we may express dG as
dG = (μ g − μ ) dN g .
(2.8.25b)
At thermodynamic equilibrium, dG = 0 and, as dN g = 0, thermodynamic
equilibrium between two phases of the same substance then requires that the
chemical potentials associated with those phases be equal, i.e., in the present case,
μ = μ g .
Returning to Fig. 2.9, let us focus upon the T = 286.25 K isotherm and examine
Eq. (2.8.24) for that portion lying between the two (yellow) open circle symbols,
both of which correspond to a fixed pressure P 0 and which, for convenience, we
shall designate as the initial, i, and final, f, values. Let us integrate both sides of
Eq. (2.8.24) over this interval, noting that the first integral on the right-hand side
vanishes (because dT ≡ 0) and carrying out an integration by parts of the second
integral, to obtain
μ f
μ i
dμ = (P v)
f
i
−
v f
v i
P dv
or
μ − μ g = P 0 (v f − v i ) −
v f
v i
P dv
=
v f
v i
(P 0 − P ) dv .
(2.8.26)
Hence, under thermodynamic equilibrium conditions for which μ = μ g , the areas
above and below the horizontal line (P = P 0 ) in Fig. 2.9 must be equal in order for
the equilibrium condition μ = μ g for phase co-existence to be satisfied.
111
with μ ≡ G/N being the chemical potential for a pure substance in a specified
phase. In the event, for example, that we have both liquid and gaseous phases of
the substance present at the same point (v, P ) of the P v-diagram, the total Gibbs
energy is given by G = G + G g and the total number of atoms/molecules is given
by N = N + N g , in which the subscripts , g denote the liquid and gas phases,
respectively. The total differential of G at the specified point (v, P ) in the phase
diagram can then be expressed as
dG =
∂G
∂N
N g
dN +
∂G g
∂N g
N
dN g .
(2.8.25a)
Now, as the total number of atoms/molecules of the substance is fixed, dN ≡ 0,
and dN = −dN g . Given that the partial derivatives in expression (2.8.25a) for dG
define the chemical potentials μ and μ g for the two phases of the substance under
consideration, we may express dG as
dG = (μ g − μ ) dN g .
(2.8.25b)
At thermodynamic equilibrium, dG = 0 and, as dN g = 0, thermodynamic
equilibrium between two phases of the same substance then requires that the
chemical potentials associated with those phases be equal, i.e., in the present case,
μ = μ g .
Returning to Fig. 2.9, let us focus upon the T = 286.25 K isotherm and examine
Eq. (2.8.24) for that portion lying between the two (yellow) open circle symbols,
both of which correspond to a fixed pressure P 0 and which, for convenience, we
shall designate as the initial, i, and final, f, values. Let us integrate both sides of
Eq. (2.8.24) over this interval, noting that the first integral on the right-hand side
vanishes (because dT ≡ 0) and carrying out an integration by parts of the second
integral, to obtain
μ f
μ i
dμ = (P v)
f
i
−
v f
v i
P dv
or
μ − μ g = P 0 (v f − v i ) −
v f
v i
P dv
=
v f
v i
(P 0 − P ) dv .
(2.8.26)
Hence, under thermodynamic equilibrium conditions for which μ = μ g , the areas
above and below the horizontal line (P = P 0 ) in Fig. 2.9 must be equal in order for
the equilibrium condition μ = μ g for phase co-existence to be satisfied.