1.1 Phase Equilibria and Phase Behavior
5
does not alter G at such combinations of P and T that satisfies the condition μ 1 =
μ 2 . In other words, the locus of points where P and T satisfy the condition μ 1 =
μ 2 represents the phase boundary between the two phases (1) and (2). On one side
of such phase boundary, μ 1 > μ 2 and Phase (2) is the thermodynamically stable
phase, whereas μ 1 < μ 2 on the other side of such phase boundary and Phase (1) is
the thermodynamically stable phase.
Is there a way to construct a phase diagram without knowing the functional form
of μ(P, T )? By definition, μ 1 is equal to μ 2 on the phase boundary, so
μ 1 (P, T ) = μ 2 (P, T )
(1.1.10)
Meanwhile, since dG = V dP − SdT [3, 4],
dμ 1 = dG 1 /N 1 = (V 1 /N 1 ) · dP − (S 1 /N 1 ) · dT = v 1 dP − s 1 dT
(1.1.11)
dμ 2 = dG 2 /N 2 = (V 2 /N 2 ) · dP − (S 2 /N 2 ) · dT = v 2 dP − s 2 dT
(1.1.12)
where v is the molecular volume and s is the entropy per molecule. From Eq. (1.1.10),
dμ 1 = dμ 2 as long as the change is along the phase boundary. Thus,
v 1 dP − s 1 dT = v 2 dP − s 2 dT
(1.1.13)
(v 1 − v 2 )dP = (s 1 − s 2 )dT
(1.1.14)
(dP/dT ) = (s 1 − s 2 )/(v 1 − v 2 ) = N A (s 1 − s 2 )/N A (v 1 − v 2 )
(1.1.15)
where N A is the Avogadro number. Thus, Eq. (1.1.15) states that the local slope
(dP/dT ) on a phase boundary is equal to the change in the molar entropy divided by
the change in the molar volume between the two phases (1) and (2) at that P and T:
(dP/dT ) = S molar //V molar
(1.1.16)
How can one find S molar ? A great property of the first-order phase transition is
that the transformation takes place at a constant temperature T and the change in the
entropy is the latent heat of the transition between the two phases (1) and (2), L 12 ,
divided by that temperature:
(dP/dT ) = L 12 /(T V molar )
(1.1.17)
As can be seen from Eq. (1.1.17), the greater the change in V molar (as in a liquid–
gas transition) the smaller the slope, dP/dT, in the phase diagram. When the change
in V molar is negative (as in the ice–liquid water transition), the slope, dP/dT, in the
phase diagram is also negative.
5
does not alter G at such combinations of P and T that satisfies the condition μ 1 =
μ 2 . In other words, the locus of points where P and T satisfy the condition μ 1 =
μ 2 represents the phase boundary between the two phases (1) and (2). On one side
of such phase boundary, μ 1 > μ 2 and Phase (2) is the thermodynamically stable
phase, whereas μ 1 < μ 2 on the other side of such phase boundary and Phase (1) is
the thermodynamically stable phase.
Is there a way to construct a phase diagram without knowing the functional form
of μ(P, T )? By definition, μ 1 is equal to μ 2 on the phase boundary, so
μ 1 (P, T ) = μ 2 (P, T )
(1.1.10)
Meanwhile, since dG = V dP − SdT [3, 4],
dμ 1 = dG 1 /N 1 = (V 1 /N 1 ) · dP − (S 1 /N 1 ) · dT = v 1 dP − s 1 dT
(1.1.11)
dμ 2 = dG 2 /N 2 = (V 2 /N 2 ) · dP − (S 2 /N 2 ) · dT = v 2 dP − s 2 dT
(1.1.12)
where v is the molecular volume and s is the entropy per molecule. From Eq. (1.1.10),
dμ 1 = dμ 2 as long as the change is along the phase boundary. Thus,
v 1 dP − s 1 dT = v 2 dP − s 2 dT
(1.1.13)
(v 1 − v 2 )dP = (s 1 − s 2 )dT
(1.1.14)
(dP/dT ) = (s 1 − s 2 )/(v 1 − v 2 ) = N A (s 1 − s 2 )/N A (v 1 − v 2 )
(1.1.15)
where N A is the Avogadro number. Thus, Eq. (1.1.15) states that the local slope
(dP/dT ) on a phase boundary is equal to the change in the molar entropy divided by
the change in the molar volume between the two phases (1) and (2) at that P and T:
(dP/dT ) = S molar //V molar
(1.1.16)
How can one find S molar ? A great property of the first-order phase transition is
that the transformation takes place at a constant temperature T and the change in the
entropy is the latent heat of the transition between the two phases (1) and (2), L 12 ,
divided by that temperature:
(dP/dT ) = L 12 /(T V molar )
(1.1.17)
As can be seen from Eq. (1.1.17), the greater the change in V molar (as in a liquid–
gas transition) the smaller the slope, dP/dT, in the phase diagram. When the change
in V molar is negative (as in the ice–liquid water transition), the slope, dP/dT, in the
phase diagram is also negative.