5.3 Phase Equilibria in One-Component Systems
209
Exercise 5.3
For most substances a solid–liquid coexistence curve has a positive slope, but in the water phase
diagram the ice I–liquid curve has a negative slope. Explain this phenomenon. In the 3 He phase
diagram a horizontal region occurs in the solid–normal liquid curve and a region with a negative
slope also occurs. Interpret these two phenomena given that the explanation of the negative slope
is not the same as with the liquid water–ice I curve.
In order to have a representation of the function P P(T ) corresponding to the
equilibrium of two phases, we integrate the Clapeyron equation written in the form
dP
∆S m
∆V m
dT
(5.3-9)
For a solid–liquid or a solid–solid phase transition, we assume that the phases have
nearly constant volume so that ∆V m is nearly constant. For a sufficiently small range
of temperature we can assume that ∆S m is nearly constant, so that
P 2 − P 1 ≈
∆S m
∆V m
(T 2 − T 1 )
(5.3-10a)
where P 1 is the pressure corresponding to temperature T 1 and P 2 is the pressure corresponding to temperature T 2 . We can also write this equation in the form
P P(T ) P 1 +
∆S m
∆V m
(T − T 1 ) P 1 +
∆H m
T 1 ∆V m
(T − T 1 )
(5.3-10b)
Equation (5.3-10) represents a linear function. In the water phase diagram several of
the solid–solid equilibrium curves are nearly linear, and this equation should be an
adequate approximation for these curves.
E X A M P L E 5.3
Estimate the pressure on a system of liquid and solid water if the equilibrium melting temperature is equal to −0.100 ◦ C. The density of ice is 0.917 g cm −3 , the density of liquid water is
1.000 g cm −3 , and the molar enthalpy change of fusion is 6008 J mol −1 .
Solution
∆V m (18.01 g mol −1 )
1
1.00 g cm −3 −
1
0.917 g cm −3
1 m 3
10 6 cm 3
−1.63 × 10 −6 m 3 mol −1
P 2 − P 1 ≈
6008 J mol −1
(273.15 K)
−1.63 × 10 −6 m 3 mol −1
− 0.100 K
≈ 1.35 × 10 6 J m −3 1.35 × 10 6 N m −2 1.35 × 10 6 Pa 13.3 atm
P 2 ≈ 14.3 atm
209
Exercise 5.3
For most substances a solid–liquid coexistence curve has a positive slope, but in the water phase
diagram the ice I–liquid curve has a negative slope. Explain this phenomenon. In the 3 He phase
diagram a horizontal region occurs in the solid–normal liquid curve and a region with a negative
slope also occurs. Interpret these two phenomena given that the explanation of the negative slope
is not the same as with the liquid water–ice I curve.
In order to have a representation of the function P P(T ) corresponding to the
equilibrium of two phases, we integrate the Clapeyron equation written in the form
dP
∆S m
∆V m
dT
(5.3-9)
For a solid–liquid or a solid–solid phase transition, we assume that the phases have
nearly constant volume so that ∆V m is nearly constant. For a sufficiently small range
of temperature we can assume that ∆S m is nearly constant, so that
P 2 − P 1 ≈
∆S m
∆V m
(T 2 − T 1 )
(5.3-10a)
where P 1 is the pressure corresponding to temperature T 1 and P 2 is the pressure corresponding to temperature T 2 . We can also write this equation in the form
P P(T ) P 1 +
∆S m
∆V m
(T − T 1 ) P 1 +
∆H m
T 1 ∆V m
(T − T 1 )
(5.3-10b)
Equation (5.3-10) represents a linear function. In the water phase diagram several of
the solid–solid equilibrium curves are nearly linear, and this equation should be an
adequate approximation for these curves.
E X A M P L E 5.3
Estimate the pressure on a system of liquid and solid water if the equilibrium melting temperature is equal to −0.100 ◦ C. The density of ice is 0.917 g cm −3 , the density of liquid water is
1.000 g cm −3 , and the molar enthalpy change of fusion is 6008 J mol −1 .
Solution
∆V m (18.01 g mol −1 )
1
1.00 g cm −3 −
1
0.917 g cm −3
1 m 3
10 6 cm 3
−1.63 × 10 −6 m 3 mol −1
P 2 − P 1 ≈
6008 J mol −1
(273.15 K)
−1.63 × 10 −6 m 3 mol −1
− 0.100 K
≈ 1.35 × 10 6 J m −3 1.35 × 10 6 N m −2 1.35 × 10 6 Pa 13.3 atm
P 2 ≈ 14.3 atm