Geology Reference
In-Depth Information
ANSWERS TO EXERCISES
Chapter 2
Δ S = +ve (always true for melting).
Δ V = −ve. Therefore d P /d t = Δ S V = −ve.
The negative slope of the melting curve indicates
that  the melting temperature falls as pressure is
increased.
2.3 Δ S = 202.7 + (2 × 82.0) − 241.4 − 41.5 = 83.8 J K −1 mol −1
Δ V = 32.6 × 10 −6 m 3 mol −1
Δ S V = 2.57 × 10 6 J K −1 m −3 = 2.57 × 10 6 Pa K −1
2.1 Point X Phases present are calcite + quartz + CO 2
gas.
φ== (
)
33
.
C
CaOSiO CO
,
,
.
2
2
3
+=+→=
F
3 2
F
2
.
Point Y, ϕ = calcite + quartz + wollastonite + CO 2 =
4 → F = 1. Temperature and P CO 2 can vary indepen-
dently at point X without changing the equilibrium
assemblage.
2.2 The lower density of ice indicates that at 0 °C,
V ice > V water . For the reaction:
We know one point on the reaction boundary (10 5 Pa
at 520 °C). At 520 + 300 °C the pressure on the
boundary will be 300 × 2.57 × 10 6 Pa = 7.710 × 10 8 Pa.
As all phases are anhydrous, a straight reaction
boundary is expected between these two points. The
volume of the grossular + quartz assemblage is less
than the anorthite + 2(wollastonite) assemblage, and
therefore it will be found on the high-pressure side.
HO HO
ice
2
2
water
 
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