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110 4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
P4B.16 �e �rst derivative of chemical potential with respect to temperature is given
by equation [4B.1a–128], (∂µ/∂T)p = −Sm.�e second derivative is therefore
( ∂
2µ
∂T2
)
p
= −(∂Sm
∂T
)
p
Entropy is de�ned by dS = dqrev/T ([3A.1a–80]) and, from section Section 3B.3
on page 90, at constant pressure dqrev = dH = CpdT .�erefore
dSm =
Cp ,mdT
T
and hence (∂Sm
∂T
)
p
=
Cp ,m
T
Consequently the curvature of the chemical potential with temperature is
(∂2µ/∂T2)p = −Cp ,m/T
Heat capacity is invariably positive, so this expression implies a negative cur-
vature (since T cannot be negative).
For water, the curvatures of the liquid and gas lines at the normal boiling point
(373.15 K) are:
liquid line: (∂
2µ(l)
∂T2
)
p
=
−Cp ,m(l)
T
= −75.3 JK−1mol−1
373.15K
= −0.202 JK−2mol−1
gas line: (∂
2µ(g)
∂T2
)
p
=
−Cp ,m(g)
T
= −33.6 JK−1mol−1
373.15K
= −0.0900 JK−2mol−1
�e magnitude of the curvature is therefore greater for the liquid.
Solutions to integrated activities
I4.2 (a) �e expressions are plotted on the graph shown in Fig. 4.3. Note that the
liquid-vapour line is only plotted for T3 ≤ T ≤ Tc because the liquid phase
does not exist below the triple point and there is no distinction between
liquid and vapour above the critical point.�e solid-liquid line is plotted
for T ≥ T3.
(b) �e standard melting point is the temperature corresponding to a pres-
sure of 1 bar on the solid-liquid boundary. Setting p = 1bar in the equa-
tion for the solid-liquid boundary and substituting in the value of p3
gives:
1 = 0.4362 × 10−6 + 1000(5.60 + 11.727x)x
�is equation is rearranged to the standard quadratic form
11727x2 + 5600x − 0.9999995638 = 0
which on solving for x gives x = 1.78... × 10−4 or x = −0.477.... �en,
since x = T/T3 − 1 where T3 = 178.15K, it follows that
T = 178.15(1.78... × 10−4 + 1) = 178.18K
or T = 178.15(−0.477... + 1) = 93.11K