Page 237 - Physical Chemistry

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Chapter 7 steepness of the slope of the P-versus-T fusion line (Fig. 7.1b) means that unless P P 1
2
One-Component Phase Equilibrium is very large, the change in melting-point temperature T will be quite small. More-
and Surfaces fus
over, the properties of solids and liquids change only slowly with pressure (Sec. 4.4).
Therefore, unless P P is quite large, we can approximate S, H, and V
2 1 fus fus fus
as constant. To integrate (7.23), we can assume either that S/ V is constant or
fus fus
that H/ V is constant. For small changes in freezing point, these two approxi-
fus fus
mations give similar results and either can be used. For substantial changes in freez-
ing point, neither approximation is accurate for solid–liquid transitions. However, the
equilibrium lines for many solid–solid transitions (Sec. 7.4) on P-versus-T phase dia-
grams are observed to be nearly straight over wide temperature ranges. The constant
slope dP/dT S/ V for such solid–solid transitions means that taking S/ V
trs trs trs trs
is often a good approximation here.
If we approximate S/ V as constant, then (7.23) becomes
fus fus
¢ S ¢ H
P P fus 1T T 2 fus 1T T 2 solid–liq. eq., T T small
1
2
2
1
2
1
2
1
¢ V T ¢ V
fus 1 fus
(7.24)
If we approximate ¢ H and ¢ V as constant, Eq. (7.23) gives
fus
fus
¢ H T
P P fus ln 2 solid–liq. eq., T T small (7.25)
2
2
1
1
¢ V T
fus 1
EXAMPLE 7.5 Effect of pressure on melting point
Find the melting point of ice at 100 atm. Use data from Prob. 2.49.
Since this is a solid–liquid equilibrium, Eq. (7.24) applies. [Afrequent student
error is to apply Eq. (7.19) to solid–liquid equilibria.]For1gof ice, H
fus
3
3
333.6 J and the densities give V V V 1.000 cm 1.091 cm
fus liq solid
3
0.091 cm . Let state 1 be the normal melting point. Then P P 100 atm
2 1
1atm 99 atm, and (7.24) becomes
333.6 J
99 atm ¢T
3
1273.15 K21 0.091 cm 2
3
¢T 7.38 K cm atm>J
3
We now use two values of R to convert cm atm to joules, so as to eliminate
3
cm atm/J.
3
1
cm atm 8.314 J mol K 1
¢T 7.38 K 0.75 K
1
3
J 82.06 cm atm mol K 1
Hence, T 273.15 K 0.75 K 272.40 K. The pressure increase of 99 atm
2
has lowered the melting point by only 0.75 K to 0.75°C.
Exercise
Repeat this problem assuming that H/ V is constant. (Answer: 272.40 K.)
fus
fus
Exercise
At the normal melting point of NaCl, 801°C, its enthalpy of fusion is 28.8 kJ/mol,
3
3
the density of the solid is 2.16 g/cm , and the density of the liquid is 1.73 g/cm .
5
3
What pressure increase is needed to raise the melting point by 1.00°C?
(Answer: 39 atm.)