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EXAMPLE 7.4 Change of vapor pressure with temperature Section 7.3
The Clapeyron Equation
The normal boiling point of ethanol is 78.3°C, and at this temperature H
vap m
38.9 kJ/mol. To what value must P be reduced if we want to boil ethanol at
25.0°C in a vacuum distillation?
The boiling point is the temperature at which the liquid’s vapor pressure
equals the applied pressure P on the liquid. The desired value of the applied
pressure P is thus the vapor pressure of ethanol at 25°C. To solve the problem,
we must find the vapor pressure of ethanol at 25°C. We know that the vapor pres-
sure at the normal boiling point is 760 torr. The variation of vapor pressure with
temperature is given by the approximate form (7.19) of the Clapeyron equation:
2
d ln P dT H RT . If the temperature variation of H is neglected, in-
m vap m
tegration gives [Eq. (7.21)]
P 2 ¢H m 1 1
ln a b
P 1 R T 2 T 1
Let state 2 be the normal-boiling-point state with T (78.3 273.2) K
2
351.5 K and P 760 torr. We have T (25.0 273.2) K 298.2 K and
2 1
3
760 torr 38.9 10 J>mol 1 1
ln a b 2.38
1
P 1 8.314 J mol K 1 351.5 K 298.2 K
760 torr>P 10.8, P 70 torr
1
1
The experimental vapor pressure of ethanol at 25°C is 59 torr. The substantial
error in our result is due to nonideality of the vapor (which results mainly from
hydrogen-bonding forces between vapor molecules) and to the temperature vari-
ation of H ; at 25°C, H of ethanol is 42.5 kJ/mol, substantially higher
vap m vap m
than its 78.3°C value. For an improved calculation, see Prob. 7.25.
Exercise
The normal boiling point of Br is 58.8°C, and its vapor pressure at 25°C is
2
0.2870 bar. Estimate the average H m of Br in this temperature range.
vap
2
(Answer: 30.7 kJ/mol.)
Exercise
Use data in Table 7.1 to estimate the boiling point of Ar at 1.50 atm. (Answer:
91.4 K.)
Exercise
Use Fig. 7.6 to find the slope of a ln P-versus-1/T plot for the vaporization of
H O near 35°C. Then use this slope to find H of H O at 35°C. (Answers:
m
2
vap
2
5400 K, 45 kJ/mol.)
Solid–Liquid Equilibrium
For a solid–liquid transition, Eq. (7.19) does not apply. For fusion (melting), the Clap-
eyron equation (7.18) and (7.17) reads dP dT S V H (T V).
fus fus fus fus
Multiplication by T and integration gives
2 dP 2 ¢ S dT 2 ¢ H dT (7.23)
fus
fus
fus
fus
1 1 ¢ V 1 T ¢ V
The quantities S ( S S ), H, and V change along the solid–liquid
fus liq solid fus fus
equilibrium line due to changes in both T and P along this line. However, the
fus fus