Page 147 - Essentials of physical chemistry

Gibbs’ Free Energy and Equilibria 109

volume is approximately (373.15=273.15) (22,414 mL=mol) ¼ 30620 mL=mol using Charles’ law.
Thus, we arrive at what is known as the Clapeyron equation, which can be used as is to study
transitions in fusion of minerals but

dP DH vap DH vap
dT ¼ T bp (DV vap ) ffi T bp V vap

is not quite what we need here for liquids. We pause here to note that B. P. E. Clapeyron (1799–
1864) was a French engineer who made several contributions to thermodynamics and was actually
the person who plotted the Carnot cycle as a PV diagram shown in a previous chapter. Interestingly,
Carnot himself actually used the concept of ‘‘caloric’’ in his derivation but Clapeyron put the Carnot
cycle into the form we have shown it in this text.
The Clapeyron equation was extended to a more usable form by a German physicist R. G.
dP
Clausius (1822–1888) by splitting the differential and integrating the new form using the ideal
dT
dP DH vap
gas law for the vapor at low pressure (a very good approximation) to obtain ¼ ,
dT T RT
P

dP d ln P DH vap dT
which can be rearranged to P ¼ ¼ DH vap and then further to d ln P ¼ ¼
dT dT RT 2 R T 2

DH vap 1 1 1 2
d since d ¼ d(T ) ¼ T dT (a clever step indeed!).
R T T

Graphically, this leads to a plot similar to the case of the temperature dependent equilibrium
constant shown above in the section on equilibrium. We see that we can plot a logarithm of the
pressure against reciprocal Kelvin temperature and expect to find a straight line with a negative
slope.
ð ð
P 2 T 2
P 2 DH vap dT DH vap 1 1
This can be integrated as d ln P ¼ ln ¼ ¼ ,so
P 1 R T 2 R T 2 T 1
P 1 T 1
the final working equation becomes the very useful Clausius–Clapeyron equation

P 2 DH vap 1 1
ln ¼ :
P 1 R T 2 T 1
Note this so-called Clausius–Clapeyron equation has five variables, so that a number of possible
problems can be formulated for quiz questions and in a practical sense it is a very useful equation to
find DH vap from P, T data or a boiling point if one knows DH vap .
In Table 6.1, we see values of temperature at which the vapor pressure is at certain values.
The final values at 100 kPa are not quite the normal boiling points because 1 atm ¼ 101.325 kPa.
This table shows the modern way to represent these types of data, which reveals the vapor
pressures of solids at low temperature, but often only provides two data points for P, T in
the range of room temperature. Further, the use of only two points can only yield a perfect line
when one plots the ln (P)versus(1=T). In order to show that the approximations made
in deriving the Clausius–Clapeyron equation are good but not perfect, we also present in
Figure 6.3, the older style data for the vapor pressure of liquid water (H 2 O) in mmHg and
1 atm boiling point (see Table 6.2 [2])

DH vap 1 DH vap 1
,
ln(P) ¼ ln(760) þ
R T(K) R 373:15