Page 133 - Modern physical chemistry

124 Relationships between Phases

Example 6.4

Fitting equation (6.43) to vapor pressure data on benzene over the temperature range
40° to 100° C, with P in torrs, one finds that LIR is 3884 K and Cis 17.6254. Calculate the
vapor pressure of benzene at 60° C.
Here we have
lnP = 3884 K + 17.6254.
T

When
T=333.15 K
we obtain

lnP = _ 3884 K + 17.6254 = 5.9670
333.15 K
and
T=390toIT.

6.9 Variation of Vapor Pressure with Total Pressure

Adding to the pressure on a pure condensed phase at a given temperature tends to
squeeze molecules out of the phase and so acts to increase the vapor pressure of the
given substance. The addition may be effected by mixing with the vapor a second gas
that does not dissolve appreciably in the condensed phase.
Let the initial substance be A and the added one be B. Then for a given amount of A,
equation (6.26) applies at the boundary between the two phases. For an infinitesimal increase
in total pressure P, we have
[6.44]

as in equation (6.27).
Now, the pressure acting on the condensed form of A is P while the pressure of A in
the gaseous phase is the partial pressure P A • We consider the gas to be an ideal mixture;
so A there behaves as if B were not present. We also consider the movement along the
boundary to be at a fixed temperature. Condition (5.82) with dT equal to zero applies to
each phase. So equation (6.44) becomes

[6.45]
Let VCl) be the molar volume of A in the condensed phase while V(2) is the volume of
gas holding one mole of A. Also consider as an approximation that the gas is ideal. Equa-
tion (6.45) reduces to
dP A
"(1) dP = RT P . [6.46]
A
When the change in vapor pressure P A and the change in molar volume VCl) are small, the
coefficients of the differentials are approximately constant and (6.46) yields the formula

[6.47]