70
TABLE 6-1
Temperature and Pressure Data
f
20
30
40
50
60
70
80
90
100
P
175
31 8
55.3
92.5
1494
2337
355 1
5258
760
1/7
0003412
0 003301
0003195
0 003095
0 003004
0002915
0 002832
0 002755
0 002681
logP
1 243
1 502
1 743
1 966
2 174
2369
2550
2721
2881
We know from theoretical principles, however, (and your text may explain
this) that the vapor pressure of a liquid is related to its heat of vaporization
(A// v ), which is a physical constant characteristic of the liquid, to the gas constant (R = 1.987 cal/mole deg), and to the Kelvin temperature (T), by the
equation
X F
+ *
(6
'
6)
From this expression, it follows that a straight-line graph should be obtained if
we lety = logP and* = I/T; then the equation will be of the form
y = mx + b
(6-1)
A TJ
with the slope composed of a collection of constants, m = - . ™?D »
an ^
tne
intercept, b = B.
PROBLEM:
Using the data of Table 11-1, plot a graph from which you can calculate the heat of
vaporization of water, A//,.
SOLUTION:
Table 6-1 shows the data given in Table 11-1, along with the values of log/
3 and 1/T
calculated from the data. If you need help with the logarithms, see pp 13-15. (K =
°C + 273.2.)
Figure 6-5 shows the plot of log P against 1/7. The slope of this straight-line
graph is
°50
° = -2.22 x HP deg
-0.000225