Ponto de Ebulição e Fusão

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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 111
0 100 200 300 400 500 600 700
0
10
20
30
40
50
Liquid
Vapour
Solid
T/K
p/
ba
r
Figure 4.3
�e 93.11K solution is rejected since it lies below T3 where the liquid, and
therefore the solid-liquid boundary, does not exist.�e standard melting
point is therefore estimated to be 178.18K .
(c) �e standard boiling point is the temperature at the point on the liquid-
vapour phase boundary corresponding to p = 1bar. Substituting this
value of p into the equation for the liquid-vapour boundary and noting
that ln 1 = 0 gives
0 = −10.418/y+21.157−15.996y+14.015y2−5.0120y3+4.7334(1−y)1.70
Solving numerically gives
y = 0.645... and so T = y × Tc = 0.645... × 593.95 = 383.54K
(d) Use the Clapeyron equation for the liquid-vapour boundary [4B.8–133]:
dp
dT
=
∆vapH
T∆vapV
which rearranges to ∆vapH = T∆vapV × dp
dT
To �nd dp/dT , use d ln x = dx/x so that dp/dT = p × d ln p/dT . �e
expression for ln p is inserted and di�erentiated, and then evaluated at
the standard boiling point found above. For the evaluation of d ln p/dT it
does not matter that the expression has p in bar not Pa because the slope