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108 4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
andM is the molar mass. It follows that ∆T = q/nlCm
∆T = q
nlCm
= q
(m/M)Cm
=
q
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
− pH2OV
RT
× ∆vapH ×
M
mCm
= − (3.17 × 103 Pa) × (50 × 10−3m3)
(8.3145 JK−1mol−1) × ([25 + 273.15]K)
× (44.0 × 103 Jmol−1)
× 18.0158 gmol−1
(250 g) × (75.5 JK−1mol−1)
= −2.68... K
So the �nal temperature will be (25 − 2.68...) ○C = 22.3 ○C
P4B.10 (a) �e normal boiling point is the temperature at which the vapour pressure
is equal to 1 atm, or 760Torr. From the data, this is at 227.5 ○C .
(b) �e integrated formof theClausius–Clapeyron equation, given by [4B.10–
133], ln(p/p∗) = −(∆vapH/R)(1/T − 1/T∗), is rewritten as
ln p
p∗
= −
∆vapH
R
1
T
+
∆vapH
RT∗
�is implies that a plot of ln(p/p∗) against 1/T should be a straight line
of slope −∆vapH/R and intercept ∆vapH/RT∗; p∗ may be taken to be
760Torr (1 atm).�e plot is shown in Fig. 4.2.
θ/○C p/Torr T−1/K−1 ln(p/p∗)
57.4 1 0.003 025 −6.633
100.4 10 0.002 677 −4.331
133.0 40 0.002 462 −2.944
157.3 100 0.002 323 −2.028
203.5 400 0.002 098 −0.642
227.5 760 0.001 997 0.000
�e data fall on a good straight line, the equation of which is
ln(p/p∗) = (−6.446 × 103) × (T−1/K−1) + 12.91
�e slope is equal to −∆vapH/R, so:
∆vapH = −slope × R = −(−6.446 × 103 K) × (8.3145 JK−1mol−1)
= 53.6 kJmol−1
P4B.12 (a) �e expression dH = CpdT + Vdp implies that for a phase change
d∆trsH = ∆trsCpdT + ∆trsVdp