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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 109
0.0018 0.0022 0.0026 0.0030
−7
−6
−5
−4
−3
−2
−1
0
1
T−1/K−1
ln
(p
/p
∗
)
Figure 4.2
�e Clayperon equation [4B.4a–131], dp/dT = ∆trsS/∆trsV , is rearranged
to give ∆trsVdp = ∆trsSdT and this is used to replace ∆trsVdp in the
previous equation: d∆trsH = ∆trsCpdT + ∆trsSdT . Replacing ∆trsS by
∆trsH/T gives the required expression:
d∆trsH = ∆trsCpdT +
∆trsH
T
dT
(b) Startingwith the second expression, application of the quotient rule d(u/v) =
(vdu − udv)/v2 to the le� hand side and d ln x = dx/x to the right hand
side gives
d(∆trsH
T
) = ∆trsCpd lnT hence Td∆trsH − ∆trsHdT
T2
=
∆trsCpdT
T
�is expression is rearranged for d∆trsH to obtain the �rst expression:
d∆trsH = ∆trsCpdT +
∆trsH
T
dT
P4B.14 �e variation of vapour pressure with applied pressure is given by [4B.2–130],
p = p∗eVm∆P/RT . In this case the change in pressure (due to the depth) is ∆P =
ρgd. Since ρ = M/Vm, equation [4B.2–130] becomes
p = p∗eVm(M/Vm)gd/RT = p∗eMgd/RT
At a depth of 10 m the e�ect on the vapour pressure is
p/p∗ = exp((18.0158 × 10−3 kgmol−1) × (9.81ms−1) × (10m)
(8.3145 JK−1mol−1) × ([25 + 273.15]K)
) = 1.00071
�e fractional increase in vapour pressure is therefore 0.00071 or 0.071% .