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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 95
P3E.4 (a) p = (RT)/Vm
( ∂S
∂V
)
T
= ( ∂p
∂T
)
V
= ( ∂
∂T
RT
Vm
)
V
= R
Vm
( ∂
∂T
T)
V
= R
Vm
.
(b) p = (RT)/(Vm − b) − a/V 2m
( ∂S
∂V
)
T
= ( ∂p
∂T
)
V
= ( ∂
∂T
[ RT
Vm − b
− a
V 2m
])
V
= R
Vm − b
( ∂
∂T
T)
V
− 0 = R
Vm − b
.
(c) p = (RTe−a/(RTVm))/(Vm − b)
( ∂S
∂V
)
T
= ( ∂p
∂T
)
V
= ( ∂
∂T
RTe−a/(RTVm)
Vm − b
)
V
= (1 + a
RT2Vm
) R
Vm − b
e−a/(RTVm) .
�e change in entropy for an isothermal expansion is
∆S = ∫
S(Vf)
S(Vi)
dS = ∫
Vf
Vi
( ∂S
∂V
)
T
dV
Comparing the (∂S/∂V)T for the perfect and van der Waals gases implies that
the change in entropy due to expansion (i.e. positive change in V ) is always
larger for the van der Waals gases: (∂S/∂V)T ,vdW > (∂S/∂V)T ,perfect. �is is
because R/(Vm − b) > R/Vm for the excluded volume constant b is positive.
�e result is expected as the relative volume expansion for the van der Waals
gases is always slightly greater than that of the perfect gas due to the �xed �nite
excluded volume.�erefore, the entropy is expected to bemore sensitive to the
changes in volume, especially at the very small volume.
P3E.6 Using Maxwell’s relation (∂S/∂p)T = − (∂V/∂T)p and (∂S/∂T)p = Cp/T the
given equation becomes
dS = ( ∂S
∂p
)
T
dp + ( ∂S
∂T
)
p
dT = −(∂V
∂T
)
p
dp +
Cp
T
dT
= 1
T
(CpdT − [ 1
V
(∂V
∂T
)
p
]TVdp)
�e expansion coe�cient is de�nded as α = (1/V)(∂V/∂T)p , therefore
TdS = CpdT − αTVdp
�e entropy change is de�ned in [3A.1a–80], dS = dqrev/T , hence qrev = ∫
f
i TdS.
For an isothermal reversible process q = qrev and dT = 0.�erefore, assuming