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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 565
E16C.7(b) �e Einstein–Smoluchowski equation [16C.15–713], D = d2/2τ, relates the dif-
fusion coe�cient D to the jump distance d and time τ required for a jump.
Approximating the jump length as the molecular diameter, then d ≈ 2a where
a is the e�ective molecular radius.�is is estimated using the Stokes–Einstein
equation [16C.4b–708], D = kT/6πηa, to give 2a = 2kT/6πηD.
Combining these expressions and using the value η = 0.386 × 10−3 kgm−1 s−1
for the viscosity of heptane gives
τ = d
2
2D
= 1
2D
( 2kT
6πηD
)
2
= 1
18D3
( kT
πη
)
2
= 1
18 × (3.17 × 10−9 m2 s−1)3
((1.3806 × 10−23 JK−1) × (298 K)
π × (0.386 × 10−3 kgm−1 s−1)
)
2
= 2.01 × 10−11 s = 20.1 ps
E16C.8(b) �e root mean square displacement in three dimensions is given by equation
[16C.13b–712], ⟨r2⟩1/2 = (6Dt)1/2, where D is the di�usion coe�cient and t is
the time period.�erefore t = ⟨r2⟩/6D.
For an iodine molecule in benzene, D = 2.13 × 10−9 m2 s−1
t(1.0 mm) = ⟨r2⟩
6D
= (1.0 × 10−3 m)2
6 × (2.13 × 10−9 m2 s−1)
= 78 s
t(1.0 cm) = ⟨r2⟩
6D
= (1.0 × 10−2 m)2
6 × (2.13 × 10−9 m2 s−1)
= 7.8 × 103 s = 2.2 h
For a sucrose molecule in water, D = 0.5216 × 10−9 m2 s−1
t(1.0 mm) = ⟨r2⟩
6D
= (1.0 × 10−3 m)2
6 × (0.5216 × 10−9 m2 s−1)
= 320 s
t(1.0 cm) = ⟨r2⟩
6D
= (1.0 × 10−2 m)2
6 × (0.5216 × 10−9 m2 s−1)
= 3.2 × 104 s = 8.9 h
Solutions to problems
P16C.2 �e thermodynamic force F is given by [16C.3b–706]
F = −RT
c
( ∂c
∂x
)
T ,p
Substituting c(x) = c0e−ax
2
into the above expression gives
F = − RT
c0e−ax2
(−2ac0xe−ax
2
) = 2axRT