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564 16MOLECULES INMOTION
E16C.3(b) �e thermodynamic force F is given by [16C.3b–706]
F = −RT
c
( ∂c
∂x
)
T ,p
Substituting c(x) = c0 − βc0x2 into the above expression gives
F = − RT
c0 − βc0x2
(−2βc0x) =
2βxRT
1 − βx2
�e constant α is found by noting that c = c0/2 at x = 20 cm = 0.20 m. Hence
c0/2 = c0 − βc0 × (0.20 m)2 and therefore β = 12.5 m−2.
At T = 298 K and x = 8 cm the force is
F = 2 × (12.5 m−2) × (8 × 10−2 m) × (8.3145 JK−1mol−1) × (298 K)
1 − (12.5 m−2) × (8 × 10−2 m)2
= 5.4 kNmol−1
A similar calculation at x = 16 cm gives F = 7.3 kNmol−1 .
E16C.4(b) �e thermodynamic force F is given by [16C.3b–706]
F = −RT
c
( ∂c
∂x
)
T ,p
Substituting c(x) = c0e−αx2 into the above expression gives
F = − RT
c0e−αx2 (−2αc0xe
−αx2) = 2αxRT
�e constant α is found by noting that c = c0/2 at x = 10 cm = 0.10 m. Hence
c0/2 = c0e−α(0.10 m)2 and therefore α = ln 2/(0.10 m)2 = 69.3 m−2
�e thermodynamic force at T = 291 K and x = 10.0 cm is
F = 2(69.3m−2)×(0.10m)×(8.3145 JK−1mol−1)×(291K) = 33.5 kNmol−1
E16C.5(b) �e root mean square displacement in three dimensions is given by [16C.13b–
712], ⟨r2⟩1/2 = (6Dt)1/2, where D is the di�usion coe�cient and t is the time
period. Hence,
t = ⟨r2⟩
6D
= (1.0 × 10−2 m)2
6 × (4.05 × 10−9 m2 s−1)
= 4.1 × 103 s
E16C.6(b) �e Stokes–Einstein equation [16C.4b–708], D = kT/6πηa, relates the di�u-
sion coe�cient D to the viscosity η and the radius a of the di�using particle,
which is modelled as a sphere. Recall that 1 cP = 10−3 kgm−1 s−1.
a = kT
6πηD
= (1.3806 × 10−23 JK−1) × (298 K)
6π × (1.00 × 10−3 kgm−1 s−1) × (1.055 × 10−9 m2 s−1)
= 0.207 nm