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554 16MOLECULES INMOTION
2.4 × 10−2 JK−1m−1 s−1.
r = JzA = −κA
∆T
∆z
= −(2.4 × 10−2 JK−1m−1 s−1) × (2.00 m2) × [(0) − (70)] K
0.0500 m
= 67 W
E16A.7(b) �e viscosity η is given by [16A.11c–696], η = pMD/RT . In turn the di�usion
constant is given by [16A.9–694], D = 1
3 λυmean, where λ is the mean free
path length λ = kT/σ p [16A.1a–690], and υmean is the mean speed υmean =
(8RT/πM)1/2 [16A.1b–690]. �e �rst step is to �nd an expression for η as a
function of temperature
η = pMD
RT
= pM
RT
kT
3σ p
(8RT
πM
)
1/2
= M
3σNA
(8RT
πM
)
1/2
= 1
3σNA
(8RM
π
)
1/2
T1/2
= 1
3 × (8.8 × 10−19 m2) × (6.0221 × 1023mol−1)
× (8 × (8.3145 JK−1mol−1) × (0.078... kgmol−1)
π
)
1/2
T1/2
= (8.08... × 10−7 kgK−1/2m−1 s−1) × (T/ K)1/2
where 1 J = 1 kgm2 s−2 has been used to arrive at the units on the �nal line. Us-
ing this expression the following table is drawnup (recall that 10−7 kgm−1 s−1 =
1 µP)
T/K η/(kgm−1 s−1) η/(µP)
273 1.34 × 10−5 134
298 1.40 × 10−5 140
1000 2.56 × 10−5 256
E16A.8(b) In the solution to Exercise E16A.7(b) it is shown that
η = 1
3σNA
(8RMT
π
)
1/2
hence σ = 1
3ηNA
(8RMT
π
)
1/2
Recalling that 10−7 kgm−1 s−1 = 1 µP, the cross section is computed as
σ = 1
3 × (1.66 × 10−5 kgm−1 s−1) × (6.0221 × 1023mol−1)
× (8 × (8.3145 JK−1mol−1) × (0.02802 kgmol−1) × (273 K)
π
)
1/2
= 0.424 nm2