This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ane (in mol/s). (10) Solution: The steady state diffusion rate is given by the following equation: J = PA ( C A − C B ) To solve for J, we must first solve for the permeability, P: kD P=
Δx ⎛
−23 J ⎞
2
⎜ 1.38 × 10
⎟ ( 298 K )
⎝
KT
4.363 × 10 −21 J
K⎠
−10 m
where D =
. =
=
= 4.363 × 10
Ns ⎞
⎛
6π rη
s
−12 Ns
6π 5 × 10 −10 m ⎜ 0.001 2 ⎟ 9.424 × 10
⎝
m
m⎠
Hence: ⎛
m2 ⎞
( 0.0015 ) ⎜ 4.363 × 10 −10 ⎟
s⎠
⎝
kD
m
P=
=
= 0.000654 −9
Δx
10 m
s Now that we know P, we can find J: J = PA ( C A − C B ) ( ) ( ) 2
m
⎛ 1 mol ⎞ ⎛ 1000 L ⎞ 4π ⋅ ( 5 × 10 −7 m ) [100 mM − 2 mM ] ⎜
⎝ 1000 mmol ⎟ ⎜ 1 m 3 ⎟
⎠⎝
⎠
s
mol
J = 2.013 × 10 −13
s J = 0.000654 B. Suppose the glucose was replaced inside and outside by a second molecule V, whose diffusion coefficient is twice that of glucose’s but w...
View
Full
Document
 Spring '13
 SanjayKumar

Click to edit the document details