34 10 3 3 s s j1

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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 N-s ⎞ ⎛ 6π rη s −12 N-s 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...
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