   Chapter 9, Problem 61P

Chapter
Section
Textbook Problem

Transport Phenomena61. Sucrose is allowed to diffuse along a 10.-cm length of tubing filled with water. The tube is 6.0 cm2 in cross-sectional area. The diffusion coefficient is equal in 5.0 × 10−10 m2/s, and 8.0 × 10−14 kg is transported along the tube in 15 s. What is the difference in the concentration levels of sucrose at the two ends of the tube?

To determine
The difference in concentration levels of sucrose at the two ends of the tube.

Explanation
From Fick’s law, the difference in concentration levels between ends of the tube is Diffusionrate=masstime=DA(C2C1L)C2C1=(Diffusionrate)LDA where (C2C1)/L called as concentration gradient.

Given info: The diffusion coefficient is 5.0×1010m2/s , cross-sectional area of the tube is 6.0cm2 , length of the tube is 10cm , and the diffusion rate is 8.0×1014kg/15s .

The formula for the difference in concentration levels is,

C2C1=(Diffusionrate)LDA

• L is length of the tube.
• D is diffusion coefficient.
• A is cross-sectional area of the tube.

Substitute 5.0×1010m2/s for D , 6.0cm2 for A , 8

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