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egoroff_w [7]
3 years ago
6

The diffusion of oxygen (O2) through living tissue is often first approximated as the diffusion of

Chemistry
1 answer:
kotegsom [21]3 years ago
4 0

Explanation:

We assume that O_{2} is represented by A and H_{2}O is represented by B respectively.

According to Wilke Chang equation as follows.

       D_{AB} = \frac{7.4 \times 10^{-8} \times (\phi_{B} M_{B})^{1/2} \times T}{V^{0.6}_{A} \times \mu_{B}}

       D_{O_{2} - H_{2}O} = \frac{7.4 \times 10^{-8} \times (\phi_{H_{2}O} M_{H_{2}O})^{1/2} \times T}{V^{0.6}_{O_{2}} \times \mu_{H_{2}O}}

where,   T = absolute temperature = (273 + 37)K = 310 K

       \phi_{H_{2}O} = an association parameter for solvent water = 2.26

    M_{H_{2}O} = Molecular weight of water = 18 g/mol

    \mu = viscosity of water (in centipoise) = 0.62 centipoise

    V_{O_{2}} = the molar volume of oxygen = 25.6 cm^{3}/g mol

Hence, putting the given values into the above formula as follows.

   D_{O_{2} - H_{2}O} = \frac{7.4 \times 10^{-8} \times (\phi_{H_{2}O} M_{H_{2}O})^{1/2} \times T}{V^{0.6}_{O_{2}} \times \mu_{H_{2}O}}

                       = \frac{7.4 \times 10^{-8} \times (2.26 \times 18)^{1/2} \times 310 K}{(25.6)^{0.6}_{O_{2}} \times 0.692}

                       = 3021.7 \times 10^{-8} cm^{2}/s

Thus, we can conclude that the diffusion of O_{2} in water by the Wilke-Chang correlation  at 37^{o}C.

             

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