Question

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

Answer 1

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 $37^{o}C$.