Respuesta :
Explanation:
We assume that [tex]O_{2}[/tex] is represented by A and [tex]H_{2}O[/tex] is represented by B respectively.
According to Wilke Chang equation as follows.
[tex]D_{AB} = \frac{7.4 \times 10^{-8} \times (\phi_{B} M_{B})^{1/2} \times T}{V^{0.6}_{A} \times \mu_{B}}[/tex]
[tex]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}}[/tex]
where, T = absolute temperature = (273 + 37)K = 310 K
[tex]\phi_{H_{2}O}[/tex] = an association parameter for solvent water = 2.26
[tex]M_{H_{2}O}[/tex] = Molecular weight of water = 18 g/mol
[tex]\mu[/tex] = viscosity of water (in centipoise) = 0.62 centipoise
[tex]V_{O_{2}}[/tex] = the molar volume of oxygen = 25.6 [tex]cm^{3}/g mol[/tex]
Hence, putting the given values into the above formula as follows.
[tex]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}}[/tex]
= [tex]\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}[/tex]
= [tex]3021.7 \times 10^{-8} cm^{2}/s[/tex]
Thus, we can conclude that the diffusion of [tex]O_{2}[/tex] in water by the Wilke-Chang correlation at [tex]37^{o}C[/tex].
