Question

(a) Explain why the diffusion coefficient of ${\mathbf{CH}}_{\mathbf{3}}\mathbf{OH}$is greater than that of

sucrose in Table 23-1.

(b) Make an order-of-magnitude estimate of the diffusion coefficient of water

vapor in air at 298K.

Short answer

The solution is

a) ${\mathrm{CH}}_3\mathrm{OH}$is a smaller molecule than sucrose, which is a larger molecule, it diffuses

faster.

b) the order of magnitude of ${\text{10}}^{-5}{\mathrm{m}}^2/\mathrm{s}$.

Step-by-step solution

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We will solve the following problems in this task:

a) Describe why the diffusion coefficient of ${\mathrm{CH}}_3\mathrm{OH}$in Table 23-1 is greater than the diffusion coefficient of sucrose.

b) At 298K , make an order-of-magnitude estimate of the diffusion coefficient of water vapour in air.

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a) Representative diffusion coefficients at 298K are provided in Table 23-l:

- Sucrose has a diffusion coefficient $0.52\times {10}^{-9}{\mathrm{m}}^2/\mathrm{s}$, whereas ${\mathrm{CH}}_3\mathrm{OH}$has a diffusion coefficient of $1.6\times {10}^{-9}{\mathrm{m}}^2/\mathrm{s}$.

- As the size of the diffusing particle and the viscosity of the fluid both rise, the friction coefficient increases.

- Some larger molecules diffuse slower than small ones due to their vast radius.

- As a result, ${\mathrm{CH}}_3\mathrm{OH}$is a smaller molecule than sucrose, which is a larger molecule, it diffuses faster.

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b) At 298K , the diffusion coefficient of liquid water is $2.3\times {10}^{-9}{\mathrm{m}}^2/\mathrm{s}$.

Table 23-1 illustrates that diffusion in liquids is ${10}^4$times slower than diffusion in gases, which is owing to the increased viscosity of liquids, according to the Pg6l8.

$(2.3\times {10}^{-9}{\mathrm{m}}^2/\mathrm{s})\times {10}^4=2.3\times {10}^{-5}{\mathrm{m}}^2/\mathrm{s}$

- Since we're dealing with water vapour in air (which is a gas), the diffusion coefficient is: - which gives us an order of magnitude of${10}^{-5}{\mathrm{m}}^2/\mathrm{s}$.