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Diffusion Coefficient

Diffusion Coefficient (D) quantifies the rate at which particles move through a fluid. It is a crucial parameter in understanding how particles disperse over time.

The diffusion coefficient measures how fast particles spread out in a fluid. It's like observing how quickly a drop of ink disperses in a glass of water.

Key Points:

  • Rate of Spread: The diffusion coefficient indicates how quickly particles move through the fluid. Higher diffusion coefficients mean faster spreading.
  • Dependence on Temperature and Viscosity: The diffusion coefficient is directly proportional to temperature and inversely proportional to the viscosity of the fluid.

Mathematical Explanation

The Stokes-Einstein equation relates the diffusion coefficient to the hydrodynamic diameter (\(R_H\)) of the particles:

\[ D = \frac{k_B T}{6 \pi \eta R_H} \]

Where:

  • \(k_B\) is the Boltzmann constant (\(1.380 \times 10^{-23} \, \text{J/K}\)),
  • \(T\) is the absolute temperature (in Kelvin),
  • \(\eta\) is the viscosity of the fluid (in Pa·s),
  • \(R_H\) is the hydrodynamic radius of the particle (in meters).

Worked Example

A suspension of nanoparticles in water is maintained at \(298 \, \text{K}\) (25°C). The viscosity of water is \(0.89 \times 10^{-3} \, \text{Pa} \cdot \text{s}\).

The diffusion coefficient \(D\) is measured to be \(1.43 \times 10^{-12} \, \text{m}^2/\text{s}\).

Using the Stokes-Einstein equation

\[ R_H = \frac{k_B T}{6 \pi \eta D} = \frac{1.38 \times 10^{-23} \, \text{J/K} \times 298 \, \text{K}}{6 \pi \times 0.89 \times 10^{-3} \, \text{Pa} \cdot \text{s} \times 1.43 \times 10^{-12} \, \text{m}^2/\text{s}} \approx 6.1 \times 10^{-9} \, \text{m} = 61 \, \text{nm} \]

Self-Assessment Questions

  1. How does the viscosity of a fluid affect the diffusion coefficient of particles suspended in it?

    Answer

    The viscosity of the fluid affects how easily the particles can move; higher viscosity means more resistance to motion, resulting in a lower diffusion coefficient.

  2. Why is the diffusion coefficient temperature-dependent?

    Answer

    The diffusion coefficient is temperature-dependent because higher temperatures increase the kinetic energy of the fluid molecules, leading to more vigorous collisions and faster particle movement.

  3. Given a fluid with viscosity \(\eta = 1.0 \times 10^{-3} \, \text{Pa} \cdot \text{s}\) and temperature \(T = 300 \, \text{K}\), calculate the diffusion coefficient for particles with a hydrodynamic radius of \(50 \, \text{nm}\).

    Answer

    Using the Stokes-Einstein equation

    \[ D = \frac{k_B T}{6 \pi \eta R_H} = \frac{1.38 \times 10^{-23} \, \text{J/K} \times 300 \, \text{K}}{6 \pi \times 1.0 \times 10^{-3} \, \text{Pa} \cdot \text{s} \times 50 \times 10^{-9} \, \text{m}} \approx 1.1 \times 10^{-12} \, \text{m}^2/\text{s} \]

  4. If the diffusion coefficient is \(1.33 \times 10^{-12} \, \text{m}^2/\text{s}\), what is the hydrodynamic radius of the particles?

    Answer

    Rearranging the Stokes-Einstein equation to solve for \(R_H\)

    \[ R_H = \frac{k_B T}{6 \pi \eta D} = \frac{1.38 \times 10^{-23} \, \text{J/K} \times 300 \, \text{K}}{6 \pi \times 1.0 \times 10^{-3} \, \text{Pa} \cdot \text{s} \times 1.33 \times 10^{-12} \, \text{m}^2/\text{s}} \approx 5.5 \times 10^{-9} \, \text{m} = 55 \, \text{nm} \]