
# Mass Transport: Collisions and Diffusion

Shaun Williams, PhD

## Collision Parameters

### Chemical Environments

• Chemistry study solids and liquids
• Densities of over $$10^{22}$$ particles per cubic centimeter
• They also study interstellar nebular
• Densities of $$\lt 10^3$$ particles per cubic centimeter
• We need to identify key parameters that can be used to define any system

### Molecular Collisions

• The most common parameters used to describe collisions between two molecules
• Distribution of molecular velocities
• Average collision energy
• Effective cross-sectional area ("collision cross section")
• Average collision frequency
• Average distance a molecule travels between collisions ("mean free path")

### Molecular Speeds

• The distribution of molecular speeds in our canonical sample obey a Maxwell-Boltzmann distribution
• Any statistical average as three characteristic values:
• The mean (average)
• The mode (having the highest probability)
• The median (half population above and half below)

### Various Maxwell-Boltzmann Speeds

• From chapter 3: $$\expect{v} = \sqrt{\frac{8k_\mathrm{B}T}{\pi m}}$$
• We can determine the mode \begin{align} 0 &= \frac{d\mathcal{P}_v(v)}{dv} = \left( \frac{2m^3}{\pi k_\mathrm{B}^3T^3} \right) \left[ \begin{split} & 2ve^{-\frac{mv^2}{2k_\mathrm{B}T}} \\ & +v^2 e^{-\frac{mv^2}{2k_\mathrm{B}T}} \left( -\frac{m}{2k_\mathrm{B}T} \right) \left( 2v \right) \end{split} \right] \\ v_\chem{mode} &= \sqrt{\frac{2k_\mathrm{B}T}{m}} \\ v_\chem{med} &= 1.538 \sqrt{\frac{k_\mathrm{B}T}{m}} \end{align}

### Average Relative Speed

• The average relative speed is the crucial quantity for molecular collisions
• For a generic molecules A and B $$\expect{v_\chem{AB}} = \sqrt{\frac{8k_\mathrm{B}T}{\pi \mu}}$$
• If the two molecules are the same then $$\expect{v_\chem{AA}} = \sqrt{\frac{16k_\mathrm{B}T}{\pi m}} = \sqrt{2} \expect{v}$$

### Root Mean Squared Velocity

• Particles are equally likely to travel in any direction
• This means that $$\expect{\vec{v}}=0$$
• Due to this we define the rms velocity \begin{align} v_\chem{rms} &\equiv \left( \expect{v^2} \right)^\bfrac{1}{2} = \left( 3 \expect{v_x^2} \right)^\bfrac{1}{2} \\ &= \sqrt{\frac{3k_\mathrm{B}T}{m}} \end{align}

### Collision Energy

• The collision energy is the energy of a pair of colliding particles relative to the potential energy when the two particles are separated by a large distance
• We may believe that we need to worry about the frame of reference we use
• The collision energy is the same regardless of whether the frame of reference is moving or not

### Frame of Reference

$$K = 2\left( \frac{1}{2} mv^2 \right) = mv^2$$
$$K = \left( \frac{1}{2} m\left( 2v \right)^2 \right) = 2mv^2$$

### So what do we use?

• The kinetic energy relevant to collision strength is measured in the center of mass reference frame
• The frame in which the center of mass is at rest
• For two particles A and B, the collision energy is $$\frac{\mu v_\chem{AB}^2}{2}$$ $$\expect{E_\chem{AB}} = \frac{\mu \expect{v_\chem{AB}}^2}{2} = \frac{\mu \left( 3k_\mathrm{B}T \right)}{2\mu} = \frac{3k_\mathrm{B}T}{2}$$
• Note: from chapter 3: $$\expect{v_x^2} = \frac{k_\mathrm{B}T}{m}$$

### Collision Cross Section

• Expresses the maximum separation between interacting molecules in the form of an area.

### Selected Collision Cross Sections, $$\sigma \left( \AA^2 \right)$$

Gas $$\sigma (\AA^2)$$ Gas $$\sigma (\AA^2)$$
$$\chem{He}$$ 22 $$\chem{Ne}$$ 25
$$\chem{H_2}$$ 27 $$\chem{Ar}$$ 36
$$\chem{O_2}$$ 36 $$\chem{N_2}$$ 37
$$\chem{NH_3}$$ 43 $$\chem{CO_2}$$ 45
$$\chem{CH_4}$$ 52 $$\chem{H_2O}$$ 62
$$\chem{C_2H_4}$$ 71 $$\chem{C_2H_2}$$ 72
$$\chem{C_2H_6}$$ 84 $$\chem{SO_2}$$ 90
$$n-\chem{C_4H_{10}}$$ 148 $$\chem{CHCl_3}$$ 156

### Collision Cross Section of Dissimilar Molecules

\begin{align} \sigma_\chem{AB} &= \pi d_\chem{AB}^2 = \pi \left( \frac{d_\chem{A}+d_\chem{B}}{2} \right)^2 \\ &= \frac{\pi}{4} \left( d_\chem{A}^2 + 2d_\chem{A}d_\chem{B} + d_\chem{B}^2 \right) \\ &= \frac{\pi}{4} \left( \frac{\sigma_\chem{A}}{\pi} + \frac{2\sqrt{\sigma_\chem{A}\sigma_\chem{B}}}{\pi} + \frac{\sigma_\chem{B}}{\pi} \right) \\ &= \frac{1}{4} \left( \sigma_\chem{A} + 2\sqrt{\sigma_\chem{A}\sigma_\chem{B}} + \sigma_\chem{B} \right) \end{align}

### Average Collision Frequency

• The number of particles in the path of a reference particle is $$N_\chem{coll} = \rho \sigma l$$
• The collision frequency is related to time it takes the reference particle to travel the distance $$l$$ $$\gamma = \frac{N_\chem{coll}}{\Delta t_\chem{coll}} = \frac{\rho \sigma l}{\bfrac{l}{\expect{v_\chem{AA}}}} = \rho \sigma \expect{v_\chem{A}}$$

### Mean Free Path

• The average distance a particle travels between collisions
• Given by the average speed divided by the average collision frequency $$\lambda = \frac{\expect{v}}{\gamma} = \frac{1}{\sqrt{2}\rho \sigma}$$

### Example 5.1

What are $$\rho$$, $$\lambda$$, $$\expect{v_\chem{AA}}$$, and $$\lambda$$ for $$\chem{N_2}$$ at $$1.00\,\mathrm{bar}$$ and $$298\,\mathrm{K}$$? What is the average collision energy in units of $$\chem{cm^{-1}}$$?

### Changes in Trajectory

• So far we evaluated equation independent of the change in the particles trajectory after each collision
• We need to know how far, averaged over all these collisions, the molecules will travel
• We will fall back on statistical mechanics to determine how far, in any direction, a given molecule will travel after $$N_\chem{coll}$$ collisions

## The Random Walk

### Randomness

• Oddly, randomness makes our task more manageable
• We have a classic random walk problem
• For random walk problems
1. An atoms travels along a single line, suffering a collision at regular intervals
2. For each collision, the atom either continues or reverses direction
3. After $$N$$ collisions, how far has the atom traveled?

### Flipping a Coin

• Let’s treat the molecule collisions as flipping a coin $$N$$ times
• Heads means moving forward one step
• Tails means moving backward one step
• Our progress is the number of times we get heads, $$i$$, minus the number of times we get tails, $$j$$ $$k=i-j\;\text{and}\;N=i+j$$
• The coin flipping in random so many different $$k$$’s

### Doing Some Math

$$\mathcal{P}(k)=\frac{N!}{2^Ni!j!}$$

 \begin{align} i &= N-j=N-(i-k) \\ &= N-i+k \\ i &= \frac{N+k}{2} \end{align} \begin{align} j &= N-i=N-(j+k) \\ &= N-i-k \\ j &= \frac{N-k}{2} \end{align}

$$\mathcal{P}(k) = \frac{N!}{2^N \frac{N+k}{2}!\frac{N-k}{2}!}$$

### More Math

• Due to the size of $$N$$, this equation become unwieldy so we use some approximations $$\ln \mathcal{P}(k) = \ln N! - N\ln 2 - \ln \frac{N+k}{2}! - \ln \frac{N-k}{2}!$$
• We can simplify using Stirling’s approximation $$\ln N! \approx N\ln N - N$$
• Therefore $$\ln \mathcal{P}(k) \approx N\ln N - \frac{1}{2} \left[ \begin{split} & (N+k)\ln (N+k) \\ & +(N-k) \ln (N-k) \end{split} \right]$$

### Skipping Some More Math

• After performing another approximation and doing quite a bit of algebra we arrive at $$\ln \mathcal{P}(k) \approx -\frac{k^2}{2N}$$
• Undoing the logarithm and normalizing the result we find that $$\mathcal{P}(k) \approx \sqrt{\frac{2}{\pi N}} e^{-\frac{k^2}{2N}}$$

### Similar Equations

• We can rewrite our equation in terms of distance \begin{align} \mathcal{P}_x(x) &= \frac{1}{\sqrt{2\pi}L} e^{-\frac{x^2}{2L^2}} \\ \mathcal{P}_V(x,y,z) &= \mathcal{P}_x(x)\mathcal{P}_y(y)\mathcal{P}_z(z) = \frac{1}{\sqrt{8\pi^3}L^3} e^{-\frac{x^2+y^2+z^2}{2L^2}} \\ \mathcal{P}_r(r) &= \frac{4\pi}{\sqrt{8\pi^3}L^3} e^{-\frac{r^2}{2L^2}} r^2 \end{align}

## Transport without External Forces

### Wandering Particles

• Particle tend to wander in containers
• A substance is isolated on one side of a container by a wall
• When the wall is removes the substance will begin to move throughout the entire container
• If we wait long enough, the substance will be fully mixed throughout the container
• This is due to random elastic collisions - diffusion

### Diffusion

• Assume the bath (substance we are diffusing into) is a liquid or a gas
• We want the density as a function of time
• The width $$L$$ of the distribution is $$\sqrt{N_\chem{coll}}\lambda$$ in each direction
• From this we can find that $$L^2=\lambda^2 \gamma t$$ $$\mathcal{P}_r(r,t) = \frac{4\pi}{\sqrt{8\pi^3}\lambda^3} e^{-\frac{r^2}{2\lambda^2 \gamma t}} r^2$$

### Diffusion Equation

• We typically write the previous equation in terms of the diffusion constant \begin{align} D &= \frac{\lambda^2\gamma}{2} = \frac{\expect{v_\chem{AA}}}{2\rho\sigma} = \frac{L^2}{2t} \\ \mathcal{P}_r(r,t) &= \frac{\pi}{2\left(\pi Dt\right)^\bfrac{3}{2}} e^{-\frac{r^2}{4DT}} r^2 \end{align}
• For molecule B diffusing through molecule A $$D_\chem{B:A} \equiv \frac{\expect{v_\chem{AB}}}{2\rho_\chem{A}\sigma_\chem{AB}} = \frac{1}{2}\lambda \expect{v_\chem{B}}$$

### Distance Travelled

• We can use our probability in a root mean square distance travelled during diffusion $$\expect{x^2} = \int_0^\infty r^2 \mathcal{P}_r(r,t)\,dr = 6Dt$$
• This yields Einstein’s equation for diffusion $$r_\chem{rms} = \expect{r^2}^\bfrac{1}{2} = \sqrt{6Dt}$$

### Fick's Laws

• The flux, $$J(z)$$, is the rate at which mass flows through a given cross-sectional area
• Fick’s first law gives $$J(z_0) = -D \left. \left( \frac{d\rho}{dz} \right) \right|_{z_0}$$
• Fick’s second law gives the change in the number density $$\frac{d\rho}{dt} = D\frac{d^2\rho}{dz^2}$$

### Example 5.2

Find the flux $$J(Z)$$ and the flow rate $$\frac{d\rho}{dt}$$ if the concentration is given by the linear gradient $$\rho(Z) = \rho_0 cZ$$

### Diffusion in Solids

• Molecules can diffuse through and across the surface of solids
• Note: $$r$$ is now the distance along the surface $$\mathcal{P}_r(r,t) = \frac{1}{2Dt} e^{-\frac{r^2}{4Dt}} r$$

### Viscosity

• Fluids flow much slower than the particle velocity
• There are attractive forces between particles
• Inter-particle repulsive force allows particles to exchange momentum
• These correspond to viscosity, the resistance to flow
• We will start with hard spheres that only have the repulsive force

### Simple Viscosity

• The number of collisions that take place in an area of the wall in some time is \begin{align} N_\chem{coll} &= \frac{\text{number of colliding particles in volume}}{\text{time per collision}} \Delta t \\ &= \frac{\rho A \frac{\lambda}{2}}{\frac{\lambda}{\expect{v_x}}} \Delta t = \frac{\rho A \expect{v_x} \Delta t}{2} \end{align}

### How Does the Momentum Change per Collision?

• Remember from physics that $$\Delta p_i = m\Delta v_i$$ $$\Delta v_i = \Delta x \frac{\Delta v_i}{\Delta x} \approx \Delta x \frac{dv_i}{dx}$$
• We can set the change in $$x$$ to the distance between collisions, which is the mean free path \begin{align} \Delta v_i &\approx \lambda \frac{dv_i}{dx} \\ \Delta p_i &= m\lambda \frac{dv_i}{dx} \end{align}

### Viscosity is a Force

• Viscosity is a force so \begin{align} F &= ma = m\frac{dv}{dt} = \frac{d(mv)}{dt} = \frac{dp}{dt} \approx \frac{\Delta p}{dt} \\ F_\chem{viscosity} &\approx \frac{\Delta p}{\Delta t} = \left( \frac{\rho A\expect{v_x}}{2}\right) \left(m\lambda\frac{dv}{dx}\right) \equiv \eta A\frac{dV}{dx} \end{align}
• $$\eta$$ is the viscosity

### Solving the Viscosity

• For flow in one direction we have $$\eta = \frac{1}{2}\rho \expect{v_x} m\lambda = \frac{1}{2}\rho \sqrt{\frac{8k_\mathrm{B}T}{\pi m}}m \left(\frac{1}{\rho \sigma}\right)=\sqrt{\frac{2mk_\mathrm{B}T}{\pi \sigma^2}}$$

### Stokes' Law

• For a solvent with viscosity, $$\eta$$, the speed $$v$$ of a sphere of radius $$r$$ (large compared to the molecular dimensions of the solvent) will meet a resisting force of $$F_\chem{viscosity} = -6\pi \eta rv$$

### Poiseuille's Formula

• For a viscous fluid flowing through a tube of radius $$r$$ with pressure drop $$\Delta P$$ over a distance $$l$$, the volume flow rate is $$\frac{dV}{dt} = \frac{\pi r^4 \Delta P}{8\eta l}$$

### Example Diffusion Constants and Viscosities

Sample $$D_\chem{obs}$$ $$D_\chem{calc}$$ Sample $$D_\chem{obs}$$ $$D_\chem{calc}$$
$$\chem{H_2}\text{ in }\chem{N_2} \\ \text{293.15 K, 1.01 bar}$$ $$0.772$$ $$1.06$$ $$\chem{H_2} \\ \text{300 K, 0.01 bar}$$ $$9.0\times 10^{-5}$$ $$10.8\times 10^{-5}$$
$$\chem{Ar}\text{ in }\chem{N_2} \\ \text{293.15 K, 1.01 bar}$$ $$0.190$$ $$0.312$$ $$\chem{Ar} \\ \text{300 K, 1.00 bar}$$ $$22.9\times 10^{-5}$$ $$36.3\times 10^{-5}$$
$$\chem{Ar}\text{ in }\chem{N_2} \\ \text{573.15 K, 1.01 bar}$$ $$0.615$$ $$0.853$$ $$\chem{Ar} \\ \text{600 K, 1.00 bar}$$ $$39.0\times 10^{-5}$$ $$50.8\times 10^{-5}$$
$$\chem{H_2O}\text{ in }\chem{N_2} \\ \text{293.15 K, 1.01 bar}$$ $$0.242$$ $$0.290$$ $$\chem{H_2O} \\ \text{300 K, 0.10 bar}$$ $$10.0\times 10^{-5}$$ $$14.3\times 10^{-5}$$
$$\chem{H_2O}\text{ in }\chem{acetone(liq)} \\ \text{298 K}$$ $$4.56\times 10^{-5}$$ $$4.7\times 10^{-4}$$ $$\chem{H_2O(liq)} \\ \text{298 K}$$ $$8.90\times 10^{-3}$$ $$14.3\times 10^{-5}$$

## Transport with External Forces

### Sedimentation

• Typically gravity doesn’t play a role with small molecules and atoms
• Large molecules can be effected however
• Sedimentation uses gravity to separate compounds by different masses in solution
• Ultracentrifuges effectively increase the effective force of gravity

### Frictional Force

• For molecules that obey Stokes’ law, we can write a more general friction force law $$F_\chem{viscosity}=-fv$$
• $$f$$ is the frictional force
• Ultracentrifuges use angular velocity to increase the force $$F_\chem{cent} = (m-m_0) \omega^2 r \equiv m\omega^2 rb,\, b\equiv \left(1-\frac{m_0}{m}\right)$$
• $$b$$ is the buoyancy correction

### Simplifying

• At a steady state speed, $$v_\chem{ss}$$ $$v_\chem{ss} = \frac{m\omega^2 rb}{f}$$
• Another definition of the diffusion constant $$D=\frac{k_\mathrm{B}T}{f}$$
• This makes the steady state velocity $$v_\chem{ss} = \frac{m\omega^2 rbD}{k_\mathrm{B}T}$$

### Electrophoresis

• Electrostatic force is opposed by viscosity
• The mobility of the molecules through the gel goes as $$v \propto \ln m$$
• So mixtures can be separated based on molecular masses
• Both electrophoresis and sedimentation are somewhat fallible because the mobility of particles is effected by shape

### Convection and Chromatography

• Convection is the net flow of a gas or liquid sample, usually through some other fluid medium
• The medium must be macroscopically distinguishable from the sample
• For a tube laying parallel to the $$z$$ axis $$\expect{v_\chem{conv}} \equiv \frac{dz}{dT}$$
• The distribution of sample in a chromatography column $$\mathcal{P}_z(z,t) = \left(4\pi Dt\right)^{-\bfrac{1}{2}} e^{-\frac{\left(z-\expect{v_\chem{conv}}t\right)^2}{4Dt}}$$

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