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
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: \( \expect{v_x^2} = \frac{k_\mathrm{B}T}{m} \)
Intermolecular Potential Energy Surface with \(E_\chem{AB}\)
Collision Cross Section
Expresses the maximum separation between interacting molecules in the form of an area.
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
An atoms travels along a single line, suffering a collision at regular intervals
For each collision, the atom either continues or reverses direction
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
$$ \mathcal{P}(k)=\frac{N!}{2^Ni!j!} $$
$$ \begin{align}
i &= N-j=N-(i-k) \\
&= N-i+k \\
i &= \frac{N+k}{2}
\end{align} $$
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 \( \ln N! \approx N\ln N - N \)
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
$$ \mathcal{P}(k) \approx \sqrt{\frac{2}{\pi N}} e^{-\frac{k^2}{2N}} $$
\(\mathcal{P}(k)\) Versus \(k\) for \(N=100\) and \( N=1000\)
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
$$ r_\chem{rms} = \expect{r^2}^\bfrac{1}{2} = \sqrt{6Dt} $$
System for Fick's Laws
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
Momentum Transfer
System for Elementary Viscosity Law
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
Schematic of Ultracentrifuge
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
$$ \mathcal{P}_z(z,t) = \left(4\pi Dt\right)^{-\bfrac{1}{2}} e^{-\frac{\left(z-\expect{v_\chem{conv}}t\right)^2}{4Dt}} $$