
# Statistical Mechanics and Molecular Interactions

Shaun Williams, PhD

## Molecular Interactions

### Gas Phase

• Average distance between molecules is usually so large that intermolecular interactions can be neglected
• During compression, these interactions become more important
• Unless they are highly reactive, intermolecular forces tend to be weak

### Intermolecular Potential Energy

• The intermolecular potential energy is zero when there are no interactions between the particles
• Repulsions correspond to positive contributions
• Attractions correspond to negative contributions
• The most important interaction is the repulsion between particles
• When molecules are crammed together the like charges repel each other

### Repulsion

• The distance at which this repulsion becomes important is small
• Determined by the atomic radii
• $$u_\chem{repulsion}\approx Ae^{-\frac{2cR}{a_0}}$$
• If the two particles can react, then this represents the energy barrier that must be overcome

### Attractions

• Attractions between non-reacting molecules are significant over a much smaller temperature range
• The thermal energy $$k_\mathrm{B}T$$ can easily cancel the negative attractive contribution
• Lowering the temperature tends to bring molecules closer together, first forming clusters, then droplets, then the formation of a condensed phase

### Dipole-Dipole Interaction

• One of the strongest of the attractive intermolecular forces is the dipole-dipole interaction
• The potential energy for the two rigid dipoles $$\mu_\chem{A}$$ and $$\mu_\chem{B}$$ interacting over distance $$R$$ is $$u_{2-2} = -\frac{2\mu_\chem{A}\mu_\chem{B}}{\left( 4\pi \varepsilon_0 \right)R^3}\left[ \begin{split} & \cos \theta_\chem{A} \cos \theta_\chem{B} \\ & - \frac{1}{2} \sin\theta_\chem{A} \sin\theta_\chem{B} \cos\left( \phi_\chem{A}-\phi_\chem{B} \right) \end{split} \right]$$
• Note: $$\phi=\phi_\chem{B}-\phi_\chem{A}$$

### Dipole-Dipole Interactions Continued

• If the two dipole moments both lie on the z axis $$u=-\frac{2\mu_\chem{A}mu_\chem{B}}{\left( 2\pi \varepsilon_0 \right) R^3}$$
• In the liquid and gas phases, the dipole moments are not fixed in place, but rotate constantly
• We now need to determine how the average potential is governed

### Beginning the Derivation

• In the gas phase, the molecules rotate through orientations of all types (attractive and repulsive)
• We need to average over all angles $$\theta_\chem{A}$$, $$\theta_\chem{B}$$, and $$\phi$$ using the classical average value theorem $$\expect{u} = \int \mathcal{P}(u)u\,d\tau$$
• The canonical distribution tells us that $$\mathcal{P}(u)$$ will have the form $$\mathcal{P}(u) = Ae^{-\frac{u}{k_\mathrm{B}T}}$$

### Normalizing the Function

• $$A$$ is the normalization constant \begin{align} \int \mathcal{P}(u)\,d\tau &= A\int e^{-\frac{u}{k_\mathrm{B}T}}\, d\tau =1 \\ A &= \frac{1}{\int e^{-\frac{u}{k_\mathrm{B}T}}\,d\tau} \end{align}
• We can rewrite the average value equation $$\expect{u} = A\int e^{-\frac{u}{k_\mathrm{B}T}} u\,d\tau = \frac{\int e^{-\frac{u}{k_\mathrm{B}T}}u\,d\tau}{\int e^{-\frac{u}{k_\mathrm{B}T}}\,d\tau}$$

### Simplifying the Equation

• We need to use the following volume element to average away all the variables except $$R$$ $$d\tau' = \sin \theta_\chem{A} \,d\theta_\chem{A} \sin \theta_\chem{B} \,d\theta_\chem{B}\,d\phi$$
• Because $$u \ll k_\mathrm{B}T$$ we can use a Taylor Series expansion $$\text{for }\left| x\right| \ll 1:\; e^x \approx 1+x$$
• This reduces our average potential energy to a simpler form
• After using a Taylor Series Expansion we have $$\expect{u}_{\theta,\phi} = \frac{\int e^{-\frac{u}{k_\mathrm{B}T}}u\,d\tau'}{\int e^{-\frac{u}{k_\mathrm{B}T}}\,d\tau'} \approx \frac{\int \left( 1-\frac{u}{k_\mathrm{B}T} \right)u\,d\tau'}{\int \left( 1-\frac{u}{k_\mathrm{B}T} \right)\,d\tau'} = \dots = -\frac{2\mu_\chem{A}^2\mu_\chem{B}^2}{3R^6k_\mathrm{B}T}$$

### Induced Dipole

• So far we have dealt with a permanent dipole moment on two molecules
• An electric field $$\varepsilon$$ will induce a dipole moment $$\mu_\chem{induced} = \alpha \varepsilon$$
• $$\alpha$$ is the polarizability of the molecule
• Dipole-induced dipole potential energy $$u(R)=-\frac{4\mu_\chem{A}^2\alpha_\chem{B}}{\left( 4\pi \varepsilon_0 \right)R^6}$$

### Dispersion Force

• Non-polar molecules a weak attractive force called the dispersion force $$u_\chem{disp} \equiv E_\chem{disp} \approx -\frac{\alpha^2 \Delta E}{2R^6}$$
• Where $$\Delta E$$ is the separation between the ground and lowest excited state of the molecule.
• Notice that all the attraction interactions are proportional to $$-\frac{1}{R^6}$$

### Model #1

• The most sophisticated model is the Lennard-Jones potential \begin{align} u_\chem{LJ}(R) &= \varepsilon\left[ \left( \frac{R_e}{R} \right)^{12} - 2 \left( \frac{R_e}{R} \right)^6 \right] \\ \text{or} \\ u_\chem{LJ}(R) &= 4\varepsilon\left[ \left( \frac{R_\chem{LJ}}{R} \right)^{12} - \left( \frac{R_\chem{LJ}}{R} \right)^6 \right] \end{align}

### Model #2

• The square well potential contains
• An impenetrable region
• An attractive region
• Interactions are turned off at long range
• $$u_\chem{sq}(R) = \begin{cases} \infty & \text{if R\le R_\chem{sq}} \\ -\varepsilon & \text{if R_\chem{sq}\lt R \le R'_\chem{sq}} \\ 0 & \text{if R>R'_\chem{sq}} \end{cases}$$

### Model #3

• The simplest model
• Particles modeled as hard spheres
• $$u_\chem{hs} = \begin{cases} \infty & \text{if R\le R_\chem{hs}} \\ 0 & \text{if R \gt R_\chem{hs}} \end{cases}$$

### Solids

• The strength of the forces holding atoms together in a solid are great compared to the forces pushing atoms apart
• There are not any translations or rotations
• Those motions have been converted into vibrations

### Liquids

• The most difficult phase to describe
• Molecules bound tightly enough to interact strongly, but not strongly enough to keep them fixed in position
• Descriptions of liquids usually rely on average properties of the structure
• Foremost is the pair correlation function $$\mathcal{G}(R)$$ $$\frac{\text{# of molecules at distance R}}{\text{# of molecules at distance R in random distribution}}$$

## Pressure of a Non-Ideal Gas

### Pressure of a Non-Ideal Gas

• Use methods developed in previous chapters to solve for the pressure of a non-ideal gas
• We need to solve the partial derivative $$P=-\left( \frac{\partial E}{\partial V} \right)_{S,N} = \left( \frac{\partial E}{\partial S} \right)_{V,N} \left( \frac{\partial S}{\partial V} \right)_{E,N}$$
• The first term defines temperature
• The dependence on the intermolecular forces must be due to the last term

### Entropy in a Non-Ideal Gas

• The entropy is determined by the ensemble size $$\Omega$$
• We need to determine how $$\Omega$$ is affected by the intermolecular forces
• The degree of freedom is directly affected is translational motion
• The partition function $$\mathcal{P}(E)=\Omega(E)\mathcal{P}(i) = \frac{\Omega(E)e^{-\frac{E}{k_\mathrm{B}T}}}{Q(T)}$$

### Now Some Highlights of the Derivation

• The $$N$$-particle translational partition function for the ideal gas $$Q_\chem{trans}(T,V) = \frac{1}{N!}\left( \frac{2\pi mk_\mathrm{B}T}{h^2} \right)^\bfrac{3N}{2} V^N$$
• The probability of the system being in any one translational state $$i$$ that has energy $$E$$ is $$\mathcal{P}(i) = \frac{e^{-\frac{E}{k_\mathrm{B}T}}}{Q_\chem{trans}(T,V)}=\frac{1}{\Omega}$$

### The Pressure of a Non-Ideal Gas

• After a lot of work we find that \begin{align} P &= \frac{Nk_\mathrm{B}T}{V}-\frac{N^2k_\mathrm{B}T}{2V^2} \mathcal{I}(T) \\ &= \frac{nRT}{V}-\frac{n^2RT}{2V^2}\mathcal{N}_\mathrm{A} \mathcal{I}(T) \end{align}
• where $$\mathcal{I}(T)\equiv V \int_0^1 \int_0^1 \int_0^1 \left( e^{-\frac{u(R)}{k_\mathrm{B}T}}-1 \right)\,dm_x\,dm_y\,dm_z$$

### The Virial Expansion

• One of the two most common forms of the non-ideal gas law
• Writes the pressure as power series in the density $$P=RT\left[ \frac{1}{V_m} + B_2(T)\frac{1}{V_m^2} \right]$$ where $$B_2(T)\approx -\frac{1}{2}\mathcal{N}_\chem{A} \mathcal{I}(T) = -2\pi \mathcal{N}_\mathrm{A} \int_0^\infty \left( e^{-\frac{u(R)}{k_\mathrm{B}T}} -1 \right)R^2\,dR$$

### Virial Expansion Continued

• There are higher order terms in the virial expansion $$P = RT \left[ \frac{1}{V_m} + B_2(T)\frac{1}{V_m^2} + B_3(T)\frac{1}{V_m^3} + \dots \right]$$
• Ignoring terms past the second order we find that $$V_m = \frac{RT}{2P}\left[ 1 + \sqrt{1+\frac{4B_2(T)P}{RT}} \right]$$

### Virial Expansion Examples

System $$B_2\, @\, 273\,\mathrm{K} \\ (\mathrm{L\,mol^{-1}})$$ $$V_m\, @\, 1\,\mathrm{bar} \\ (\mathrm{L})$$ $$V_m\, @\, 10\,\mathrm{bar} \\ (\mathrm{L})$$
Helium 0.0222 22.733 2.293
Argon -0.0279 22.683 2.243
Ideal Gas 22.711 2.271

### The van der Waals Coefficients

• The second most common form of a non-ideal gas law
• Basically is a rewrite of the $$B_2(T)$$ coefficients
• Doing some math we can arrive at \begin{align} & B_2(T) \approx b-\frac{a}{RT} \\ & \text{where} \\ & b = \frac{2\pi \mathcal{N}_\chem{A} R_\chem{LJ}^3}{3} \text{ and } a=\frac{16\pi \mathcal{N}_\chem{A}^2\varepsilon R_\chem{LJ}^3}{9} \end{align}
• Plugging these values in and doing algebra we arrive at the traditional form $$\left( P+\frac{a}{V_m^2}\right) \left(V_m-b\right) = RT$$

### Lennard-Jones Parameters, van der Waals Coefficients, and Second Virial Coefficients

Gas $$\bfrac{\varepsilon}{k_\mathrm{B}} \\ (\mathrm{K})$$ $$R_\chem{LJ} \\ (\mathrm{\AA})$$ $$a \\ (\mathrm{L^2\,bar\,mol^{-1}})$$ $$b \\ (\mathrm{L\,mol^{-1}})$$ $$B_2(298\,\mathrm{K}) \\ (\mathrm{L\,mol^{-1}})$$ $$B_2(298\,\mathrm{K})\text{(calc)} \\ (\mathrm{L\,mol^{-1}})$$
$$\chem{He}$$ 10 2.58 0.0346 0.0238 0.012 0.020
$$\chem{Ne}$$ 36 2.95 0.208 0.01672 0.011 0.022
$$\chem{Ar}$$ 120 3.44 1.355 0.03201 -0.016 -0.004
$$\chem{Kr}$$ 190 3.61 2.325 0.0396 -0.051 -0.042
$$\chem{H_2}$$ 33 2.97 0.2453 0.02651 0.015 0.023
$$\chem{N_2}$$ 92 3.68 1.370 0.0387 -0.004 0.011
$$\chem{O_2}$$ 113 3.43 1.382 0.03186 -0.016 -0.001
$$\chem{CO}$$ 110 3.59 1.472 0.03948 -0.008 0.001
$$\chem{CO_2}$$ 190 4.00 3.658 0.04286 -0.126 -0.057
$$\chem{CH_4}$$ 137 3.82 2.300 0.04301 -0.043 -0.016
$$\chem{C_2H_2}$$ 185 4.22 4.516 0.05220 -0.214 -0.062
$$\chem{C_2H_4}$$ 205 4.23 4.612 0.05821 -0.139 -0.080
$$\chem{C_2H_6}$$ 230 4.42 5.570 0.06499 -0.181 -0.115
$$\chem{C_6H_6}$$ 440 5.27 18.82 0.1193 -1.454 -0.542

### Example 4.1

Estimate the van der Waals coefficient for helium and argon using their Lennard-Jones parameters.

### Conversion to a Liquid

• As we compress a gas the particles get closer together
• Once the spacing becomes less than the average molecular size we expect the fluid to behave as a liquid
• We now need to look at the pair correlation function

## The Pair Correlation Function

### Pair Correlation Function

• The average number of molecules at distances between $$R_a$$ and $$R_b$$ is given as $$\bar{N}_{R_a,R_b} = 4\pi \rho \int_{R_a}^{R_b} \mathcal{G}(R)R^2\,dR$$
• We can evaluate a function similar to the pair correlation function, $$\mathcal{P}_R(_{R12})$$
• The position probability distribution function per unit volume $$\mathcal{P}_{V^N}(x_1,\dots,z_N) = \frac{e^{-\frac{U(x_1,\dots,z_N)}{k_\mathrm{B}T}}}{Q'_U(T,V)}$$

### Position Probability Distribution Function

• After doing quite a bit of calculus we arrive at $$\mathcal{P}_R(R)=\frac{4\pi R^2V\int_0^\infty \cdots \int_0^\infty e^{-\frac{U(x_1,\dots,z_N)}{k_\mathrm{B}T}}\,dx_3\dots dz_N}{Q'_U(T,V)}$$
• It turns out that $$4\pi \rho \int_{R_a}^{R_b} \mathcal{G}(R)R^2\,dR = \bar{N}_{R_a,R_b} = \left(N-1\right) \int_{R_a}^{R_b} \mathcal{P}_R(R)\,dR$$
• We can equate the integrands $$\left(N-1\right) \mathcal{P}_R(R)\,dR = 4\pi \rho \mathcal{G}(R)R^2\,dR$$

### Manipulating This Expression

\begin{align} \mathcal{G}(R) &= \frac{N-1}{4\pi \rho R^2}\mathcal{P}_R(R) = \left( \frac{V(N-1)}{4N\pi R^2} \right) \mathcal{P}_R(R) \\ &= \left(\frac{V}{4\pi R^2}\right) \mathcal{P}_R(R) \\ &= \left(\frac{V}{4\pi R^2}\right) \frac{4\pi R^2V \int_0^a \cdots \int_0^c e^{-\frac{U}{k_\mathrm{B}T}}\,dx_3\dots dz_N}{Q'_U(T,V)} \\ &= \frac{V^2 \int_0^a \cdots \int_0^c e^{-\frac{U}{k_\mathrm{B}T}}\,dx_3\dots dz_N}{Q'_U(T,V)} \end{align}

### More Manipulations

\begin{align} \mathcal{G}(R) &= V^2 \frac{\int_0^a \cdots \int_0^c e^{-\frac{U}{k_\mathrm{B}T}}\,dx_3\dots dz_N}{\int_0^a \cdots \int_0^c e^{-\frac{U}{k_\mathrm{B}T}}\,dx_1\dots dz_N} \\ &= V^2 \frac{V^{N-2}e^{-\frac{u}{k_\mathrm{B}T}}\left[ 1+\frac{\mathcal{I}(T)}{V} \right]^{\frac{N(N+1)}{2}-1}}{V^N\left[ 1+\frac{\mathcal{I}(T)}{V} \right]^{\frac{N(N+1)}{2}}} \\ &= e^{-\frac{u(R)}{k_\mathrm{B}T}}\left[ 1+\frac{\mathcal{I}(T)}{V} \right]^{-1} \approx e^{-\frac{u(R)}{k_\mathrm{B}T}} \end{align}

## Bose-Einstein and Fermi-Dirac Statistics

### Indistinguishable Particles

• The wavefunction of a system of indistinguishable particles must be
• Antisymmetric to exchange for half-integer spin particles - Fermions
• Symmetric to exchange for integer spin particles – Bosons
• For an atom, if the total of # electrons, protons, and neutrons is even then it is a Boson
• $$\chem{{}^4He}$$

### Helium-4

• Because it is a Boson lets consider a group of helium-4 atoms
• Let’s look at their translations
• They can all occupy the same quantum state
• This is non-classical because it means they all have the same position and velocity

### Counting States

• To evaluate the partition function we need to count states
• For fermions, this counting leads to Fermi-Dirac statistics
• For Bosons, this counting leads to Bose-Einstein statistics

### Consider Two Particles Limited to Five Values

$$I_b$$
$$I_a$$ 1 2 3 4 5
1 (11) (12) (13) (14) (15)
2 (21) (22) (23) (24) (25)
3 (31) (32) (33) (34) (35)
4 (41) (42) (43) (44) (45)
5 (51) (52) (53) (54) (55)

### Some States Are Available for Bosons But Not Fermions

• For Fermions $$\Omega = \frac{g!}{N!(g-N)!}$$
• For Bosons $$\Omega =\frac{(g+N-1)!}{N!(g-N)!}$$
• In these equations, there are $$g$$ possible states for $$N$$ particles
• At high temperatures, the spin statistics becomes irrelevant

### Bose-Einstein Condensates

• There is a large disparity in the lab between liquid $$\chem{{}^3He}$$ and $$\chem{{}^4He}$$
• $$\chem{{}^4He}$$:
• At low temps, the number of accessible states is small, the chance that multiple atoms occupy the same quantum state is high ($$@\, 2.17\,\mathrm{K}\text{ and } 1\,\mathrm{bar}$$)
• Liquid expands on cooling
• Thermal conductivity becomes discontinuous
• $$\chem{{}^3He}$$:
• Exchange repulsion resists formation of the condensed phase more
• $$3.2\,\mathrm{K}\,\chem{{}^3He}$$
• $$4.22\,\mathrm{K}\,\chem{{}^4He}$$
• Because $$\chem{{}^4He}$$ atoms do not repel each other strongly, liquid $$\chem{{}^4He}$$ becomes a superfluid at temps below $$2.17\,\mathrm{K}$$

### Superfluid $$\chem{{}^4He}$$

• Superfluids flow without significant resistance
• Even upward against gravity
• Superconductivity – also attributable to a similar manifestation of Bose-Einstein statistics
• In this case, pairs of electrons (called Cooper pairs) are the Bosons of the system
• Bose-Einstein statistics begin to dominate when $$\rho_\mathrm{ps} = \rho \lambda_\chem{dB}^2 \gt 2.612$$

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