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: \(\phi=\phi_\chem{B}-\phi_\chem{A}\)
The Dipole-Dipole Interaction
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
$$ \begin{align}
u_\chem{LJ}(R) &= \varepsilon\left[ \left( \frac{R_e}{R} \right)^{12} - 2 \left( \frac{R_e}{R} \right)^6 \right] \\
\text{or} \\
u_\chem{LJ}(R) &= 4\varepsilon\left[ \left( \frac{R_\chem{LJ}}{R} \right)^{12} - \left( \frac{R_\chem{LJ}}{R} \right)^6 \right]
\end{align} $$
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 \(\mathcal{G}(R)\)
$$ \frac{\text{# of molecules at distance $R$}}{\text{# of molecules at distance R in random distribution}} $$
Pair Correlation Function
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
$$ V_m = \frac{RT}{2P}\left[ 1 + \sqrt{1+\frac{4B_2(T)P}{RT}} \right] $$
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
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
$$ \left(N-1\right) \mathcal{P}_R(R)\,dR = 4\pi \rho \mathcal{G}(R)R^2\,dR $$
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
Graphical Representation of Two Boson Wavefunctions
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
$$ \rho_\mathrm{ps} = \rho \lambda_\chem{dB}^2 \gt 2.612 $$