Statistical Mechanics Applications

Notes Chemistry

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Langmuir adsorption

To illustrate the difference in choices of ensembles.

Canonical ensemble

Find the partition function from the number of ways to distribute

$\Omega(N)=\frac{M!}{N!(M-N)!}\\ \implies Q=\frac{M!}{N!(M-N)!}\exp(\beta N\epsilon)$

With Stirling’s approximation, chemical potential can be calculated

$\mu=-\epsilon+k_B T(\ln N-\ln(M-N)).$

Rearranging from here,

$\rho=\frac{1}{1+\exp(\beta(\mu+\epsilon))}.$

Grand ensemble

The grand partition function, on the other hand, can be simplified by binomial theorem!

$\begin{align} \Xi&=\sum_{N=0}^MQ\exp(N\beta\mu)\\ &=\sum_{N=0}^M\frac{M!}{N!(M-N)!}\exp(\beta N(\epsilon+\mu))\\ &=(1+\exp(\beta (\epsilon+\mu))) \end{align}$

And the average number is the derivative

$\begin{align} \langle N\rangle&=\left( \frac{\partial \ln \Xi}{\partial \beta\mu} \right)_{M,T}\\ &=\frac{M\exp[\beta(\mu+\epsilon)]}{1+\exp[\beta(\mu+\epsilon)]}\\ \langle \rho\rangle&=\frac{1}{1+\exp(\beta(\mu+\epsilon))}. \end{align}$

Somewhat faster than using canonical ensemble.

Widom’s particle insertion method

Chemical potential is not an intuitive quantity as temperature or pressure. The following analyses allows a more intuitive understanding of the chemical potential.

From the fundamental equations and bridge relationship,

$\mu=\left( \frac{\partial A}{\partial N} \right)_{T,V}=-k_BT\left( \frac{\partial \ln Q}{\partial N} \right)_{T,V}.$

In the limit of large $N$ , where it can be treated as continuous,

$\mu=-k_BT\ln\left( \frac{Q(N+1,V,T)}{Q(N,V,T)} \right).$

For an ideal gas,

$Q=\frac{V^N}{\Lambda^{3N}N!}\\ \mu^\mathrm{id}=-k_BT\ln\frac{V}{\Lambda^3 (N+1)}.$

We can now split the chemical potentials of all interacting particles into ideal and excess parts.

$\begin{align} \beta\mu^\mathrm{ex}&=\mu-\mu^\mathrm{id}\\ &=-\ln\left( \frac{\int \exp[-\beta U(\mathbf{r}^{N+1})]\,d\mathbf{r}^{N+1}}{V\int \exp[-\beta U(\mathbf{r}^{N})]\,d\mathbf{r}^{N}} \right)\\ &=-\ln\left(\frac{1}{V}\int \frac{\int \exp[-\beta \Delta U(\mathbf{r}^{N};\mathbf{r}_{N+1})]\exp[-\beta U(\mathbf{r}^{N})]\,d\mathbf{r}^{N}}{\int \exp[-\beta U(\mathbf{r}^{N})]\,d\mathbf{r}^{N}} \,d \mathbf{r}^{N+1}\right)\\ &=-\ln(\langle \exp[-\beta \Delta U] \rangle_N) \end{align}$

The last step is valid if the system is homogeneous, so $\Delta U$ does not depend on $\mathbf{r}_{N+1}$ .

This expression is key to calculate $\mu$ in a computer simulation.

Therefore, $\mu^\mathrm{ex}$ is related to the average Boltzmann factor associated with the random insertion of an additional particle.

Hard-sphere gas

Potential is infinity when two molecules overlap.

$\langle \exp[-\beta \Delta U] \rangle_N=P_\mathrm{overlap}\times0+P_\mathrm{no\ overlap}\times 1$

Therefore, $\mu^\mathrm{ex}=k_B T \ln P_\mathrm{no\ overlap}$ , which is very intuitive! The more dilute, the higher $P_\mathrm{no\ overlap}$ , the lower chemical potential; vice versa.

Virial expansion

The virial expansion is an expansion of the compressibility factor in a power series of $\rho$ ,

$Z=1+B_2(T)\rho+B_3(T)\rho^2+...$

To find the coefficients, try linking $\rho$ to $\mu^\mathrm{ex}$ and use particle insertion method.

$\begin{align} \left(\frac{\partial \mu}{\partial \rho}\right)_T&=\left(\frac{\partial \mu}{\partial P}\right)_T\left(\frac{\partial P}{\partial \rho}\right)_T&&\mathrm{chain\ rule}\\ &=\frac{1}{\rho}\left(\frac{\partial P}{\partial \rho}\right)_T && \mathrm{Gibbs-Duhem}\\ &=\frac{k_B T}{\rho}[1+2B_2\rho+...]&&\mathrm{Virial\ for\ }P\\ \left(\frac{\partial \mu^\mathrm{ex}}{\partial \rho}\right)_T&=\frac{k_B T}{\rho}[2B_2\rho+...]\\ \mu^\mathrm{ex}(\rho)-0&=k_B T(2B_2\rho+\frac{3}{2}B_3\rho^2+...) \end{align}$

By far, the only assumption is that the virial expansion holds. For low density, we can truncate it to $B_2$ term only,

$B_2\approx\frac{\mu^\mathrm{ex}}{2k_B T \rho}.$

Now use Widom. By considering the probability inside and outside the interaction volume, we have

$\begin{align} \mu^\mathrm{ex}&=-k_B T \ln\left(1+\rho\int\exp[-\beta \phi(r)]-1 \,d\mathbf{r}\right)\\ &\approx k_B T \rho\int1-\exp[-\beta \phi(r)]\,d\mathbf{r}\\ B_2(T)&=\frac{1}{2}\int1-\exp[-\beta \phi(r)]\,d\mathbf{r}\\ &=\frac{1}{2}\int -f(r)\,d\mathbf{r}. \end{align}$

The abbreviation $f(r)$ is the Mayer $f$ -function.

Hard sphere

$\phi(r)=\begin{cases} \infty && r<\sigma,\\ 0 &&r\geq \sigma. \end{cases}$

Hence, $B_2(T)=2\pi\sigma^3/3$ .

Note that there is no temperature dependence, called athermal, meaning there is no energy scale in the hard-shpere model.

Square well

$\phi(r)=\begin{cases} \infty && r<\sigma,\\ -\epsilon && \sigma<r<\lambda\sigma\\ 0 &&r\geq \lambda\sigma. \end{cases}$

Hence,

$\begin{align} B_2(T)&=B_2^{HS}+\frac{1}{2}\int_\sigma^{\lambda\sigma}[\exp(\beta\epsilon)-1]r^2\,dr\\ &=\frac{2\pi\sigma^3}{3}[1-(\lambda^3-1)][\exp(\beta\epsilon)-1]. \end{align}$

Now there is a temperature dependence, comparing to the scale of $\epsilon$ . At the point $B_2$ changes sign, Boyle temperature, the system is close to ideal.

Flory-Huggins theory of polymer solutions

Consider polymer chain on a lattice. Assumptions:

No interpenetration => no double occupancy
All sites are either occupied by polymer or solvent => no empty site
Mean field approximation.

Chemical potential

We want to again use Widom’s method. Need to calculate two parts

the probability that the insertion is not an overlap
the energy changed during this insertion.

More assumptions:

The inserting polymer is not connected
Segments and solvents are randomly distributed.

With $N$ polymers, lengths $r$ , total $M$ sites, $\eta$ fraction occupancy, $z$ average non-bonded neighbouring segment, $\Delta \epsilon=\epsilon_{SP}-(\epsilon_{SS}+\epsilon_{PP})/2$ .

$\mu^\mathrm{ex}=-rk_B T \ln(1-\eta)-2rz\eta\Delta\epsilon+\mathrm{const.}$

Note:

It is necessary that $\Delta \epsilon>0$ for phase separation to be possible
any terms independent of $\eta$ is collected into constant term.

Also,

$\mu^\mathrm{id}=k_BT\ln(N/M)=k_BT\ln(\eta)+\mathrm{const.}$

By using the Flory-Huggins parameter $\chi=z\Delta \epsilon/kT$ ,

$\mu=k_B T (\ln \eta-r\ln(1-\eta)-2r\chi\eta)$

is the F-H chemical potential of a polymer solution.

Critical point

The critial points are 1st and 2nd derivatives of $P$ w.r.t. $V$ are 0. Here, $M$ is effectively $V$ , and by using Gibbs-Duhem, the extrema of $(\partial \mu/\partial \eta)$ is the same as $(\partial P/\partial V)$ .

$\eta_\mathrm{crit}=\frac{1}{1+\sqrt{r}}\\ \chi_\mathrm{crit}=\frac{(1+\sqrt{r})^2}{2r}$

Unlike regular binary solutions, the polymer solution

does not have a symmetric phase diagram - $\eta_\mathrm{crit}$ moves increasingly to the left as the polymer length increase
short limit of $\chi$ is 2; long limit is 0.5

Another way to look at the asymmetric nature is to investigate $U$ and $S$ .

The potential energy change for mixing is

$\Delta U=Mk_B T\chi\eta(1-\eta)$

which still causes demixing, but is still symmetric.

The entropic contribution is

$\Delta S=-k_BM\left[(1-\eta)\ln(1-\eta)+\frac{n}{r}\ln(\eta)\right].$

Here we can see that longer polymer leads to less entropy, and the entropy is asymmetric.

Flory-Huggins parameter

Compare to the virial expansion (noting $\rho=\eta/r$ ), at low densities

$B_2=r^2(\frac{1}{2}-\chi).$

Ideal behaviour occurs when $\chi=1/2$ ;
Good solvent - smaller energy penalty - $\chi<1/2$ , $B_2>0$ , increase osmotic pressure;
Poor solvent - $\chi>1/2$ , reduced osmotic pressure, phase separation.

Experimental determination of $B_2$ measures the osmotic pressure.

Landau theory

Order of phase transition

At phase transition, $T$ , $P$ and $\mu$ , and hence $G$ of the two coexisting phases are equal.

1st order phase transition:

The 1st derivative of free energy is discontinuous;
Sudden volumn change, non-zero enthalpy of transition;
Need to overcome a free energy barrier.

2nd order phase transition:

The 1st derivative is continuous, but the 2nd derivative is not;
Continuous, and no barrier.

Parameterization

Use order parameter to describe free energy as a polynomial.

The order parameter is 0 in disordered phase, and non-zero in ordered phase.
Can be symmetric (e.g. magnetization) with only even power terms, or not (e.g. solid-fluid transition).
Coefficients are continuous functions of temperature.
Linear term always vanishes (since derivative at 0 need to be 0 at (at least) high T).
Highest power has a +ve coefficient, for stability.

Techniques

Sketch out the shapes at temperatures above, at and below coexistence.
Clearly keep track of the expected number of roots in free energy and in 1st derivative.
Use $\Delta=b^2-4ac=0$ to simplify calculation whenever double roots are expected.
Always verify the order by checking the continuity of derivatives.
- Use $\frac{d A}{d T}=\frac{\partial A}{\partial M}\frac{\partial M}{\partial T}+\frac{\partial A}{\partial T}$ to simplify.

Limitations

Assumption that the polynomial expansion is valid?
Fluctuations around the minimum are ignored.

Diffusion

The Brownian motion of molecules is responsible for diffusion.

Fick’s laws

First law: flux proportional to concentration gradient,

$j_x(x,t)=-D\frac{\partial c(x,t)}{\partial x}.$

By considering the accumulation of material in a slab, and take small limit of $x$ and $t$ ,

$\frac{\partial c}{\partial t}=-\frac{\partial j_x}{\partial x}.$

Second law, aka diffusion equation: combining the above two equations,

$\frac{\partial c(x,t)}{\partial t}=D\frac{\partial^2 c(x,t)}{\partial x^2}.$

Deduce the property of the solution

If the initial condition is $c(x,0)=\delta(x)$ , the exact solution is a Gaussian, widening with time

$c(x,t)=\frac{1}{\sqrt{4\pi Dt}}\exp(-\frac{x^2}{4Dt}).$

However, to find out the time-dependence of the mean squared width does not require the knowledge of this exact form, as the following.

$\begin{align} \langle x^2(t)\rangle&=\sigma^2-\langle x(t)\rangle^2\\ &=\int x^2c(x,t)\,dx-0\\ \int x^2\frac{\partial c(x,t)}{\partial t} \,dx &=D\int x^2\frac{\partial^2 c(x,t)}{\partial x^2}\,dx &&\mathrm{from\ Fick's\ 2nd}\\ \frac{\partial}{\partial t}\int x^2c(x,t)\,dx &=D\int x^2\frac{\partial^2 c(x,t)}{\partial x^2}\,dx\\ \frac{\partial \langle x^2(t)\rangle }{\partial t}&=D\{[x^2\frac{\partial c(x,t)}{\partial x}]_{-\infty}^\infty-2D\int x\frac{\partial c(x,t)}{\partial x}\,dx\}\\ &=2D\{ -[xc(x,t)]_{-\infty}^\infty +\int c(x,t)\, dx \}\\ &=2D. \end{align}$

Random walk model

Lattice spacing is $l$ , equal probability left or right, frequency $\Gamma$ .
Calculate expectation values $\langle x^2 \rangle$ by listing out all possibilities - note the need to use double sum.

$\langle x^2(t) \rangle=\Gamma t l^2\\ \implies D=\frac{\Gamma l^2}{2}.$

Green-Kubo relation

Into continuous space, and using $x(t)=\int_0^t v(t')\, dt'$ , we can evaluate the mean square displacement as

$\begin{align} \langle x^2(t) \rangle&=\langle \left(\int_0^tv_x(t')\,dt'\right)^2 \rangle\\ &=\int_0^t\int_0^t \langle v_x(t')v_x(t'') \rangle\,dt'\,dt''\\ &=2\int_0^tdt' \int_0^{t'}\langle v_x(t')v_x(t'') \rangle\,dt'' &&(*)\\ &=2\int_0^tdt' \int_0^{t'}\langle v_x(t'-t'')v_x(0) \rangle\,dt''. \end{align}$

The term $\langle v_x(t')v_x(t'') \rangle$ is called the velocity auto-correlation function, which measures the average correlation between the velocity of a particle at times $t'$ and $t''$ .
The line $(*)$ comes from splitting the limit of integration and changing the order.
The last line is valid because the auto-correlation function is a equilibrium property, and hence is invariant under an arbitrary time shift.

Therefore, we can write the diffusion coefficient in the Green-Kubo relation

$D=\int_0^\infty \langle v_x(\tau)v_x(0) \rangle\,d\tau.$

Langevin equation

With GK relation, we can start to estimate diffusion coefficients, provided a few more assumptions.

Under an external force $F$ , the average drift velocity is $\langle v \rangle=F/\gamma$ . $\gamma$ is the friction coefficient, and $u=1/\gamma$ is the mobility.
It can be estimated by Stokes expression $\gamma=6\pi\eta a$ .
Random collisions with neighbours is caused by a random force $F^R(t)$ , not correlated with the velocity.

Now the equation of motion of the particle is

$m\dot{v}(t)=F^R(t)-\gamma v(t)$

which is the Langevin equation.

Multiply by $v(0)$ and take average, the $F^R$ term vanishes, and results in a ODE $m\langle \dot{v}_x(t)v_x(0) \rangle+\gamma\langle v_x(t)v_x(0) \rangle=0$ .

This can be solved to get $\langle v_x(t)v_x(0)\rangle=\langle v^2_x(0) \rangle\exp[-\gamma t/m]$ . Maxwell distribution gives $m\langle v^2_x(0) \rangle=k_B T$ .

Hence from GK, $D=\frac{k_B T}{6\pi\eta a}$ . This is Stokes-Einstein relation.

Einstein-Smoluchowski relation

Same goal as Langevin, but more general, using flux. At equilibrium, the diffusion flux exactly balances the flux due to external force.

The flux is the number density times the average (over these particles) velocity: $j=\rho\langle v\rangle$ .

Flux due to external force:

$\begin{align} j_\mathrm{ext}&=\rho\langle v_\mathrm{ext}\rangle\\ &=\rho uf_\mathrm{ext}\\ &=-\rho u\frac{\partial U(x)}{\partial x}. \end{align}$

Diffusion flux (by Fick’s 1st):

$j_\mathrm{D}=-D\frac{\partial \rho(x)}{\partial x}.$

Therefore, at equilibrium,

$0=-\rho u\frac{\partial U(x)}{\partial x}-D\frac{\partial \rho(x)}{\partial x}\\ \mathrm{Also,\ }\rho(x)=\rho_0\exp[-U(x)/k_BT].$

Satisfied if $u=D/k_BT$ . This is the Einstein-Smoluchowski relation.

Polymers

Closely linked to diffusion.

Size

The freely jointed chain consists of segments, which is larger than monomers (Kuhn length). Analysis of the end-to-end distance $\langle \mathbf{R}_\mathrm{ee}\rangle$ is similar to the random walk in diffusion, except that it is vector addtion.

$\langle \mathbf{R}_\mathrm{ee}\rangle=\mathbf{0}\\ \begin{align} \langle \mathbf{R}_\mathrm{ee}^2\rangle&=\langle (\sum_i\mathbf{l}_i)\cdot(\sum_j\mathbf{l}_j)\rangle\\ &=\sum_{ij}\langle \mathbf{l}_i\cdot \mathbf{l}_j \rangle\\ &=\sum_{i}l^2+\sum_{i\neq j}l^2 \langle \cos\theta\rangle\\ &=Nl^2 \end{align}$

The expectation of $\theta$ is 0 since segments are uncorrelated.

The ratio between RMS end-to-end distance and contour length is therefore $1/\sqrt{N}$ .

Probability distribution of size

Considering 1D case, then it is not too hard to work out the distribution of end-to-end length.

Abbreviate $(N_R-N_L)=n$ . The probability distribution is

$P(N,n)=\frac{N!}{\frac{N+n}{2}!\frac{N-n}{2}!}\left(\frac{1}{2}\right)^N.$

Taking natural log for easier manipulation.

$\begin{align} \ln P&=N\ln N -\frac{1}{2}[(N+n)\ln(N+n)+(N-n)\ln(N-n)]\\ &=-\frac{1}{2}[(N+n)\ln(1+\frac{n}{N})+(N-n)\ln(1-\frac{n}{N})]\\ &=-\frac{1}{2}[(N+n)\left\{ \frac{n}{N}-\frac{1}{2}\left(\frac{n}{N}\right)^2 \right\}+(N-n)\left\{- \frac{n}{N}-\frac{1}{2}\left(\frac{n}{N}\right)^2 \right\}\\ &=-\frac{n^2}{2N} \end{align}$

Note: need to take up to quadratic term in the Taylor expansion because the final result is quadratic.

Therefore, the distribution is a Gaussian,

$P\propto \exp \left(-\frac{n^2}{2N}\right).$

For normalization, it is important to note that $n$ changes in steps of 2 (since $N_R$ and $N_L$ change simultaneously). Hence, a half is needed to ‘properly’ normalize.

$P=\sqrt{\frac{1}{2\pi N}}\exp \left(-\frac{n^2}{2N}\right)$

Thermodynamics of polymers

Entropic force

Having calculated via $\Omega$ , might as well go towards entropy. In the expression for $P(N,n)$ , the $(1/2)^N$ term does not depend on $n$ , so

$\begin{align} S(n)&=k_B\ln \Omega=k_B\ln P +\mathrm{const.}\\ &=-\frac{k_B n^2}{2N}+\mathrm{const.}\\ S(X_\mathrm{ee})&=-\frac{k_BX_\mathrm{ee}^2}{2Nl^2}+\mathrm{const.} \end{align}$

In this context, the first law becomes

$dE=TdS+FdX_\mathrm{ee}.$

At constant temperature, for ideal polymer,

$T\left( \frac{\partial S}{\partial X_\mathrm{ee}} \right)_T+F=\left( \frac{\partial E}{\partial X_\mathrm{ee}} \right)_T=0.$

Solving for $F$ gives a ‘restoring force’, which is completely entropic in origin:

$F_\mathrm{restoring}=-F=-\frac{X_\mathrm{ee} k_B T}{Nl^2}.$

Force, without Stirling

Formally write the ‘internal energy’ as $E=-FX_\mathrm{ee}=-Fl\sum_i s_i$ , where $s_i$ is $\pm 1$ . This is a separable energy, so it is possible to write partition functions.

$q=\sum_{s=\pm1}\exp[-\beta Fl s]=2\cosh(\beta F l)\\ Q=q^N.$

Then,

$\begin{align} \langle X_\mathrm{ee}\rangle&=\frac{\langle E\rangle}{-F}\\ &=\frac{1}{F}\frac{\partial \ln Q}{\partial \beta}\\ &=Nl\tanh(\beta Fl)\\ F&=\frac{k_B T}{l}\mathrm{artanh}\left(\frac{X_\mathrm{ee}}{Nl}\right). \end{align}$

Adiabatic stretching

Equivalent to saying constant $S$ . Differentiate $E(T, X_\mathrm{ee})$ w.r.t. $X_\mathrm{ee}$ at constant $S$ gives

$\left( \frac{\partial E}{\partial T} \right)_{X_\mathrm{ee}}\left( \frac{\partial T}{\partial X_\mathrm{ee}} \right)_S+\left( \frac{\partial E}{\partial X_\mathrm{ee}} \right)_T=\left( \frac{\partial E}{\partial X_\mathrm{ee}} \right)_S.$

Therefore,

$\left( \frac{\partial T}{\partial X_\mathrm{ee}} \right)_S=\frac{k_B T X_\mathrm{ee}}{C_X Nl^2}>0.$

Heats up when stretched, cools down when released!

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Author：2LalA猪

Permalink：https://butteraddict.fun/2020/04/Statistical-Mechanics-Applications/

Published on：25th April 2020, 6:23 pm

Updated on：9th June 2020, 3:19 pm

License：CC BY 4.0

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