Review session -- updated

Sophia will have a review session Friday at 10am.

PS4 notes

There might be some confusion about the difference between partial molar properties in mixtures and intensive properties in pure components.

A partial molar something is defined by the expression given at the top problem 5, and it involves a derivative with respect to the number of molecules of a component. It is not the total quantity divided by the number of molecules in the mixture.

On the other hand, an intensive property is the quantity divided by the number of molecules, for a single component system.

Maxwell relations: be careful

You'll be tempted to start swapping around derivatives to get Maxwell relations, but remember not all derivatives can be transformed using this approach. For example,

(dS/dT)_(V,N) cannot be transformed using a Maxwell relation. It is not equal to (dP/dV)_(T,N), if that is what you were thinking.

Recall:
(dS/dT) = (d2A/dT2)
(dP/dV) = (d2A/dV2)

Therefore, these are not the same second derivative of a free energy.

If it had instead been (dS/dV)_(T,N) then we could have said that equals (dP/dT)_(V,N).

Hint on PS3 problem 5

The relationship we derived in class between Cp and Cv may be useful in both parts (b) and (c).

Hint on PS3, problem 4

In converting from A(T,V,N) to E(S,V,N), it may simplify your life to use the following relationship:

-E/T^2 = d(A/T)/dT (constant V and N)

You can use this equation to first get E(T,V,N) and then you can make the appropriate substitution to get E(S,V,N).

Be careful of using Mathematica to do all of the manipulations here. Sometimes it doesn't simplify to the best answer.

Calculus manipulations handout correction

The definition of constant-pressure heat capacity should involve a derivative at constant P, not constant V.

PS3

A couple of notes:

For problem 8.1 in MDF, do two things: (1) express Z in terms of the Gibbs free energy in integrated form (e.g., Z = G + ...). (2) write Z in differential form (e.g., Z = x dy + ...)

For problem 3, don't forget that the chemical potential is also the Gibbs free energy per particle, for a single component system.

PS3, problem 4 correction

The expression for the Helmholtz free energy should start with:

3 N k T ln[ ... ]

The N was missing in the handout.

Also, in part (b), you are calculating the intensive heat capacities (i.e. 1/N dE/dT and 1/N dH/dT).

Hint on problem 4 on PS2

In computing the entropy change of the environment, it is helpful to think of it as a constant T and P bath. We can write for the environment, at constant N,

dS = 1/T dE + P/T dV

Since the environment is a big bath, small changes of dE and dV do not affect its temperature or pressure. Therefore, we can integrate this equation to get:

DeltaS = (DeltaE + P * DeltaV) / T = DeltaH / T

Thus, if you know how much enthalpy the supercooled liquid gives up when it freezes, this must be the amount that is added to the environment, and so you can compute DeltaS for the environment this way.

Regarding problem 4 grading on PS1

If you used factorials in this problem to account for indistinguishibility of particles, i.e (V/v)^N / N! instead of just (V/v)^N, and got half a point taken off, see Sophia and you can get that half point back if you want. The N! doesn't affect the problem, because the number of particles is constant throughout (only volumes are changing). The true quantum mechanical result is that the N! should be in there, we just hadn't really gotten to that yet in class.

PS2: sign error in problem 4

In parts (b) and (c), it should be dh = de + P dV (plus instead of minus).

Some questions answered

"If 2 systems are allowed to exchange E (but not N or V) heat 'flows'. But how does it actually flow? In other words, how is energy exchanged microscopically? Via collisions? Via rearrangement of potential energies?"

This is exactly how energy moves around from one place to another. Molecules are constantly bombarding each other. Each molecule exerts a potential energy "field" around it which is felt by its neighbors. Even in dilute gases, molecules occasionally whiz past each other long enough to feel these potential energies, or occasionally even get close enough to be involved in what might be considered a collision (it's hard to define rigorously exactly when a collision occurs, since in reality, most atoms are kind of 'soft'). It is through these interactions that molecules exchange kinetic and potential energies.

Imagine, in the simplest scenario, that someone breaks a racked set of billiard balls. In that case, the kinetic energy of the cue ball is divided up and transferred to the kinetic energies of all the other balls. This is not all that different than bringing a molecule at a high temperature in contact with molecules at a lower temperature.

"What is the fundamental definition of a phase? In other words, what is different (or the same) between a system of 2 phases and a situation where two systems are just sitting next to each other?"

You're anticipating our discussion of phase equilibrium later on! Basically, a phase is a homogeneous system: no spatial variations in density, energy, temperature, etc. We can consider two phases in equilibrium as being a subset of the situation where two systems sit next to each other. The idea of phase equilibrium, however, is that these two phases sit next to each other on their own, without barriers, membranes, etc. That is, water spontaneously separates into a phase of ice and a phase of liquid at 0 degrees C.

Irreversible sand problem

From class today we were calculating the work done when the sand is raised irreversibly by the piston. Here is perhaps a clearer way to look at that problem.

Before the piston is raised, it has some potential energy U1. After the piston is raised, it has a new potential energy U2, given in this case by U1 - mgz, where m is the mass of the piston, g is the gravitational potential, and z is how high it was raised. The difference in energy (U2-U1) is equal to whatever work energy the gas put into the piston, W (the gas does work on the system).

W = (U2 - U1)
= -mgz

But since Pext = mg/A:

W = -Pext * A * z
= -Pext * Delta V

Zotero

This is not at all related to class, or even to thermodynamics, but this is such a brilliant discovery I can't help but share it. People, you have to check out Zotero (http://www.zotero.org/).

This software builds right into your Firefox browser, and it enables you to oh-so-easily save and organize scientific information, particularly journal references and PDF articles. You can tag entries with your own keywords, it does all kinds of search, and it is quickly becoming an Endnote-type replacement. Best of all, it's open source.

No, I'm not related to the Zotero folks. But I find myself using this software more and more every day, and it has really simplified a lot of things. (Insert "back in my day" grad school story here.)

Stirling's approximation

In the last homework, we made frequent use of Stirling's approximation for the log of a factorial. In fact, this approximation occurs again and again in statistical mechanics. You may have been wondering where it comes from. Here is a simple derivation. For large N we can write the derivative of ln(N!) as:

d ln(N!) / dN ~ [ln(N!) - ln((N-1)!)] / (N - (N-1))

= ln[N!/(N-1)!]

= ln N

Now, integrating we have:

ln(N!) = N ln(N) - N + const

Usually we take the constant to be zero, since it is negligible if N is very large. In general, Stirling's approximation works very well in statistical mechanical theory because we are using very large values of N.

Also, you might have noticed that Stirling's approximation simplifies the expression for number of combinations of M items picked from N. Notice that N! = N^N / e^N according to Stirling's formula. Then we can write

N!/M!(N-M)! ~ N^N / M^M (N-M)^(N-M)

Notice that the exponential terms canceled out of this expression. A little manipulation puts this formula in a nicer form:

N!/M!(N-M!) = [x^x * (1-x)^(1-x)]^(-N)

where x = M/N. Just about any time you use the combination expression, it will greatly simplify your life to convert to this kind of expression. If you don't see where this comes from, try the derivation yourself. It's just a little algebraic manipulation.

PS1 Problem 5a

There is a small typo in the diagram. It should say N_B "B" particles rather than N_A "B" particles.

One more point from lecture today

Just want to clarify something more rigorously for you mathematically inclined folks:

We were talking about how to calculate Omega_total for the case in which one system exchanges energy with another. We wrote the following down:

Omega_total = Sum Omega_1(E1 - dE) Omega_2(E2 + dE)

What we were saying in class was that Omega_total is strongly peaked around its maximum value. By strongly peaked, what we mean specifically is that we can approximate the above equation just using the maximum term, rather than the complete sum:

Omega_total = Omega_1(E1 - dE*) Omega_2(E2 + dE*)

where dE* is the value of dE that maximizes this equation. This is actually called the "maximum term method."

A question...

"Does the equal a priori rule only apply to classical systems, and not quantum ones?"

Actually, no! Everything we have been discussing in recent classes--the rule of equal a priori probabilities and the concepts of microstates, macrostates, and ensembles--are equally valid in quantum mechanics. It just turns out that the way you write these things out is a little bit more difficult to conceptualize since, instead of thinking of configurations as microstates, we have to start thinking of quantum-mechanical states. The classical viewpoint here will just make things easier to think about physically, but realize that it does not provide the basis for these things, which are instead very general.

Regarding problem 5b on PS1

Some of the math may get a little tedious in this problem. Let me make a recommendation. Once you write down the density of states expression, substitute the expression x*N for the number of defects n, where x is the fractional number of defects n/N. Then simplify the expression that way, and indeed a lot of simplification should happen.

In the rest of the problem, keep things in terms of x, rather than E or n, to keep things easy. You can still take the energy derivative in a pretty easy way. Since x = n/N = E/(epsilon*N) you can express d/dE of something as just 1/(epsilon*N) * d/dx of that something.

And yes, there are two problem 5s. Just seeing if you were paying attention. :)

Regarding problem 4 on PS1

We will touch upon this more Friday, but recall that ideal gases consist of structureless, volumeless particles. That means that if one ideal gas particle occupies a small volume cell, it doesn't exclude other particles from occupying that same volume cell. This might affect your way of counting configurations in this problem.

Some questions about todays lecture

(1) "This times averages = microstate averages part... what are we referring to by <A>, and is the delta just 1 for the correct N,V,E and 0 for the incorrect N,V,E?"

Yes, exactly. That delta just means that we only sum up A for those configurations that have the right N, V, E.

A is any property we want to calculate that comes from microscopic configurations. It could be something like "the average number of hydrogen bonds a water molecule makes with its neighboring molecules" or "the average rate at which molecules bounce off of the walls of the container (which could be used to calculate the force, and hence, the pressure)". It is just a general property that, given a list of velocities and positions of all the atoms, we could have a recipe to calculate it.

So all we are saying is that one way to measure A would be to, say, check the instantaneous molecular configuration every minute, calculate A for that configuration, and then tabulate it. We would do this for a long time, so we get a long list of A values. Then we take the average of them. This is the time average.

The alternative way we can do this, at equilibrium, is find out what the N, V, and E values are for that same system. Then we just take *all* of the microscopic configurations that have those values, in a very systematic way, and we average A over those. This average will be equal to the time average because the rule of equal a priori probabilities says that we are equally likely to pick any one of these microstates at any moment in time.


(2) "Can you say S = kB ln Omega(N,V,E) = S(N) + S(V) + S(E)?"

You can only say this in the case that we can write Omega(N,V,E) = Omega(N)Omega(V)Omega(E). In general this is not the case. The reason is, the number of ways of arranging N molecules in a volume V with energy E is usually not separable. For example, the system today in class, where we had a bunch of toy molecules that could each either be in an energy=0 or energy=1 state. If there are N molecules, and we want to find out how many ways there are to have a total energy of E, then the number of ways is N!/E!(N-E)!. There is no way to separate this into two functions f(E)f(N).

PS1

A couple of clarifications:

Problem 2 - you can assume constant volume conditions

Problems 4 and 5 - you can neglect both kinetic and potential energies.

Lecture today

Just realized there was a mistake in the summary I wrote on the board at the beginning of lecture today in the integral of (N k_B / V) dV. There should be a prefactor of N in the result: S = N k_B ln V + const(N,V).