From WolfWikis

LECTURE 3

Partition functions

Once the exponential nature of f(E) and the identity of β = 1/kT are known it is easy to show that the probability p_j that a system is in state j is indeed

The partition function Q is generally a function of N,V and β (or: T).

We can use the partition function to calculate many properties of the ensemble. We have seen above that statistics can be used to calculate moments of distributions. Well we have a discrete distribution of energies here, lets calculate the first moment. It is the average energy:

<E> = Σ p_jE_j

Let's take a partial derivative of lnQ versus β with N and V constant:

Interestingly we can get the first moment of E by manipulating the partition function Q!

We can also take the derivative versus T instead and get:

(See 17-3 for derivation)

Many other quantities can be computed from Q by this route.

We make a little jump to section 17-6 at this point

Molecular partition functions versus system partition functions

A system such as a gas can consist of a large number of subsystems -e.g. molecules.

How is the partition function of the system built up from those of the subsystems? This depends on whether the subsystems are distinguisable or indistinguishable. Energy is additive so that:

E_tot(N,V) = ε₁(V) + ε₂(V) + ….

Each of the molecules can have their energy distributed over their respective energy states -e.g. its vibrations-. This means that each ε₁ is already a summation over the states of the molecule. Let us assume that we can somehow distinguish all the molecules as: a,b,c,d… and denote the energy state they are in by i,j,k:

El(N,V) = ε_ia (V) + ε_jb+

A good example would be the molecules in a molecular crystal. They only move around a fixed site and so we can distinguish by how far molecule 'a' is from a given corner of the crystal.

The systems partition function becomes:

IF the particles are independent and indistinguishable we can split up the summation into a product of molecular partition functions:

Q(N,V,β) = q_a(V,T).q_b(V,T)q_c(V,T)….

Each of this is a sum over the energy states of the molecule:

So far I have done little effort to distinguish between the partition function of a molecular system q as opposed to the whole ensemble Q (the gas e.g.). If the entities that we called 'systems' are distinguishable we can simply multiply their partition functions, just like probabilities are multipliable. So we get:

Q= q^N for N distinguishable systems

If the molecules sit embedded in a crystal we can do that because we can e.g. distinguish them by their location. In a gas that is not possible. This means that if we swap two molecules this should not add any energy levels. There are a total of N! permutations possible and to avoid counting realizations double we get:

Q= q^N / N! (indistinguishable systems)

If all the molecules in the crystal are the same, say methane we simply get:

Q(N,V,β) = q(V,β)^N

In a gas the situation is different because we have no way to distinguish one molecule from the other. If we have N molecules we can perform N! permutations (switches) that should not affect the outcome.

It can be shown that for such a system we must correct for this effect (make sure we do not count levels twice) and the partition function becomes:

Translational partition function of a monatomic gas

We are ready now to determine the translational partition function.

We take the particle-in-a-box energy levels and postulate that gas molecules can occupy these levels independently and indistinguishably (I.e. we assume an ideal gas). We also postulate that the quantum numbers n₁, n₂ and n₃ operate independently for each particle and use a cubic container with side 'a'.

The partition function for one atom becomes:

We can write out the triple summation as a product of three summations, which are identical. This leads to:

This summation is a bit nasty but in general we have billions and billions of energy levels and their ΔE is really small. In that case we are almost back in the classical continuous case and can replace the summation by an integral

I have fibbed a bit by changing the lower boundary from 1 to zero, but if there are billions of levels that one really does not matter.

The integral is still a nasty one but at least the full integral from 0 to infinity is well known. (It is the integral over a full Gaussian curve)

For convenience I abbreviated all the constants to α. Substituting back we find:

(Strictly speaking we can only apply this only if the number of available states is large compared to the number of particles. For most gases this is the case. Exceptions are e.g. He at low temperatures and the electrons in a metal. There we must apply a different kind of statistics. This is a quantum effect and will be dealt with later)

Let's immediately use our result to calculate the average energy <ε> of a molecule:

For the whole gas we know that:

This result is very similar to the result of the (older) classical kinetic gas theory that said that the observed energy of an ideal gas should read as U=3/2nRT.

We postulate therefore that the observed energy of a macroscopic system should equal the statistical average over the partition function as shown above.

In other words: if you know the particles your system is composed of and their energy states you can use statistics to calculate what you should observe on the whole ensemble.

Thus statistics gives you the bridge between the atomistic theory and the phenomenological one.

Another important point here is that for an ideal gas the energy U only depends on T. Now that the know how, it is easy to calculate the heat capacity (at constant volume):

As you can see by differentiation the value of this quantity is 3/2nR and should be constant with temperature for an ideal monatomic gas.

Breakdown of Boltzmann statistics

From quantum mechanics -treated in CH433- it can be shown that the above reasoning is only valid if the number of available states is much larger than the number of particles. For that to be the case the following relationship should hold:

If this inequality is not fulfilled a different type of statistical distribution needs to be applied. Table 17.1 in the book gives an idea when this happens. The electron gas inside a metal is a clear example of breakdown. Of the physical gases only the lightest two start to deviate a little at low temperatures (See chapter 17.7)

Other partition functions

The above derivation for the translations of a gas also holds for a polyatomic gas. Just make sure you use the total mass of the molecule. A monatomic gas has three degrees of freedom per molecule:

1. movement in x-direction
2. movement in y-direction
3. movement in z-direction

A polyatomic gas has other levels that you can 'stuff' energy into. The molecule can rotate and vibrate and if enough energy is available you could also excite the electrons involved in the σ and π- bonds.

In reasonable approximation the partition function of the molecule would become:

q_tot = q_{translational}q_vibrationalq_rotationalq_electronic

We will only scratch the surface of the additional degrees of freedom and their partition functions.

Electronic

At room temperature the system is usually in its ground electronic state. This means that the electronic partition function q_electronic = 1 . Usually we do not have to worry about these degrees of freedom. If we do there are usually just a few levels to worry about. This includes their degeneracy g. If there is a single energy level at a certain energy g=1, if there are two g=2 etc. We must simply multiply the Boltzmann factor with this number.

If there is more than one level to worry about we could follow the same procedure as we did for the translational states:

1. define the energy states and their degeneracies
2. compose the partition function q for the molecule and Q for the gas
3. use the (β or T) derivative of lnQ to determine <E>
4. use the T derivative of U ≈ <E> to find the contribution to the heat capacity.

Vibrational

The vibrational partition function's <E> and contribution to the heat capacity can be derived with exactly the same procedure. We already know the energy levels and can stuff them into the Boltzmann formulism and execute the summation that defines the partition function.

The derivation for the vibration states of the harmonic oscillator E_vib = h ν. (v+½) (h-nu times (vee plus one half)) can be found on page 740. It involves a geometric series that can be easily found from doing a long division of 1/(1-x).

The book uses a characteristic temperature Θ

This is a good way to express the stiffness of the vibrating bond in units of the Boltzmann constant. Because the stiffness obviously depends on what bond your are talking about, this is a good way to do the same thing we did for the critical temperature of the non-ideal gases. Look at fig 18.3 The vibrational heat capacity is shown as function of the reduced temperature T/Θ to get a general picture valid for all diatomic gases. Compare table 18.2 to see on what different absolute scales we have to think for different gases. Clearly the vibrational contribution to the heat capacity depends on temperature. For many molecules (especially light ones) the vibrational contribution only kicks in at quite high temperatures.

The value of Θ_vib is determined mostly by

1. the strength of a bond (the stronger the higher Θ)
2. the (effective) mass of the molecule (the lighter the higher Θ)

Molecules with low Θ often dissociate at higher temperatures, although the harmonic oscillator model is not sufficient to describe that phenomenon.

Rotational

The rotational partition function has one complication. The quantum number is J and the energy :::E=BJ(J+1)

However there is usually more than one level at a given value of J. This is a form of degeneracy g and you'll hear more of that in CH433. The number of levels goes as 1,3,5,7,9 etc or g=2J+1.

Therefore the summation to find the partition function q_rot(T) (see 18.31) contains an extra factor g=2J+1. This factor is mathematically very handy, because it happens to be the derivative of J(J+1). That makes it possible to change variables to X=J(J+1) and that makes the integration very easy. Notice in the derivation on page 744 that, just like in the translation case, we replace the summation by an integral. Because we are dealing with far fewer levels this is less justified in the rotational case.

How justified it is depends on the gas and the temperature we consider. Roughly speaking we should be at a temperature T>> Θ_rot.

The rotational constant B is directly linked to the moment of inertia I = μR², where μ is the reduced mass of the molecule (1/μ = 1/m₁ + 1/m₂) and R is the bond length.

Again we can scale the behavior of different systems to one and the same picture by introducing a characteristic temperature

As we did for the translations we can calculate the moment <ε> . For vibrations we get a relatively complicated function of temperature (18.25). For rotations the moment is simply equal to Nk_bT (see page 745) and this means that the rotational contribution to the molar C_v of a diatomic is simply R. As the vibrational and electronic contributions to the heat capacity are typically negligible at room temperature we get:

1. monatomic C_v= C_v^trans = 3/2 nR
2. diatomic C_v= C_v^trans+C_v^rot = 3/2 nR +nR = 5/2nR