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CH 431/Lecture 8

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LECTURE 8

CH 431
  Lecture 1  
  Lecture 2  
  Lecture 3  
  Lecture 4  
  Lecture 5  
  Lecture 6  
  Lecture 7  
  Lecture 8  
  Lecture 9  
  Lecture 10  
  Lecture 11  
  Lecture 12  
  Lecture 13  
  Lecture 14  
  Lecture 15  
  Lecture 16  
  Lecture 17  




Two items left on our wish list

The development of the new state function entropy has brought us much closer to a complete understand ing of how heat and work are related:

  1. the spontaneity problem
    we now have a criterium for spontaneity for isolated systems
  2. the asymmetry between work->heat (dissipation) and heat->work (power generation)
    at least we can use the new state law to predict the limitations on the latter.

Two problems remain:

  1. we would like a spontaneity criterium for all systems (not just isolated ones)
  2. we have a new state function S, but what is it?

Entropy on a microscopic scale

Let us start with the latter. Yes we can use S to explain the odd paradox between w and q both being forms energy on the one hand, but the conversion being easier in one direction than the other, but we have introduced the concept entropy purely as a phenomenon on tis own. Scientifically there is nothing wrong with such a phenomenological theory except that experience tells us that if you can understand the phenomenon itself better your theory becomes more powerful.

To understand entropy better we need to leave the macroscopic world and look at what happens on a molecular level and do statistics over many molecules. First, let us do a bit more statistics of the kind we will need.

Permutations

If we have n distinguishable objects, say playing cards we can arrange them in a large number of ways. For the first object in our series we have n choices, for number two we have n-1 choices (the first one being spoken for) etc. This means that in total we have

W=n(n-1)(n-2)….4.3.2.1 = n! choices.

The quantity W is usually called the number of realizations in thermodynamics.

The above is true if the objects are all distinguishable. If they fall in groups within which they are not distinguishable we have to correct for all the swaps within these groups that do not produce a distinguishably new arrangement. This means that W becomes n!/a!b!c!…z! where a,b, c to z stands for the size of the groups. (Obviously a+b+c+..+z = n)

In thermodynamincs our 'group of objects' is typically an ensemble of systems, think of size Nav and so the factorial become horribly large. This makes it necessary to work with logarithms. Fortunately there is a good approximation (by Stirling) for a logarithmic factorial:

lnN! = NlnN-N

(See book for derivation)

Changing the size of the box with the particles in it

Intermezzo: Causality versus correlation.

In Europe nosey little children who are curious to find out where their newborn little brother or sister came from, often get told that the stork brought it during the night. When you look at the number of breeding pairs of this beautiful bird in e.g. Germany since 1960 you see a long decline to about 1980 when the bird almost got extinct. After that the numbers go up again -due to breeding programs mostly-. The human birth rate in the country follows a pretty much identical curve and the correlation between the two is very high (>.98 or so). (Dr H. Sies in Nature (volume 332 page 495; 1988)

Does this prove that storks indeed bring babies? answer


Image:CH431_Image92.gif
S&McQ
818

The expansion of an ideal gas against vacuum is really a wonderful model experiment, because nothing else happens but a spontaneous expansion and a change in entropy. No energy change, no heat, no work, no change in mass, no interactions, nothing. In fact, it does not even matter whether we consider it an isolated process or not. We might as well do so. Physicists and Physical Chemists love to find such experiments that allows them to retrace causality.

All this means that if we look at what happens at the atomic level, we should be able to retrace the cause of the entropy change.

As we have seen before, the available energy states of particles in a box depend on the size of the box.

Image:CH431_Image39.gif

Clearly if the side 'a' (and therefore the volume of the box) changes the energy spacing between the states will become smaller.


Therefore during our expansion against vacuum, the energy states inside the box are changing. Because U does not change the average energy <E> is constant.

Of course this average is taken over a great number of molecules (systems) in the gas (the ensemble), but let's look at just two of them and for simplicity let us assume that the energy of the states are equidistant (rather than quadratic in the quantum numbers n).

Image:CH431_Wexpansion.jpg

As you can see there is more than one way to skin a cat, or in this case to realize the same average <E> of the complete ensemble (of only two particles admittedly). Before expansion I have shown three realizations W1, W2, W3 that add up to the same <E>. After expansion however, there are more energy states available and the schematic figure shows twice as many realizations W in the same energy interval.

Boltzmann was the first to postulate that this is what is at the root of the entropy function, not so much the (total) energy itself (that stays the same!) but the number of ways the energy can be distributed in the ensemble. Note that because the ensemble average (or total) energy is identical, we could also say that the various realizations W represent the degree of degeneracy Ω of the ensemble.

S&McQ
829ff

Boltzmann considered a much larger (canonical) ensemble consisting of a great number of identical systems (e.g. molecules, but it could also be planets or so). If each of our systems already has a large number of energy states the systems can all have the same (total) energy but distributed in rather different ways. This means that two systems within the ensemble can either have the same distribution or a different one. Thus we can divide the ensemble A in subgroups aj having the same energy distribution and calculate the number of ways to distribute energy in the ensemble A as

W= A!/a1!a2!….

Boltzmann postulated that entropy was directly related to the number of realizations W, that is the number of ways the same energy can be distributed in the ensemble. (This leads immediately to the concept of order versus disorder. E.g. if the number of realizations is W=1, all systems must be in the same state (W=A! / A!0!0!0!…) which is a very orderly arrangement of energies.)

If we were to add two ensembles to each other the total number of possible arrangements Wtot becomes the product W1W2 but the entropies should be additive. As logarithms transform products into additions Boltzmann assumed that the relation between W and S should be logarithmic and wrote:

S= k.lnW

Again, if we consider a very ordered state, e.g. where all systems are in the ground state the number of realizations A!/A!= 1 so that the entropy is zero. If we have a very messy system where the number of ways to distribute energy over the many many different states is very large S becomes very large. Thus entropy is very large.

This immediately gives us the driving force for the expansion of a gas into vacuum or the mixing of two gases. We simply get more energy states to play with, this increases W. This means an increase in S. This leads to a spontaneous process.

Partition functions and entropy

S&McQ
841

We have seen above that the partition function of a system gives us the key to calculate thermodynamic functions like energy or pressure as a moment of the energy distribution. We can extend this formulism to calculate the entropy of a system once its Q is known. The derivation is shown on page 840 and involves the use of the Stirling approximation. The end result is

S= U/T + klnQ (20.42)

Entropy versus temperature

S&McQ
853

We can put together the first and the second law for a reversible process with no other work than volume work and obtain:

dU= δqrev + δwrev
δqrev= TdS
δwrev= -PdV

Thus:

dU= TdS -PdV for reversible changes

This is a very interesting expression because we no longer have any path functions in it, as U, S and V are all state functions. This means this expression must be an exact differential.

(We can generalize the expression to hold for irreversible processes, but then it become an inequality

dU≤ TdS - PdV

We will develop this later, because we still have an item on the wish list)

Natural variables

S&McQ
897

Consider:

dU= TdS -PdV

This equality expresses U in two variables U(S,V). They play a special role and are called the natural variables of U.

Entropy and heat capacity

However, there is nothing to stop us from expressing U in other variables (see example 21-1), e.g. T and V. If fact we can derive some interesting relationships if we do:

S&McQ
854
Image:CH431_Image80.gif
  1. We write U out in T and V
  2. We write U in its natural variables
  3. We rearrange 2) to find an expression for dS
  4. We substitute 1) into 3) and rearrange
  5. This is the definition of Cv
  6. We write out S in T and V

Comparing the three bottom formulas it becomes clear that

Image:CH431_Image81.gif

A less elegant expression for the partial derivative of S versus V is also found.

We can play the same game for our other state function H= U + PV

As dH= dU +d(PV)= dU + PdV + VdP

we find for the reversible case

dH = dU + PdV + VdP= TdS -PdV + PdV + VP = TdS + VdP

The natural variables of the enthalpy are therefore S and P (not: V).

A similar derivation as above departing from the enthalpy shows that the temperature change of entropy is related to the constant pressure heat capacity:

Image:CH431_Image82.gif

This means that if we know the heat capacities as a function of temperature we can calculate how he entropy changes with temperature. Usually it is a lot easier to do this on basis of data that were obtained for P constant than for V constant, so that the route with Cp is the more common one.

Absolute entropies. The third law

S&McQ
855-856

In the unlikely case that we have Cp data all the way from 0K we could write:

Image:CH431_Image83.gif

It is tempting to set the S(0) to zero and make the entropy thus an absolute quantity. As we have seen with enthalpy, it was not really possible to do so for H(0). All we did was define ΔH for a particular process, although if Cp data are available we could construct an enthalpy function (albeit with a floating zero point) by integration of Cp (instead of Cp/T !).

As we saw, we needed to be careful for phase changes (like melting etc.) This is true for both integrations.

Still for entropy S the situation is a bit different than for H. Here we can actually put things on an absolute scale. Both Nernst and Planck have proposed to do so. Nernst postulated that for a pure and perfect crystal S should indeed to go to zero as T goes to zero. This postulate is known as the third law of thermodynamics.

It is certainly true that for the great majority of materials we end up with a crystalline material at sufficiently low materials (although there are odd exceptions like liquid helium). However, it should be mentioned that a completely perfect crystal can only be grown at zero Kelvin! It is not possible to grow anything at 0K however. At any finite temperature the crystal always incorporates defects, more so if grown at higher temperatures. When cooled down very slowly the defects tend to be ejected from the lattice for the crystal to reach a new equilibrium with less defects. This tendency towards less and less disorder upon cooling is what the third law is all about.

However, the ordering process often becomes impossibly slow, certainly when approaching absolute zero. This means that real crystals always have some frozen in imperfections. Thus there is always some residual entropy. Fortunately, the effect is often too small to be measured. This is what allows us to ignore it in many cases (but not all).

We could state the Third law of thermodynamics as follows:

The entropy of a perfect crystal approaches zero when T approaches zero.
(but perfect crystals do not exist).


Another complication arises when the system undergoes a phase transition, e.g. the melting of ice. As we can write:

ΔfusH= qP

If ice and water are in equilibrium with each other the process is quite reversible and so we have:

ΔfusS = qrev/T= ΔfusH/T

This means that at the melting point the curve for S makes a sudden jump by this amount because all this happens at one an the same temperature.

Entropies are typically calculated from calorimetric data (Cp measurements e.g.) and tabulated in standard molar form. The standard state at any temperature is the hypothetical corresponding ideal gas at one bar for gases.

In table 21.3 a number of such values are shown. There are some clear trends. E.g. when the noble gas gets heavier this induces more entropy. This is a direct consequence of the particle in the box formula: It has mass in the denominator and therefore the energy levels get more crowded together when m increases: more energy levels, more entropy


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