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CH 431/Lecture 10

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LECTURE 10

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  




Partial derivatives as powerful machinery

S&McQ
888

To fully exploit the power of the state functions we need to develop some mathematical machinery by considering a number of partial derivatives.

We have seen the exact differential

dU= TdS - PdV for a reversible process

Because A≡ U-TS we wrote

dA = dU -TdS -SdT
dA = TdS -PdV -TdS -SdT = -PdV - SdT

Thus, the natural variables of A are volume and temperatures.

One way in which we can use the state functions U,H,A and G in their differential form is to make use of the fact that cross derivatives of a exact differential are identical. We can use that to derive a number of useful expressions.


The partial differentials of first and second order. Maxwell relations

S&McQ
888
899

As we said dA is an exact differential. Let's write is out in its natural variables and take a cross derivative.

Image:CH431 Image95.gif
  1. The dA expression in natural variables
  2. The partial derivatives of A of first order can already be quite interesting we see e.g. in step 2 that the first partial of A versus V (at T constant) is the negative of the pressure.
  3. Likewise in step 3 we find the (isochoric) slope with temperature gives us the negative of the entropy.
    Thus entropy is one of the first order derivatives of A. (We could argue the same thing for G, see below).
  4. When we apply a cross derivative (step 4)
  5. we get what is known as a Maxwell relation (step 5).
Question: What does 3. mean for the heat capacity? answer


A similar treatment of dG gives:

Image:CH431 Image112.gif

'Measuring' entropy

S&McQ
889

As you can see the two Maxwell relations involve a quantity that we could measure (for A: the change of pressure when heating up or cooling down isochorically) to something we cannot measure directly: entropy. (There are no entropometers!). This allows us to find entropy indirectly from a measurement we can actually do or simply from the equation of state that we (or people like Boyle, Gay-Lussac and van der Waals) distilled out of many, many experimental observations.

Let us apply the Maxwell relation on A for example to an excluded volume gas. This is a van der Waals gas without the attractive forces symbolized by the a parameter. It just has the excluded volume effect (b).

Image:CH431 Image96.gif


As you can see we can directly calculate how the entropy changes with volume, once we know the equation of state in this case.

(Question: how does this work for a complete van der Waals gas?) Δ

The Maxwell relation on G allows us to study how entropy depends on pressure. For an ideal gas we immediately obtain an expression (see also fig 22.4).

Image:CH431 Image113.gif

However we could also use the expression to correct for non-ideality. This is explained in 22-6. As the correction term in table 21.1 shows the corrections to tabulated values for the entropy of a gas like nitrogen is typically rather small.

Measuring entropy of an electrochemical reaction

S&McQ
none
cf. 887ex

The idea of measuring entropy is certainly not limited to gases, ideal or not. Take for instance a battery. In such a system an electrochemical reaction in used to do electrical work. Such work can be added to the G function by adding a pair of conjugate variables ℰ.e, where ℰ stands for the electromotive force (the voltage in Volts that the battery produces) and e stands for the amount of charged passed (the integral over current I in Coulombs). (Volts x Coulombs = Joules). We can write out dG and presume pressure constant. This then gives us a Maxwell relation as follows:

Image:CH431 Image122.gif

As you can see we could measure how the voltage of the cell changes with temperature at constant pressure and without passing any charge (i.e. the open cell voltage). This immediately gives us how the entropy changes when charge is passed. As the latter is directly linked to the stoechiometry of the electrochemical reaction, this gives us ΔreactionSmolar if we make e the appropriate number of moles of charge.

Energy U versus volume

S&McQ
890

We know that for an ideal gas energy U does not depend on volume. Let us investigate that and take it a step further.

To do so we are going to use the second and the last (Maxwell) formula for A above and maniplate A some more..

Let us investigate the effect of a isothermal change in volume on the Helmholtz free energy A:

Image:CH431 Image101.gif

The first line is simply the definition of A, to which we applied a volume derivative at T=constant. To find the third epression we use the two expressions step 2 and 5 above to substitute and rearrange a bit. What results is very interesting because the expression allows us to investigate what happens to the energy U when V changes. Let's check it for an ideal gas.

Image:CH431 Image106.gif

Here we used the ideal gas equation of state (thank you Boyle, Gay-Lussac..) to find the derivative of P versus T and we get the result we expected: U does not depend on V and this follows directly from the gas law!.

But we are in a position now to look at more complicated cases. Let us take a gas that does have excluded volumes (b) but no attraction terms

Image:CH431 Image104.gif
Image:CH431 Image105.gif

Again we see that such a gas also shows no change in U with volume!

But how about a complete van der Waals gas?

Image:CH431 Image103.gif oops, the sign of the a term is wrong. Need to upload another formula
Image:CH431 Image107.gif

As you can see, for such a gas the energy does depend on (molar) volume. In fact we could use the expression to integrate and find U as function of the molar volume. Of course we need to make sure not to send the molar volume to zero as that is not very realistic.

The above examples shown the great strength of the 'partial derivative machinery' of thermodymanics. We have derived a number of expressions that we could hardly had guessed to exist and used them to expand our understanding of gases even to the more complicated cases.

The relation between the heat capacities

S&McQ
892
p19-27
p22-11

The heat capacities CV at constant volume and CP constant pressure are related. The following derivation is rather spread out in the book (see page 892, problem 19-27 and problem 22-11)

Image:CH431_Image109.gif Image:CH431_Image115.gif

In step 10 we use a result shown above

The importance here is not so much the derivation but the end result (although it is good practice to make sure you can do the derivation yourself). As you can see the relationship CP = CV +nR for an ideal gas can be derived with rigor. We introduced it before in a rather less rigorous way. The general expression involves two new concepts:

κ the (isothermal) compressibility
α the (isobaric) expansion coefficient.

Measuring α and κ

S&McQ
none

For crystalline solids both can be measured by performing e.g. an X-ray diffraction experiment, either under heating or under applying a pressure (in a diamond anvil e.g.) and looking at the way that the positions of the diffraction peaks change. In such an experiment the regular stacking of the molecules in the lattice function as a grating for the X-rays (much like the scratches of a visible monochromator (like on a CD) does for the visible). The Martin group does this kind of measurements on a routine basis at the NSLS synchrotron X-ray source in Brookhaven NY. The peak positions are directly related to the size of the unit cell, that is to how far a moiety (in this case a S8 sulfur molecule) in the crystal lattice is away from its neighbor on the average.

The distances d can be determined from the scattering angle θ once the wavelength λ is known by using Bragg's law:

d = λ/(2.sinθ)

For condensed phases α and κ are small but important

For condensed phases both α and κ are pretty small: their volumes do not change. However their small changes are hard to stop, which makes isochoric experiments very difficult.

Question: At 25oC and 1 bar α and κ for water are about 2.5 10-4 K-1 and 5 10-5 Pa-1 resp. If we heat up a droplet of water by 10K, how much pressure must we exert to do that isochorically? answer

Image:CH431 sulfur.jpg


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