terms of T and P as:
(2.108)
Substituting eqns. 2.105 and 2.106 into this, we have:
(2.109)
The coefficient of thermal expansion is 0 at absolute zero; the choice of 1 atm for the heat capacity integration is a matter of convenience because CP measurements are conventionally made at 1 atm.
Actually, the absolute entropies of real substances tend not to be zero at absolute zero, which is to say they are not “perfectly crystalline” in the third law sense. A residual entropy, S0, which reflects such things as mixing of two or more kinds of atoms (elements or even isotopes of the same element) at crystallographically equivalent sites, must also be considered. This configurational entropy is important for some geologically important substances such as feldspars and amphiboles. Configurational entropy can be calculated as:
(2.110)
where mj is the total number of atoms in the jth crystallographic site (in atoms per formula unit) and Xi,j is the mole fraction of the ith atom (element) in the jth site (see Example 2.3). We will return to this equation when we consider multicomponent systems.
Example 2.3 Configurational entropy
Olivine is an example of a solid solution, which we will discuss at length in Chapter 3. Fe and Mg may substitute for each other in the octahedral site. Assuming that the distribution of Fe and Mg within this site is purely random, what is the configurational entropy of olivine of the composition (Mg0.8,Fe0.2)2SiO4?
Answer: To solve this problem, we need to apply eqn. 2.110. We need only consider the octahedral site containing Fe and Mg, because O and Si are the only kinds of atoms occupying the tetrahedral and anion sites. The values for X for these two sites will therefore be 1, and ln(1) = 0, so there is no contribution to configurational entropy.
For the octahedral site,
2.10 CALCULATING ENTHALPY AND ENTROPY CHANGES
2.10.1 Enthalpy changes due to changes in temperature and pressure
From eqn. 2.68, we can see that the temperature derivative of enthalpy is simply the isobaric heat capacity:
and hence:
(2.111)
Thus, the change in enthalpy over some temperature interval may be found as:
(2.112)
Cp is often a complex function of temperature, so the integration is essential. Example 2.4 illustrates how this is done.
Example 2.4 Calculating isobaric enthalpy changes
How does the enthalpy of a 1 mol quartz crystal change if it is heated from 25°C to 300°C if the temperature dependence of heat capacity can be expressed as
Answer: The first step is to convert temperature to kelvins: all thermodynamic formulae assume temperature is in kelvins. So
Performing the integral, we have:
Now that we have done the math, all that is left is arithmetic. This is most easily done using a spreadsheet. Among other things, it is much easier to avoid arithmetical errors. In addition, we have a permanent record of what we have done. We might set up a spreadsheet to calculate this problem as follows:
| Values | Formulas & Results | |||
| a_ | 46.94 | H | (a_*Temp)+(b_*Temp^2)/2+c_/Temp | |
| b_ | 0.0343 | H1 | 19301.98 | J/mol |
| c_ | 1129680 | H2 | 34498.98 | J/mol |
| Temp1 | 273 | ΔH | 15.20 | kJ/mol |
| Temp2 | 373 |
This example is from Microsoft Excel™. On the left, we have written down the names for the
