Data for the standard state of 298.15 K and 0.1 MPa. ΔHf is the molar heat (enthalpy) of formation from the elements; S° is the standard state entropy; V is the molar volume; a, b and c are constants for the heat capacity (Cp) computed as: Cp = a + bT − cT–2 J/K-mol.
where S0 is the entropy at 0 K (the configurational, or third law entropy) and ΔSφ is the entropy change associated with any phase change. Compilations for S298 are available for many minerals. Table 2.2 lists some heat capacity constants for the power series formula as well as other thermodynamic data for a few geologically important minerals. Example 2.6 illustrates how entropy and enthalpy changes are calculated.
Example 2.6 Calculating enthalpy and entropy changes
If the heat capacity of steam can be represented by a three-term power series:
with constants
Answer: We need to calculate entropy and enthalpy associated with three changes: the conversion of water to steam, raising the steam from 100°C to 200°C, and increasing the pressure from 1 atm to 3 atm. Since both S and H are state variables, we can treat these three processes separately; our answer will be the sum of the result for each of these processes and will be independent of the order in which we do these calculations.
1 Conversion of water to steam. This process will result in ΔH of 40.6 kJ. For entropy, . We converted centigrade to Kelvin, or absolute, temperature.
2 Raising the steam from 100°C to 200°C (from 373 K to 473 K) isobarically. Since heat capacity is a function of temperature, we will have to integrate eqn. 2.112 over the temperature interval:Evaluating this, we find that . The entropy change is given by:Evaluating this, we find that .
3 Increasing pressure from 1 atm to 3 atm (0.1 MPa to 0.3 MPa) isothermally. We can use eqn. 2.114 to determine the enthalpy change associated with the pressure change. On the assumption of ideal gas behavior, we can substitute 1/T for α. Doing so, we find the equation goes to 0; thus, there is no enthalpy change associated with a pressure change for an ideal gas. This is in accord with assumptions about an ideal gas: namely, that there are no forces between molecules, hence no energy is stored as potential energy of attraction between molecules.
The isothermal pressure dependence of entropy is given by eqn. 2.106. We substitute 1/T for α and RT/P for V and integrate from P1 to P2:
The whole enthalpy and entropy changes are the sum of the changes in these three steps:
2.11 FREE ENERGY
We can now introduce two free energy functions, the Helmholtz free energy and Gibbs free energy. Gibbs free energy is one of the most useful functions in thermodynamics.
2.11.1 Helmholtz free energy
We can rearrange eqn. 2.58 to read dU − TdS = − PdV. The −PdV term is the work term and the TdS term is the heat function. TdS is the energy unavailable for work. Therefore, dU − TdS is the amount of internal energy available for work, or the free energy. We define it as A, the Helmholtz free energy:
(2.119)
As usual, we are interested in the differential form (since we are more interested in changes than in absolutes):
(2.120)
or substituting eqn. 2.58 into 2.120:
(2.121)
2.11.2 Gibbs free energy
2.11.2.1 Derivation
The Gibbs free energy is perhaps misnamed. By analogy to the Helmholtz free energy, it should be called the free enthalpy (but enthalpy is an energy), because it is derived as follows:
(2.122)
and
(2.123)
or
which reduces to:
(2.124)
