This term is understood as the value of entropy in a given state (T, p) [see3.2.8]. To calculate it we proceed from the state (T = 0, pst = 101.325 kPa), and by way of a sequence of thermo- dynamic processes we arrive at the state (T ,p) while summing the entropy changes during the individual processes
where S(T1, pst) is the entropy of a substance at a sufficiently low temperature T1, such that the Debye relation (3.62) holds for T ≤ T1
S(T1, pst) = n Z T1 0 const T3 T dT = n const T3 1 3 .
∆S(s)is the entropy change on heating the substance from T1to the normal melting temperature Tfus, ∆S(s)= Z Tfus T1 C(s) p (T, pst) T dT .
∆fusS is the entropy change on melting
∆fusS =
∆fusH Tfus .
∆S(l) is the entropy change on heating a liquid substance from the normal melting temper- ature to the normal boiling temperature,
∆S(l) = Z Tvap Tfus C(l) p (T, pst) T dT .
∆vapS is the entropy change on boiling
∆vapS =
∆vapH Tvap
.
∆S(g) is the entropy change of a gaseous substance on its transition from the point (T
vap, pst) to (T, p) ∆S(g) = Z T Tvap Cp(T, pst) T dT − Z p pst ∂V ∂T ! p dp .
S Symbols: The superscripts (s),(l), (g) are used to denote quantities in a solid, liquid and gaseous phase.
Note: If a substance in the solid state transforms from one crystalline form to another (e.g. rhombic sulphur→ monoclinic sulphur), the entropy change at this phase transition has to be included in the formula.
Example
Calculate the absolute molar entropy of liquid sulphur dioxide at T = 200 K and at the stan- dard pressure 101.325 kPa. Data: T1 = 15 K, Cpm(s) = 3.77 J mol
−1K−1 at T
1, ∆S(s) = 84.2 J mol−1K−1, Tfus = 197.64 K, ∆fusH = 7403 J mol−1, Cpm(l) = 87.2 J mol−1K−1. Sulphur dioxide exists in only one crystalline form.
Solution
We determine the constant in the Debye relation (3.62) from the condition const T3 = Cpm(s) at T = T1.
Individual entropy contributions have the following values:
S(T1) = Z T1 0 const T3 T dT = const T13 3 = 3.77 3 = 1.257 Jmol −1 K−1. ∆fusS = ∆fusH Tfus = 7403 197.64 = 37.457 Jmol −1 K−1. ∆S(l) = Z T Tfus C(l) pm T dT = 87.2 ln 200 197.64 = 1.035 Jmol −1 K−1.
The absolute molar entropy of liquid sulphur dioxide is Sm(T, pst) = 1.257 + 84.200 + 37.457 + 1.035 = 123.949 J mol−1K−1.
3.5.6
Helmholtz energy
3.5.6.1 Dependence on temperature and volume
The Helmholtz energy is calculated from the definition (3.12), and from the dependence of internal energy (3.68) and entropy (3.79) on T and V
The change in the Helmhotz energy with volume at constant temperature can be calculated by integrating equation (3.35) with respect to the general prescription (3.31)
F (T, V2) = F (T, V1) −
Z V2
V1
p dV , [T ] . (3.93)
3.5.6.2 Changes at phase transitions
The changes of the Helmholtz energy at crystalline transformations, melting and boiling, ∆crystF , ∆fusF and ∆vapF during reversible phase transitions are calculated from the relation
∆F = −p∆V , [T, p, reversible phase transition] , (3.94) where ∆V is the respective change in volume.
For irreversible phase transitions, we have the inequality
∆F < −p∆V , [T, p, irreversible phase transition] . (3.95) In this case the change in the Helmhotz energy is calculated using the procedure described in 3.5.9.
Example
Calculate the change in the Helmholtz energy during reversible evaporation of 1.8 kg of liquid water at T = 373.15 K and p = 101.325 kPa. Assume that in the given state the equation of state of an ideal gas holds for water vapour, and that the volume of liquid water is negligible as compared with that of vapour.
Solution
The assigned values T and p correspond to the normal boiling point [see7.1.5] of water, i.e. to a reversible phase transition. Hence we use relation (3.94). The change in volume on evaporation, based on the specification, is:
∆V = V(g)− V(l) = V. (g) = nRT p , where the amount of substance of water is n = 180018 = 100 mol. Then
3.5.7
Gibbs energy
3.5.7.1 Temperature and pressure dependence
The Gibbs energy is calculated from the definition (3.14), and from the dependence of enthalpy (3.74) and entropy (3.83) on T and p
G(T, p) = G(T1, p1) + [H(T, p) − T S(T, p)] − [H(T1, p1) − T1S(T1, p1)] . (3.96) The change in the Gibbs energy with pressure at constant temperature can be calculated by integrating equation (3.36) with respect to the general prescription (3.31)
G(T, p2) = G(T, p1) +
Z p2
p1
V dp , [T ] . (3.97)
3.5.7.2 Changes at phase transitions
The Gibbs energy does not change during reversible phase transitions
∆G = 0 , [T, p, reversible phase transition] . (3.98) For irreversible phase transitions, we have the inequality
∆G < 0 , [T, p, irreversible phase transition] . (3.99) and the change in the Gibbs energy is calculated using the procedure described in 3.5.9.
3.5.8
Fugacity
Dependence on state variables for a homogeneous system The following equations follow from the definition (3.21)
f = p exp " Z p 0 z − 1 p dp # , (3.100) and f = RT Vm exp " z − 1 − Z Vm ∞ z − 1 Vm dVm # , (3.101)
where z is the compressibility factor. Relation (3.100) and (3.101) in particular are used to calculate fugacity from equations of state.
3.5.8.1 Ideal gas
For an ideal gas, fugacity is equal to pressure
f = p , [ideal gas] . (3.102)
3.5.8.2 Changes at phase transitions
At reversible phase transitions, fugacity does not change. If, e.g., the liquid phase is in equi- librium with the gaseous phase, we have
f(l) = f(g), [T, p] . (3.103)
Note: According to the theorem of corresponding states [see2.2.9], the fugacity coefficient φ = f /p is the function of reduced temperature Tr and reduced pressure pr. To estimate
fugacity, we use generalized diagrams of the fugacity coefficient as a function of Trand pr.
Example
Calculate the fugacity of ethane at temperature 270 K and pressure 1000 kPa. For ethane at this temperature and pressure under 1200 kPa, the equation of state z = 1 − 1.1359 · 10−4p applies, with pressure given in kPa.
Solution
We substitute for the compressibility factor z from equation (3.100) and integrate f = p exp(−1.1359 × 10−4× p) = 1000 × exp(−0.11359) = 892.7 kPa .