Ch. 6 – Water and Its Transformations

1. Ch. 6 – Water and Its Transformations Heterogeneous Systems The 4 fundamental equations we have so far are valid for closed/isolated systems (systems that do not exchange mass). We assumed that the systems were homogeneous – only one phase involved. Only 2 independent variables; mass remains constant. 1

2. Ch. 6 – Water and Its Transformations • Heterogeneous Systems: • involves more than one single phase • Concerned with the conditions of internal equilibrium between the phases • The fundamental equations must be changed to include the exchange in mass between phases (extra terms include a function µ=chemical potential)

3. Ch. 6 – Water and Its Transformations • Heterogeneous Systems involves: • Dry air, which is assumed to remain always unchanged and in a gaseous state; it behaves like a closed system. • Water, that can exist in two phases (behave like two open systems): in vapor and possible one of the condensed phases (water or ice)

4. Ch. 6 – Water and Its Transformations • Heterogeneous Systems involves: • Water, that can exist in two phases (behave like two open systems): in vapor and possible one of the condensed phases (water or ice) • Water vapor (if it exists only by itself) = can be treated as an ideal gas  water vapor obeys the equation of state; its states are determined by 2 independent variables. • Water vapor coexisting with liquid water or ice  the mixture is not an ideal gas, and equations for ideal gas do not apply

5. Ch. 6 – Water and Its Transformations Fundamental Equations for Open Systems Consider the Gibbs function for the gas phase (Z = G) where Consider a reversible process without mass exchange (closed system, i.e., dnd = dnv = 0). From our gas-only system we had, for a reversible process, 5

6. Ch. 6 – Water and Its Transformations Fundamental Equations for Open Systems Since T and p are independent variables, the coefficients of dT and dp have to be the same in both equations, so and These equalities are valid for processes with mass exchangebecause the partial derivatives are state functions themselves & independent of the process! 6

7. Ch. 6 – Water and Its Transformations • Fundamental Equations for Open Systems • We put these relationships into our first equation for G • This equation is the generalization of the fundamental equation for the change of the Gibbs function for our open system and for processes occurring reversibly with respect to mechanical (p) and thermal (T) equilibrium. • If we use the equalities G = U +pV – TS = H – TS = F + pV, we could get three other equations similar to the fundamental equations, except for the additive terms ddnd + vdnvthat will appear in all of them.

8. Ch. 6 – Water and Its Transformations • Fundamental Equations for Open Systems • Similar expressions could be written for condensed phase where

9. Ch. 6 – Water and Its Transformations • Internal Equilibrium • Assume that we have both phases isolated and in equilibrium (both at same T and p). • Now we bring them together – they stay in thermal and mechanicalequilibrium (T and p same for both phases). • What about chemical equilibrium? • Do water (ice) and vapor remain in equilibrium, or does condensation or evaporation (sublimation) take place as a spontaneous process?

10. Ch. 6 – Water and Its Transformations • Internal Equilibrium • We get the total value of the Gibbs function by adding the contribution from the gaseous and condensed phases. • Now introduce condition that total heterogeneous system is closed (dnd = 0 and dnv = dnc) and we get • where Gtot, Stot, and Vtot are the total values for the system. For example

11. Ch. 6 – Water and Its Transformations • Internal Equilibrium • The condition of internal chemical equilibrium between the two phases is v = c.(chemical potential of component k) • If we introduce the relations between Gand the other 3 characteristic functions U, H, and F we can get a set of 4 equations for the open heterogeneous system under discussion. • Applying the conditions for conservation of mass (moles), we can get another set of 4 equations for the closed system.

12. Ch. 6 – Water and Its Transformations • Fundamental Equations Generalized • We can generalize the equations to any heterogeneous system with Ccomponents and Pphases with the dry air/water system as a special case. The equations are

13. Ch. 6 – Water and Its Transformations • Fundamental Equations Generalized • If the system is closed, we have the same formulae with the substitution for the double sum in each equation of • Here, phase 1 is an arbitrarily chosen phase. • The conditions of internal chemical equilibrium are • i.e., the chemical potential of each component, is the same for all phases

14. Ch. 6 – Water and Its Transformations • Fundamental Equations Generalized • The magnitude of the double sum gives a measure of the deviation from chemical equilibrium, indicating the irreversibility of the process. • We have, then, 4 alternative definitions of the chemical potential • Note that only the last derivative is a partial molar property (since they are only defined at constant T and p).

15. Ch. 6 – Water and Its Transformations • Number of Independent Variables • For a closed system of constant composition, the number of independent variables needed to specify the state of the system is 2 (T and p, p and V, T and V). • If we have a heterogeneous system of 1 component and 2 phases (e.g., water and water vapor), we have 4 variables for the 2 phases (p, T, p, and T ). • If we impose equilibrium between the two phases, then 3 conditions on the 4 variables are imposed

16. Ch. 6 – Water and Its Transformations • Number of Independent Variables • This reduces the number of independent variables to one, p=f(T) •  = Gis the molar Gibbs function • G = G(T, p) and G = G(T , p) so the third equation implies a relation between T, p and T , p. • Fix T for equilibrium between two phases, the value of p also becomes fixed, and vice versa. • Defines curves p = f(T)along which equilibrium exists.

17. Ch. 6 – Water and Its Transformations Number of Independent Variables If we have a heterogeneous system of 1 component and 3 phases (e.g., water, water vapor and ice), we have 6 variables for the 3 phases (p, T, p, T , and p”,T”). If we impose equilibrium between the three phases, then 3 conditions on the 6 variables are imposed The first 2 equations reduce the number of independent variables from 6 to 2. The first equality in the 3rd equation reduces them by one more, and the second equality by another one again. Therefore we remain with zero independent variables  all values are fixed. defines the triple point – coexistence of all phases at equilibrium. 17

18. Ch. 6 – Water and Its Transformations • Number of Independent Variables • Previous example was for water, water vapor and ice. • What about moist air and a condensed phase? • Consider the same two variables (T and p). • Pressure of gas phase is now partial pressure of water vapor pluspartial pressure of dry air (p = pd + e). • Number of conditions is the same as before. • Now have two independent variables – Tand pd. • Gibbs (general) phase rule for Cdifferent and non reactive components, in a total of P phases, yields N degrees of freedom (independent variables).

19. Ch. 6 – Water and Its Transformations • Latent Heat • Recall we had 2 forms of the 1st Law • From these we see that • The heat absorbed by a system in a reversible process at constant volume, QV, is measured by a change in internal energy, U. • The heat absorbed by a system in a reversible process at constant pressure, Qp, is measured by a change in enthalpy, H. • For a homogeneous system

20. Ch. 6 – Water and Its Transformations • Latent Heat • Now we are to the point of being interested in heterogeneous systems where phase changes can occur. • As we saw, with a system made of only water (vapor and liquid or solid), there is only 1 independent variable, when 2 phases are in equilibrium. • Fix p or V, and T at which reversible change of phase takes place is also fixed – specific heats become irrelevant. • Latent heats are defined at constant pressure, so