Thermodynamics. Chapter 13 Phase Diagrams. NC State University

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1 Thermodynamics Chapter 13 Phase Diagrams NC State University

2 Pressure (atm) Definition of a phase diagram A phase diagram is a representation of the states of matter, solid, liquid, or gas as a function of temperature and pressure. In the Figure shown below the regions of space indicate the three phases of carbon dioxide. The curved lines indicate the coexistence curves. Note there is a unique triple point.

3 Degrees of freedom Within any one of the single-phase regions both temperature and pressure must be specified. Because two thermodynamic variables can be changed independently we say that the system has two degrees of freedom. Along any of the coexistence curves the pressure and temperature are coupled, i.e. any change in the temperature implies a change in pressure to remain on the line. Thus, along the curves there is only one degree of freedom. The triple point is a unique point in phase space and there is only one set of values of pressure and temperature consistent with the triple point. Thus, we say that at the triple point the system has zero degrees of freedom. If we follow the liquid-vapor coexistence curve towards higher temperature we find that it ends at the critical point. Above the critical point there is no distinction between liquid and vapor and there is a single fluid phase.

4 Free energy dependence along the coexistence curve In a system where two phases (e.g. liquid and gas) are in equilibrium the Gibbs energy is G = G l + G g, where G l and G g are the Gibbs energies of the liquid phase and the gas phase, respectively. If dn modes (a differential amount of n the number of moles) are transferred from one phase to another at constant temperature and pressure, the differential Gibbs energy for the process is: dg = Gg ng dng + Gl P,T nl dnl P,T The rate of change of free energy with number of moles is called the chemical potential.

5 The significance of chemical potential of coexisting phases We can write the Gibbs free energy change using the following notation: dg = m g dng + m l dnl Note that if the system is entirely composed of gas molecules the chemical potential m g will be large and m l will be zero. Under these conditions the number of moles of gas will decrease dn g < 0 and the number of moles of liquid will increase dn l > 0. Since every mole of gas molecules converted results in a mole of liquid molecules we have that: dn g = -dn l

6 Coexistence criterion In terms of chemical potential, the Gibbs energy for the phase equilibrium is: dg = m g m l dng Since the two phases are in equilibrium dg = 0 and since dn g 0 we have m g = m l. In plain language, if two phases of a single substance are in equilibrium their chemical potentials are equal. If the two phases are not in equilibrium a spontaneous transfer of matter from one phase to the other will occur in the direction that minimizes dg. Matter is transferred from a phase with higher chemical potential to a phase with lower chemical potential consistent with the negative sign of Gibb's free energy for a spontaneous process.

7 Solid-liquid coexistence curve To derive expressions for the coexistence curves on the phase diagram we use the fact that the chemical potential is equivalent in the two phases. We consider two phases a and b and write m a (T,P) = m b (T,P) Now we take the total derivative of both sides m a dp + ma P T T dt = mb dp + mb P P T T P dt The appearance of this equation is quite different from previous equations and yet you have seen this equation before. The reason for the apparent difference is the symbol m. Remember that m for a single substance is just the molar free energy.

8 The Clapeyron equation Substituting these factors into the total derivative above we have V a m dp S a m dt = V b m dp S b m dt Solving for dp/dt gives dp dt = S b a m S m V b a m V m = D trss m D trs V m = D trsh m TD trs V m This equation is known as the Clapeyron equation. It gives the two-phase boundary curve in a phase diagram with D trs H and D trs V between them. The Clapeyron equation can be used to determine the solid-liquid curve by integration. P 1 P 2 dp = D trsh m D trs V m Starting with a known point along the curve (e.g. the triple point or the melting temperature at one bar) we can calculate the rest of the curve referenced to this point. T 2 T 1 dt T

9 The liquid-vapor and solidvapor coexistence curves The Clapeyron equation cannot be applied to a phase transition to the gas phase since the molar volume of a gas is a function of the pressure. Making the assumption that V m g >> V m l we can use the ideal gas law to obtain a new expression for dp/dt. dp dt = D trsh m TV m g The integrated form of this equation P 1 dp P P 2 = T 1 T 2 = PD trsh m RT 2 D trs H m RT dt 2 yields the Clausius-Clapeyron equation. ln P 2 P 1 = D trsh m R 1 T 1 1 T 2 = D trsh m R T 2 T 1 T 1 T 2

10 Applying the Clausius- Clapeyron equation If we use DH of evaporation the C-C equation can be used to describe the liquid-vapor coexistence curve and if we use DH of sublimation this equation can be used to describe the solid-vapor curve. The pressure derived from the C-C equation is the vapor pressure at the given temperature. Applications also include determining the pressure in a high temperature vessel containing a liquid (e.g. a pressure cooker). If you are given an initial set of parameters such as the normal boiling point, for example you may use these as T 1 and P 1. Then if you are given a new temperature T 2 you can use the C-C to calculate P 2.

11 Constructing the phase diagram for CO 2 We can use the Clapeyron and Clausius-Clapeyron equations to calculate a phase diagram. For example, we can begin with the CO 2 diagram shown above. The triple point for CO 2 is 5.11 atm and K. The critical point for for CO 2 is atm and K. We also have the following data Transition D trs H o (kj/mol) T trs (K) Fusion Sublimation Note that we can calculate the enthalpy of sublimation from D vap H o = D sub H o - D fus H o = 16.9 kj/mol. r solid = 1.53 g/cm 3 and r liquid = 0.78 g/cm 3, respectively. The density r = m/v = nm/v so the molar volume is V m = V/n = M/r where M is the molar mass. In units of L/mole we have V s m = 44 g/mole/[1530 g/l] = V l m = 44 g/mole/[780 g/l] = D fus V = V l m - V s m = = L/mole

12 Constructing the phase diagram for CO 2 Starting with the triple point we use the Clausius-Clapeyron equation to calculate the liquid-vapor coexistence curve. P = 5.11exp{D vap H/R[T ]/216.15T} P = 5.11exp{2,032[T ]/216.15T} Notice that if we were to calculate the critical pressure using this formula we would obtain 77.3 atm which is about 5 atm larger than the experimental number. There are several sources of inaccuracy including mainly our neglect of the temperature dependence of the enthalpy. We can also begin a the critical point P = 72.8 exp{d vap H/R[T 304.2]/304.2T} P = 72.8 exp{2,032[t 304.2]/304.2T}

13 Pressure (atm) Constructing the liquid-vapor curve P = 5.11exp{2032[T ]/216.15T} Liquid-vapor P (atm) T (K)

14 Pressure (atm) Constructing the liquid-vapor curve P = 5.11exp{2032[T ]/216.15T} Liquid-vapor P (atm) T (K)

15 Pressure (atm) Constructing the liquid-vapor curve P = 5.11exp{2032[T ]/216.15T} Liquid-vapor P (atm) T (K)

16 Pressure (atm) Constructing the liquid-vapor curve P = 5.11exp{2032[T ]/216.15T} Liquid-vapor P (atm) T (K)

17 Pressure (atm) Constructing the liquid-vapor curve P = 5.11exp{2032[T ]/216.15T} Liquid-vapor P (atm) T (K)

18 Pressure (atm) Constructing the liquid-vapor curve P = 5.11exp{2032[T ]/216.15T} Liquid-vapor P (atm) T (K)

19 Pressure (atm) Constructing the solid-vapor curve Starting again at the triple point P = 5.11exp{D sub H/R[T ]/216.15T} P = 5.11exp{3034[T ]/216.15T} Solid-vapor P (atm) T (K)

20 Pressure (atm) Constructing the solid-vapor curve Starting again at the triple point P = 5.11exp{D sub H/R[T ]/216.15T} P = 5.11exp{3034[T ]/216.15T} Solid-vapor P (atm) T (K)

21 Pressure (atm) Constructing the solid-vapor curve Starting again at the triple point P = 5.11exp{D sub H/R[T ]/216.15T} P = 5.11exp{3034[T ]/216.15T} Solid-vapor P (atm) T (K)

22 Pressure (atm) Constructing the solid-vapor curve Starting again at the triple point P = 5.11exp{D sub H/R[T ]/216.15T} P = 5.11exp{3034[T ]/216.15T} Solid-vapor P (atm) T (K)

23 Pressure (atm) Constructing the solid-liquid curve Using the Clapeyron equation we calculate: P = [D fus H/D fus V] ln{t/216.15} P = ,967 ln{t/216.15} Solid-liquid P (atm) T (K)

24 Pressure (atm) Constructing the solid-liquid curve Using the Clapeyron equation we calculate: P = [D fus H/D fus V] ln{t/216.15} P = ,967 ln{t/216.15} Solid-liquid P (atm) T (K)

25 Pressure (atm) Constructing the solid-liquid curve Using the Clapeyron equation we calculate: P = [D fus H/D fus V] ln{t/216.15} P = ,967 ln{t/216.15} Solid-liquid P (atm) T (K)

26 Constructing the solid-liquid curve Using the Clapeyron equation we calculate: P = [D fus H/D fus V] ln{t/216.15} P = ,967 ln{t/216.15} Solid-liquid P (atm) T (K)

27 Solving Problems What is the vapor pressure of water above a lake on a day When the temperature of both water and air is 32 o C?

28 Solving Problems What is the melting point of ice underneath an ice skater Who has a mass of 100 kg assuming that the area of the Skate blades is 10-7 m 2.

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