Pressure Dependence of Gibbs’ Free Energy

1. Pressure Dependence of Gibbs’ Free Energy Methods of evaluating the pressure dependence of the Gibbs’ free energy can be developed by beginning with (see the course notes on Maxwell’s Relations for a derivation of this equation): dG = V dP - S dT Dividing by dP, while holding the temperature constant, gives a partial derivative describing how the Gibbs’ free energy varies isothermally with pressure: (G / P)T = + V Separating variables and integrating this differential equation gives an expression for calculating the change in the Gibbs’ free energy in an isothermal processes: DG = dG =  V dP The development of this expression will be a function of how the volume depends on pressure. Could you derive a similar expression describing how to calculate changes in the Helmholtz free energy for isothermal changes in volume? 32.1

2. 1.00 mole of Zn is compressed from 1.00 atm to 100.0 atm at 25.0 oC. The density of Zn is 7.14 g / cm3 at 25.0 oC. What is DG for this process? DG =  V dP = (mass / density) dP = (1.00 mole) (65.39 g / mole) / (7.14 g / cm3) x [100.0 atm - 1.00 atm] (8.314 J / 82.05 cm3 atm) What assumption was made in the last step and do you think it was reasonable? = + 91.8 J Note that the Gibbs’ free energy increases during isothermal increases in pressure. 32.2

3. What is change in the Gibbs’ free energy of 1.000 mole of liquid water when it is compressed isothermally at 4.0 oC (the roughly constant temperature of deep sea water) from 1.0000 atm to 1000.0 atm (roughly the pressure at the bottom of the Marianas Trench, the deepest point in the Pacific)? The isothermal compressibility of liquid water at 4.0 oC is +50.1368 x 10-6 atm-1. Do not assume that the liquid water is incompressible. The density of liquid water at 4.0 oC and 1.0000 atm is 0.9999750 g / cm3. Since the liquid water is viewed as compressible in this case, we need to know how the volume depends on pressure in order to evaluate the integral: DG =  V dP The dependence of volume on pressure is obtained from the definition of isothermal compressibility:   - (1/V) (V/ P)T and is given by (see the course notes on Real Gases for a derivation): V = Vo e + b P oe - b P where Pois the pressure at which Vo is known (usually 1 atm): Vo = mass / density at 1 atm and 4.0 oC = (1.000 mole) (18.02 g/mole) / 0.9999750 g / cm3 = 18.02 cm3 32.3

4. Substituting the volume dependence on pressure into the integral we obtain: DG =  V dP = 1.0000 atm1000.0 atm Vo e +  P oe - P dP = Vo e + b P oe - b P dP = Vo e + b P o(-1 / b) [e - b 1000.0 atm - e - b 1.0000 atm ] = (18.02 cm3) e + (+50.1368 10-6 atm-1) (1.0000 atm) x (-1 / +50.1368 10-6 atm-1) x [e - (+50.1368 10-6 atm -1) (1000.0 atm) - e - (+50.1368 10-6 atm -1) (1.0000 atm) ] x (8.314 J / 82.05 cm3 atm) = + 1,779 J What would the Gibbs’ free energy change have been, if the water had been considered incompressible? 32.4

5. What is the Gibbs’ free energy change for the expansion of 3.00 moles of N2 (g) from 10.00 atm to 1.00 atm at 400 K? DG =  V dP = P1 P2 (n R T / P) dP = n R T ln (P2 / P1) = (3.00 moles) (0.008314 J / mole K) (400 K) x ln (1.00 atm / 10.00 atm) = - 23.0 kJ Were we justified in using the ideal gas law to develop the answer to this question? How is the change in Gibbs’ free energy related to the reversible work done in this expansion? What mathematical difficulties would you have faced if this had been a van der Waals gas? Do you think you could have worked around them? 32.5

6. DGmelting = ? H2O (s) H2O (l) 0.0 oC, 100.0 atm DGmelting = ? H2O (s) H2O (l) 0.0 oC, 100.0 atm DG1 =  V H2O (s) dP = - 1946 cm3 atm DG3 =  V H2O (l) dP = + 1784 cm3 atm DGomelting = 0 H2O (s) H2O (l) 0.0 oC, 1.00 atm What is DG for the melting of ice at 0.0 oC and 100.0 atm? The densities of ice and liquid water are 0.9168 g/cm3 and 1.000 g/cm3, respectively, at 0.0 oC and 100.0 atm: We begin by constructing a path along which we can carry out the calculation: Why did we construct the particular path that we did? Why is DGomelting in step 2 equal to zero? The Gibbs’ free energy change for the melting of ice at 0.0 oC and 100.0 atm is therefore: DGmelting = DG1 + DGomelting + DG3 = [(- 1946 cm3 atm) + 0 + (+ 1784 cm3 atm)] x (8.314 J / 82.05 cm3 atm) = - 16.41 J Will ice melt under 100.0 atm of pressure at 0.0 oC? Relate this result to the movement of glaciers and ice skaters. 32.6

7. DGvap., equil. = 0 H2O (l) H2O (g) 298.2 K, equil. P DG3 =  V H2O (g) dP DG1 =  V H2O (l) dP DGovap., 298.2 K H2O (l) H2O (g) 298.2 K, 1.00 bar What is the equilibrium vapor pressure of liquid water at 25.0 oC? The density of liquid water at 25.0 oC is 1.000 g/cm3. Step 1: The pressure on the liquid water is changed from the equilibrium pressure to 1.00 bar isothermally at 25.0 oC: DG1 =  V H2O (l) dP = [(18.02 g / mole) / (1.000 g / cm3)] x (1.00 bar - Pequil.) (0.008314 kJ / 83.14 cm3 bar) = (0.001802 kJ / bar mole) (1.00 bar - Pequil.) Do you see the advantage of the metric system in conversion from cm3 bar to kJ? 32.7

8. Step 2: The liquid water is vaporized at 25.0 oC and 1.00 atm: DGovap, 298.2 K = DGof, 298.2 K, [H2O (g)] - DGof, 298.2 K, [H2O (l)] = (- 228.6 kJ / mole) - (- 237.2 kJ / mole) = + 8.60 kJ / mole Step 3: The pressure on the water vapor is increased from 1.00 bar to the equilibrium pressure isothermally at 25.0 oC: DG1 =  V H2O (g) dP = (R T / P) dP = (0.008314 kJ / mole K) (298.2 K) ln (Peq / 1.00 bar) = (2.479 kJ / mole) ln Peq What will be the units on Peq? Equating the Gibbs’ free energy changes along the different paths: DGvap, 298.2 K, equil. = DG1 + DGovap., 298.2 K + DG3 gives an equation that can be solved for the equilibrium pressure: 0 = (DG1 @ 0)+ 8.602 kJ / mole + (2.479 kJ / mole) ln Peq Why were we able to equate the Gibbs’ free energy change in step 1 to zero?Peq = 0.0311 bar or 24.0 mm Hg The experimental value of the equilibrium vapor pressure of water at 25.0 oC is 23.76 mm Hg. What do you think of the power of thermodynamics to make predictions about physical systems? 32.8