1. Chemistry 2402 - Thermodynamics
Lecture 9 : Free Energies
Lecture 10 : Phase Diagrams and Solubility
Lecture 11: Solubility (cont.)

2. Phase Diagrams
Liquids eke out an existence in the niche between solids and gases. P T
solid
gas
liquid
critical point
triple point
supercritical fluid
Coexistence lines represent values of P and T where two phases can coexist.

3. The slope of coexistence lines.
The variation in G is dG = -SdT +VdP
If we want to move along the coexistence line between phases I and II, then
dGI = dGII since GI = GII defines coexistence.
so that
-SIdT + VIdP = -SIIdT + VIIdP
The temperature and the pressure must, of course, be the same in any phases at coexistence. Rearranging this equation, we have

4. The Clapeyron Equation
But G = H-TS=0 and, therefore, S = H/T
Substituting this, we arrive at the Clapeyron equation
If H and V vary little with P and T then we can integrate the Clapeyron equation to give

5. Pressure melting or pressure freezing?
From the Clapeyron equation Now ΔH < 0 for freezing.
Usually, ΔV < 0 also since solids are denser than liquids. That means the dP/dT is positive as in Diagram I. When ΔV > 0 as in ice or silica, then we get Diagram II I P T
solid
gas
liquid
critical point
triple point II P T
solid
gas
liquid
critical point

6. Example Problem or How To Use This Sausage Machine
At K the enthalpy change on melting of ice is 6.0 kJ/mol and the corresponding volume change is -1.6 × 10-6 m3/mol. Estimate the temperature at which ice will melt at 1000 atm pressure (take 1 atm = 105 Nm-2). You can assume that neither the enthalpy difference nor the volume difference changes significantly with pressure.
As the enthalpy and volume differences remain constant, we can use the integrated form of the Clapeyron equation
P1 = 1.01 × 105 Nm-2, P2= 1.01 × 108 Nm-2,
ΔH = 6.0 × 103J/mol ,
ΔV = -1.6 × 10-6m3/mol,
T1 = K
Which gives T2 = 266.0K
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7. Flash Quiz! Which of the following can occur?
(assume T stays constant in every case)
For those that can, give an example from the natural world.
Gaseous hydrogen condenses as pressure is increased
Liquid iron solidifies as pressure is increased
Liquid water solidifies as pressure is increased
Normal water-ice (ice Ih) melts as pressure is increased
Gaseous water condenses as pressure is decreased.

8. Hints low H high H solid liquid gas Which of the following can occur?
(assume T stays constant in every case)
For those that can, give an example from the natural world.
Gaseous hydrogen condenses as pressure is increased
Liquid iron solidifies as pressure is increased
Liquid water solidifies as pressure is increased
Normal water-ice (ice Ih) melts as pressure is increased
Gaseous water condenses as pressure is decreased.
which has higher V?
high V

9. Answers low H high H solid liquid gas
Which of the following can occur?
(assume T stays constant in every case)
For those that can, give an example from the natural world.
Gaseous hydrogen condenses as pressure is increased
Liquid iron solidifies as pressure is increased
Liquid water solidifies as pressure is increased
Normal water-ice (ice Ih) melts as pressure is increased
Gaseous water condenses as pressure is decreased.
which has higher V?
high V
Jupiter
Earth’s inner core
Trick question: not to ice Ih, but there are high-density ice forms
Base of (some) glaciers
If you wrote “cloud formation” you assumed the gas expansion was accompanied by a temperature drop.

10. Reading P-V Phase Diagrams
liquid
solid
gas V

11. Reading P-V Phase Diagrams
critical point
s + l
liquid
l + g
solid
gas
s + g
triple line
molar V
The shaded region depicts values of the pressure and total volume for which we have coexistence of two (or three at the triple point) phases.
(The volume of the gas has been significantly reduced to fit it all in.)

12. Chemical Potential and Liquid-Vapor Equilibrium
At constant T and P, equilibrium between liquid and vapour means
dG = μliqdNliq + μvapourdNvapour = 0 or μliq = μvapour
Generally, we can’t measure chemical potentials directly but we can get close for an ideal gas using
The change in G due to a change in pressure is given by the shaded area.
G(P ) -
G(P
) = 2 1
shaded area V P P 1 2 P

13. Chemical Potential for an Ideal Gas – Introducing the Pain of Reference States
Using V = nRT/P and integrating from P1 to P2 gives
G2 = G1 + nRTln(P2 /P1)
We have to choose a reference state. In the case of gases we shall take the reference free energy Go to be the free energy of 1 mole of gas at 1 atm. We can then write
G = Go + RTln(P/atm)
and, since μ = G/N, we have
 = o + kBTln(P/atm)

14. Introducing Solutions
In the case of a mixture of ideal gases, each species acts as if it was alone (remember ideal gases don’t interact). This means we have the following chemical potential for species i
i = oi + kBTln(Pi/atm)
where Pi = xi(gas)P [ xi(gas) = Ni/N ] is the partial pressure of species i. xi is called the mole fraction.
The Cunning Plan
i(gas) = i(liq).
liquid
vapour
Use the known chemical potential of a species in the gas phase to establish its chemical potential in a solution at equilibrium with the gas.

15. Relating Partial Pressures to Concentrations
The problem is that the gas expression talks of partial pressures while concentrations are the natural way of talking about the solution.
An approximate relation was proposed by Raoult in He noted that, as long as the components in solution differed little from one another - e.g. benzene and toluene – then
Pi(gas)= xi(liq)Pi* Raoult’s Law P T
solid
gas
liquid
critical point
triple point
xi(liq) = Ni/N in solution and Pi* is the pressure at the coexistence of the liquid and vapour phases of the pure i. This pressure is called the vapour pressure of i (at the temperature of interest).

16. The Chemical Potential in an Ideal Solution (The Maths Lover’s Slide)
A solution that follows Raoult’s Law is called an ideal solution
At equilibrium i(liq) = i(gas)
For an ideal gas i(gas) = oi + kBTln(Pi/atm)
For an ideal solution i(liq) = oi + kBTln(xi(liq) Pi* /atm)
Avoiding the gas reference state, let oi + kBTln(Pi* /atm) = i*(liq) where i*(liq) is the chemical potential of a pure liquid of i at coexistence with the vapour at the vapour pressure Pi*.
So finally i(liq) = i*(liq) + kBTln(xi(liq))
we have the ideal solution chemical potential in terms of the concentration (mole fraction is directly related to concentration)

17. Changing the Reference State to a Standard State
The chemical potential of each species will have a reference state with a different pressure Pi*.
To avoid this we often assume i*(liq)  oi(liq) i.e. use 1 atm instead of Pi*
With this change we have, for the ideal solution,
i(liq) = oi(liq) + kBTln(xi(liq))

18. Sample exam questions from previous years
On the axes provided, draw a representative pressure-volume phase diagram (not to scale) of a typical pure substance which has a solid, liquid, and vapour phase. Label all regions, as well as the critical point.
At 40.0 °C, heptane has a vapour pressure of mmHg, and octane has a vapour pressure of mmHg. In a 40 °C laboratory, a mixture is prepared in a sealed container, with 100.0g of liquid heptane and 60.0g of liquid octane. Assume this mixture obeys Raoult’s Law.
If a liquid-vapour coexistence is established at that pressure, what will the mole fraction of octane in the vapour be?
Consider a gas mixture at equilibrium with a solution of the same species.
What are the conditions for coexistence in terms of the intensive properties - temperature, pressure and chemical potential.

19. Summary Next Lecture You should now
Be able to explain pressure melting and pressure freezing, and relate it’s occurrence to ΔVmelt and ΔHmelt
Be able to draw and read a typical P-V phase diagram
Understand the role of reference states when handling chemical potentials or free energies
Be able to explain the role of Raoult’s Law in the derivation of the properties of an ideal solution
Explain and use the concept of an ideal solution
Next Lecture
Solubility (cont)