The Liquid-vapor critical point in a pressure–temperature phase diagram
is at the high-temperature extreme of the liquid–gas phase boundary
(the dotted green line shows the anomalous behavior of water).
The figure shows the schematic PT diagram of a pure substance (as opposed to mixtures, which have additional state variables and richer phase diagrams, discussed below). The commonly known phases solid, liquid and vapor are separated by phase boundaries, i.e. pressure-temperature combinations where two phases can coexist.
At the triple point, even all three phases coexist. However, the liquid-vapor boundary terminates in an endpoint at some critical temperature Tc and critical pressure pc. This is the critical point.
In water, the critical point occurs at around 647 K and 22.064 MPa.
- Gold 7250 K 510MPa.
At the critical point, only one phase exists. The heat of vaporization is zero.
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Van der Walls-like phase-diagram of gold
Temperature 0.5ev<->0.5ev/kB=5802°K
kB=8.6173 10^-5eV/°K
DOI: 10.1016/j.apsusc.2005.03.019 ; [Vitiello,2005]
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GOLD
Electron configuration [Xe] 4f14 5d10 6s1
melting point: 1336.15°K
Heat of fusion (10^3 J/kg): 64.5
or 12.55 kJ·mol^−1
boiling point: 3243 °K
Heat of vaporization (10^3 J/kg): 1578
or 342 kJ·mol^−1
Density near r.t. 19.30 g·cm−3
when liquid, at m.p. 17.31 g·cm−3
Molar heat capacity 25.418 J·mol−1·K−1
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Problem:
1) Using the CRC thermodynamic data from the handout in class, make a plot of the Gibbs free energy (in [J/mole]) as a function of temperature for gold (Au) in both the solid and liquid phases over the temperature range of 1000-2000K. Both free energies must be on the same plot. Determine the equilibrium temperature between liquid and solid Au from your calculations.
2) Yet another approach to calculating the Gibbs free energy is using empirical polynomial fitting equations from thermodynamic databases. One such example is the CALPHAD database. As an example of the fitting equations I have listed the equations for solid Au in the FCC crystal structure below. Make a plot of the Gibbs free energy (in [J/mole]) as a function of temperature over the range of 1000-2000 [K] using both the CRC data from problem 1 and the CALPHAD data. Both calculations must be on the same plot. Comment as to the agreement or lack thereof between the two calculations.
3) Calculate the triple point (as T, P) for gold. (NOTE that all the thermodynamic information you need to do this calculation can be found from the CRC Handbook data. I'm not looking for the whole phase diagram, just the triple point. There are a variety of ways to approach this problem, but I'm looking for a pretty simple one.)
Solution
1) Using the CRC thermodynamic data from the handout in class, make a plot...
The CRC data for Au gives the following equations for the Gibbs free energy as a function of temperature:
Plugging into EXCEL and plotting between 1000-2000 [K] gives the following plot:
With this calculation, the crossing point is at about T = 1341 [K], which is the equilibrium melting temperature (a bit higher than the stated temperature of 1336.16 [K].)
NOTE: The CRC is pretty confusing by calling the function for Gibbs free energy "FT-H298." In fact, the enthalpy at 298 [K] is included in the equations and fitting parameters. What it means is that the thermodynamic functions are determined relative to a temperature of 298 [K].
2) Yet another approach to calculating the Gibbs free energy...
I used EXCEL again to make the calculation and the plot.
3) Calculate the triple point (as T, P) for gold.
The simple trick approach uses the fact that we discussed in class that the slope of the S-L equilibrium line on the P-T phase diagram is VERY STEEP (a nearly vertical line with pressure) so that the equilibrium melting temperature changes very little with pressure.
This means that, within a small error, the triple point temperature is given by the equilibrium melting temperature at 1 [atm], which for Au is found in the CRC handout as TM = 1336.16 [K]. The problem is half done!
To find the triple point pressure, we need to find the equilibrium pressure for the temperature of 1336.16 [K] for either the S-V or the L-V equilibrium. I decided to use the "shortcut" approach (see problem 2 on themidterm for F08 for more information.) The Clausius-Clapeyron equation gives:
In this equation, we can either use the enthalpy of vaporization for ∆H, and the vaporization temperature for T0 (which would give us the L-V equilibrium), or we could use the enthalpy and temperature of sublimation (which would give us the S-V equilibrium.) The CRC has the data for the vaporization, but not the sublimation information, so...
There's our answer for the triple point of Au: T = 1336 [K], P = 2.44 x 10-7 [atm].
