At the end of Lesson 9 a very good question arose:
If the natural derivatives of the Gibbs and Helmholtz energies [a(rho,T), g(T,P)] are conveniently measured and commonly needed, then why did most modern EOS become Helmholtz explicit instead of Gibbs?
Let’s look at the surface of state for Argon (one of the most precisely known) and see how Helmholtz and Gibbs energies behave. First in terms of density:
One of the interesting features in this graph is the discontinuity of slope and curvature occurring at the triple point. I’ve colored the melting lines red on these plots so they can be more easily visualized.
From these two plots you can see that either Helmholtz or Gibbs energies can accurately describe the vapor dome in terms of density. Let’s look at temperature:
From the temperature coordinate it’s very difficult to see the vapor dome. Let’s look at Helmholtz versus Temperature:
As you can see though, Helmholtz energy is able to create an observable vapor dome where Gibbs cannot. Let’s consider pressure:
From here it’s clear, the Gibbs energy cannot resolve independent saturated vapor and liquid lines. Let’s look at Helmholtz:
The reason that Gibbs energy cannot show separate saturated vapor and liquid lines is that g(T,P). Temperature and Pressure are not independent quantities within the vapor dome during phase change (remember the solid lines we see across the vapor dome for the pure fluids we discuss on Fluid Fridays). The Gibbs Energy of the saturated vapor and liquid are identical! This in some ways is analogous to the Maxwell Criterion and can, itself, be used to draw the vapor dome in PρT coordinates. Now imagine how difficult it would be for a genetic algorithm trying to optimize the temperature or pressure of a state near a saturation line. You would need to calculate yet another property like the Phase Identification Parameter (PIP) to determine which side of the vapor dome you were on. This inability of the Gibbs Energy to differentiate phase in the most commonly used temperature and pressure coordinates is why modern EOS are explicit in Helmholtz Energy.
So now that we know why Helmholtz energy is used, let’s get into how we can use it. Starting with our definition of the total differential from last time:
df = ∂f/∂x|y dx + ∂f/∂y|x dy. Applying the total differential to a(v,T) gives
da = ∂a/∂T|v dT + ∂a/∂v|T dv
and combining with our third fundamental property relation from last time:
da = -sdT – Pdv
we see that s=-∂a/∂T|v and P=-∂a/∂v|T and we have the start to a wonderful thermodynamic property tree:
Let’s take this further. We know that A,P,S are all thermodynamic properties, and by definition, independent of path to reach them. Therefore the properties of the exact differential are useful:
∂²f/∂x∂y = ∂²f/∂y∂x
The key property of an exact differential is that the order (path) that the derivatives are taken is irrelevant. This is also known as Schwartz’ theorem. Let’s apply this to our equation for entropy above:
-∂²a/∂v∂T = -∂/∂v(∂a/∂T|v)|T where the term in parenthesis is equal to s above such that -∂²a/∂v∂T = ∂s/∂v|T.
Similarly for pressure:
-∂²a/∂T∂v = -∂/∂T(∂a/∂v|T)|v where the term in parenthesis is equal to P above such that -∂²a/∂T∂v = ∂P/∂T|v.
From our definition of the exact differential then we see that:
-∂²a/∂v∂T = -∂²a/∂T∂v = ∂s/∂v|T= ∂P/∂T|v
This result is one of Maxwell’s Relations. Starting from our other fundamental properties we arrive at more Maxwell Relations:
u(s,v) yields ∂²u/∂s∂v = ∂T/∂v|s= -∂P/∂s|v
h(s,P) yields ∂²h/∂s∂P = ∂T/∂P|s= ∂v/∂s|P
g(T,P) yields ∂²g/∂T∂P = ∂s/∂P|T= -∂v/∂T|P
These are the four most common relations and can also be remembered from the Thermodynamic Square. Next Monday we’ll use these relations, and develop more calculus tools to continue growing our thermodynamic property tree.