Internal Energy, Enthalpy, Helmholtz free energy and Gibbs free energy are all Legebdre transformations of each other

Sometimes it pays to be persistent in thinking about things you don’t understand (if you have the time as I do). The chemical potential is of enormous interest to chemists, and yet is defined thermodynamically in 5 distinct ways. This made me wonder if the definitions were actually describing essentially the same thing (not to worry they are).

First, a few thousand definitions

Chemical potential of species i — mu(i)
Internal energy — U
Entropy — S
Enthalpy — H
Helmholtz free energy — F or A (but, maddeningly, never H)
Gibbs free energy — G
Ni — number of elements of chemical species i
Pressure — p
Volume — V
Temperature — T

Just 5 more
mu(i) == ∂H/∂Ni constant S, p
mu(i) == ∂S/∂Ni constant U, V
mu(i) == ∂U/∂Ni constant S, V
mu(i) == ∂F/∂Ni constant T, V
mu(i) == ∂G/∂Ni constant T, p

Clearly, at a given constellation of S, U, F, G the mu(i)’s won’t all be the same number, but they’re essentially the same thing. Here’s why.

Start with a simple mathematical problem. Assume you have a simple function (f) of two variables (x,y), and that f is continuous in x and y and that its partial derivatives u = ∂f/∂x and w = ∂f/∂y are continuous as well so you have

df = u dx + w dy

u and dx are conjugate variables, as are w and dy

Suppose you want to change df = u dx + w dy to

another function g such that

dg = u dx – y dw

which is basically flipping a pair of conjugate variables around

Patience, the reason for wanting to do this will become apparent in a moment.

The answer is to use what is called the Legendre transform of f which is simply

g = f – y w

dg = df – y dw – w dy

plug in df

dg = u dx + w dw – y dw – w dy == df – y dw – w dy Done.

Where does the thermodynamics come in?

Well, you have to start somewhere, so why not with the fundamental thermodynamic equation for internal energy U

dU = ∂U/∂S dS + ∂U/∂V dV + ∑ ∂U/∂Ni dNi

We already know that ∂U/Ni = mu(i)

Because of the partial derivative notation (∂) it is assumed that all the other variables say in the expression for dU e.g. V and Ni are held constant in ∂U/∂S. This will reduce the clutter in notation which is already cluttered enough.

We already know that ∂U/∂Ni is mu(i). One definition of temperature T, is as ∂U/∂S, and another for p is -∂U/∂V (which makes sense if you think about it — decreasing volume relative to U should increase pressure).

Suddenly dU looks like what we were talking about with the Legendre transformation.

dU = T dS – p dV + ∑ mu(i) dNi

Apply the Legendre transformation to U to switch conjugate variables p and V

H = U + pV ; looks suspiciously like enthalpy (H) because it is

dH = dU + p dV + V dp + ∑ mu(i) dNi

= T dS – p dV + ∑ mu(i) dNi + p dV + V dp

= T dS + V dp + ∑ mu(i) dNi

Notice how mu(i) here comes out to ∂H/dNi at constant S and P

Start with the fundamental thermodynamic equation for internal energy

dU = T dS – p dV + ∑ mu(i) dNi

Now apply the Legendre transformation to T and S and you get
F = U – TS ; looks like the Helmholtz free energy (sometimes written A, but never as H) because it is.

You get

dF = – S dT – p dV + ∑ mu(i) dNi

Who cares? Chemists do because, although it is difficult to hold U constant or S constant (and it is impossible to measure them directly) it is very easy to keep temperature and volume constant in a reaction, meaning that changes in Helmholtz free energy under those conditions is just
∑ mu(i) dNi. So here mu(i) = ∂F/∂Ni at constant T and p

If you start with enthalpy

dH = T dS + V dp + ∑ mu(i) dNi

and do the Legendre transform you get the Gibbs free energy G = H – TS

I won’t bore you with it but this gives you the chemical potential mu(i) at constant T and p, conditions chemists easily arrange all the time.

To summarize

Enthalpy (H) is one Legendre transform of internal energy (U)
Helmholtz free energy (F) is another Legendre transform of U
Gibbs free energy (G) is the Legendre transform of Enthalpy (H)

It should be clear that Legendre transforms are all reversible

For example if H = U + PV then U = H – PV

If you think a bit about the 5 definitions of chemical potential, you’ll see that it can depend on 5 things (U, S, p, V and T). Ultimately all thermodynamic variables (U, S, H, G, F, p, V, T, mu(i) ) often have relations to each other.

Examples include H = U + pV, F = U – TS, G = H -TS

Helping keep things clear are equations of state from the things you can easily measure (p,V, T). The most famous is the ideal gas law p V = nRT.

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