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Contents

Thermodynamics: Structure

Thermodynamic Identities

Problems

Problems

In thermodynamics, we often ask questions about the energy of the system. Here we will discuss the energy that we've already introduced as well as alternate formulations of the energy of a system.

The Thermodynamic Identity

Suppose that we seek the energy of a system U in terms of its usual variables, σ , V , and N . Unfortunately, we can't write a closed solution for U in terms of those three variables. But not all is lost. We can utilize the mathematical tool known as the differential. Then we get:

dU(σ, V, N) = + dV + dN

So far, this may not look helpful. But if you glance back at our previous definitions of temperature, pressure, and chemical potential, we can rewrite the above:

dU(σ, V, N) = τ  - p dV + μ dN

The result is known as the Thermodynamic Identity, and is the most basic equation in our study of thermodynamics. Notice that there is great parallel structure to the equation. All of the extensive variables appear as differentials, while the intensive variables appear alone. Note that U is still a function of just the three extensive variables, since we can think of the other three "variables" as derivable from the three extensive.

Legendre Transform

We can use another mathematical tool here to make the Thermodynamic Identity even more useful. The Legendre Transform allows us to make a variable change in our definition of U . After all, suppose we don't want the energy as a function of the three variables above, σ , V , and N .

We will utilize the Legendre Transform minimally, and not delve into the underlying mathematics. The basic idea is that you can define a new function that is related to the original by an added product of two correlated terms. Let us make this explicit by using it.

Definition of F, G, H

Suppose that F = U - στ . Then when we take the differential, we need to remember to use the product rule. We obtain:

dF = dU - σ  - τ 

Now, we can substitute in the Thermodynamic Identity to obtain:

dF = - σ  - p dV + μ dN

Notice that F is a function now of τ , V , and N . By adding the term - στ , we were able to swap two of the variables, σ and τ . We call F the Helmholtz Free Energy, and we will soon see why it is useful.

The quick mind will realize that we could define 6 such energies in total, by successively swapping all of the variables. It turns out that we'll only be interested in two more. The Enthalpy, H , swaps p and V . We write H = U + pV and obtain dH = τ  + V dp + μ dN . We also define the Gibbs Free Energy by utilizing both of these swaps. Letting G = U + pV - τσ , we obtain dG = - σ  + V dp + μ dN .

We say that the energy of any of these types is a function of the variables that appear as differentials. Remember that the terms that are not differentials can be defined in relation to those that are.

The relationships between the energies are summarized in the following figure.

Figure %: The Structure of Thermodynamics

Why Four Energies?

Understanding why we have these four different energies is crucial to saving yourself time doing thermodynamics problems. In a given problem, identify the variables that are remaining constant. Then choose the energy that has those variables in the differential, so that when you are to calculate the energy, you aren't left with many non-zero terms.

For example, in a process where the volume and number remain constant but the entropy is changing, we should use U , because the Thermodynamic Identity simplifies to dU = τ  . Becoming adept at these choices will be critical to your success at solving problems quickly.

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