Lecture 8: Thermodynamic potentials

Note

I hate thermodynamic potentials – Anonymous student, Rensselaer Polytechnic Institute, Fall 2020.

No, you don’t – Anonymous professor, Rensselaer Polytechnic Institute, Fall 2020.

Warning

This lecture corresponds to Chapter 16 of the textbook.

Summary

Attention

In this Chapter, we introduce new functions of state. Why do we want to do this? The issue is that the existing function of state related to energy we’ve encountered so far, the internal energy $U$ , has the entropy and volume as natural variable, i.e., we can write $U(S,V)$ and, while it is convenient to use the internal energy for an isentropic or isochoric processes, imposing other constraints (e.g., isothermal, isobaric) is much more difficult. For this reason, we introduce new function of states (see table below) that are built to allow for flexibility in what natural variables one wishes to constrain.

Because we have natural variables, their differentials are exact and one can use the arsenal of properties of exact differentials we summarized in Complement 1: Exact Differential. This leads to the famous Maxwell’s relationships, which provide links between the various partial derivatives with respect to natural variables.

Further, to justify the applicability of those various thermodynamic potentials, we introduce the availability function, which allows us to impose various constraints. We realize, as a key result, that the availability is the maximum amount of work one can get from a process. Furthermore, because of Clausius inequality (which, as we have seen, is formally equivalent to the second law of thermodynamics, see Lecture 5: Second Law of Thermodynamics), we see that a physical process will always evolve so as to minimize its availability. This is done, in particular, by maximizing the entropy! (I told you thermodynamics was fun!).

Finally, we provide a number of derivations of known derivatives (see the list of definitions below) and show how we can solve problems in thermodynamics by combining Maxwell’s relationships, definitions of properties, and basics multi-variable calculus (see box below on “How do we solve this?”).

Function of states and natural variables

$\arraycolsep=1.4pt\def\arraystretch{2.2} \begin{array}{lllll} \hline \textrm{Internal energy } & U & \mathrm{~d} U=T \mathrm{~d} S-p \mathrm{~d} V & T=\left(\frac{\partial U}{\partial S}\right)_{V} & p=-\left(\frac{\partial U}{\partial V}\right)_{S} \\ \textrm{Enthalpy } & H=U+p V & \mathrm{~d} H=T \mathrm{~d} S+V \mathrm{~d} p & T=\left(\frac{\partial H}{\partial S}\right)_{p} & V=\left(\frac{\partial H}{\partial p}\right)_{S} \\ \textrm{Helmholtz function } & F=U-T S & \mathrm{~d} F=-S \mathrm{~d} T-p \mathrm{~d} V & S=-\left(\frac{\partial F}{\partial T}\right)_{V} & p=-\left(\frac{\partial F}{\partial V}\right)_{T} \\ \textrm{Gibbs function } & G=H-T S ~~~& \mathrm{~d} G=-S \mathrm{~d} T+V \mathrm{~d} p ~~~& S=-\left(\frac{\partial G}{\partial T}\right)_{p}~~~ & V=\left(\frac{\partial G}{\partial p}\right)_{T}\\ \hline \end{array}$

Maxwell’s relations

$\arraycolsep=1.4pt\def\arraystretch{2.2} \begin{array}{ll} \left(\frac{\partial T}{\partial V}\right)_{S}=-\left(\frac{\partial p}{\partial S}\right)_{V} ~~~~~~& \left(\frac{\partial S}{\partial p}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{p} \\ \left(\frac{\partial T}{\partial p}\right)_{S}=\left(\frac{\partial V}{\partial S}\right)_{p} ~~~~~~& \left(\frac{\partial S}{\partial V}\right)_{T}=\left(\frac{\partial p}{\partial T}\right)_{V} \end{array}$

Note

How do we solve this?: a guide

We now have many formal tools to compute a number of effects related to thermodynamical processes operating under different constraints. We also have a lot of definitions of terms (such as susceptibilities, see below). How do we actually solve problems?

Write the thermodynamic potential in terms of particular variables.
Use Maxwell’s relations to transform partial derivatives into what you need.
Invert Maxwell’s relation using the reciprocal theorem.
Combine partial differentials using the reciprocity theorem.

$\begin{aligned} &\left(\frac{\partial x}{\partial z}\right)_{y}=\frac{1}{\left(\frac{\partial z}{\partial x}\right)_{y}} \\ &\left(\frac{\partial x}{\partial y}\right)_{z}\left(\frac{\partial y}{\partial z}\right)_{x}\left(\frac{\partial z}{\partial x}\right)_{y}=-1 \\ &\left(\frac{\partial x}{\partial y}\right)_{z}=-\left(\frac{\partial x}{\partial z}\right)_{y}\left(\frac{\partial z}{\partial y}\right)_{x} \end{aligned}$
Identify heat capacity:

$\frac{C_{V}}{T}=\left(\frac{\partial S}{\partial T}\right)_{V}$

and

$\frac{C_{p}}{T}=\left(\frac{\partial S}{\partial T}\right)_{p}$
Identify a general susceptibility (see definition below).

Key Definitions

Note

Thermodynamic potentials:

Thermodynamic potentials are functions of state introduced to account for various constraints such as isothermal, adiabatic, isochoric, and isobaric processes.

Natural variables:

In thermodynamic, natural variables are variables describing a function of state. For instance, when we write:

$dH=T \mathrm{~d} S+V \mathrm{~d} p,$

we see that pressure and entropy are the natural variables of enthalpy (written formally as $H=H(S,p)$ .

Helmholtz free energy:

This is the maximum amount of reversible work a system can do at constant temperature.

Gibbs-Helmholtz equations:

$U=-T^{2}\left(\frac{\partial}{\partial T}\right)_{V} \frac{F}{T}$

and

$H=-T^{2}\left(\frac{\partial}{\partial T}\right)_{p} \frac{G}{T}$

Maxwell’s relations:

Set of four equations that connect the partial derivatives (with respect to natural variables), due to the fact the functions of state are exact differentials. They stem from the fact that if $f$ is an exact differential then

$\mathrm{d} f=\left(\frac{\partial f}{\partial x}\right)_{y} \mathrm{~d} x+\left(\frac{\partial f}{\partial y}\right)_{x} \mathrm{~d} y$

and

$\left(\frac{\partial^{2} f}{\partial x \partial y}\right)=\left(\frac{\partial^{2} f}{\partial y \partial x}\right)$

General susceptibility:

A generalized susceptibility quantifies how much a particular variable changes when a generalized force is applied.

Isobaric expansivity:

The isobaric expansivity $\beta_{p}$ quantifies how much the volume of a system changes with temperature at constant pressure:

$\beta_{p}=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{p}$

Adiabatic expansivity:

The adiabatic expansivity $\beta_{S}$ quantifies how much the volume of a system changes with temperature in an isentropic process:

$\beta_{S}=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{S}$

Isothermal compressibility:

The isothermal compressibility quantifies how much the volume changes with pressure, at constant temperature:

$\kappa_{T}=-\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_{T}$

Adiabatic compressibility

The adiabatic compressibility quantifies how much the volume changes with pressure in an isentropic process:

$\kappa_{S}=-\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_{S}$

A full list of terms, including the ones provided here, can be found in the Index.

Learning Material

Copy of Slides

The slides for Lecture 8 are available in pdf format here: pdf

Screencast

Test your knowledge

Imagine you want to work at constant pressure $p$ , among the following choices, what thermodynamic potential should you use?
1. Enthalpy $H$
2. Internal energy $U$
3. Gibbs energy
Among the following functions, which ones is a function of state?
1. $H+W$
2. $Q+S$
3. $G+U$
To reach equilibrium under some constraints, we need to optimize some function of states. Among the ones below, which one should you maximize?
1. $H$
2. $U$
3. $S$
In thermodynamics, Maxwell’s relations have been established using what principle?
1. Thermodynamics and electromagnetic theory are unified
2. Functions of states are exact differentials, which imposes some specific mathematical conditions on the second derivatives with respect to the appropriate natural variables.
3. Maxwell relations only hold true for an ideal gas; in other words they are usually pretty crude approximations.
4. Maxwell relations are only valid for quasi-static processes, in other words, for purely reversible processes.
You need to know partial differential equations and the definitions of compressibility and expansivity to solve classical thermodynamics problems. Frankly, this is mostly a math problem. What matters more is understanding the concepts and then do the math.
1. True, at least that is what Prof. Meunier said.
2. False, at least that’s what I seem to remember after not having listened to the screencast carefully.
3. I think there is a trap and I will select this answer to make sure to not receive any credit for this question.

Hint

Find the answer keys on this page: Answers to selected test your knowledge questions. Don’t cheat! Try solving the problems on your own first!

Homework Assignment

Solve the following problems from the textbook: