Maxwell’s Relations

Introduction

Maxwell’s contributions to the fields of science span more than a single topic. The equations named after him in electrodynamics form the fundamentals of the topic, while his work on color theory is still regarded as ground-breaking. Indeed, Maxwell presented the first-ever color photography demonstration in 1861.

Maxwell’s relations in thermodynamics also carry immense significance. In this article, we will discuss what these relations are, and how they can be derived.

What Are Maxwell Relations?

Simply put, Maxwell’s relations are a set of equations that relate the derivative of different thermodynamic quantities under different scenarios. They can be arrived at using thermodynamic potentials.

Maxwell’s relations can be derived using the fact that the order of differentiation in a second-order partial differential is irrelevant. That is,

$$\mathrm{\frac{\partial}{\partial\:x}(\frac{\partial\:f}{\partial\:y})\:=\:\frac{\partial}{\partial\:y}(\frac{\partial\:f}{\partial\:x})}$$

Variables Used in Maxwell's Relation

Before we dive into the relations themselves, it would be prudent to list the variables that appear in them. All in all, there are four variables you must understand the meaning of before you can grasp the beauty of Maxwell’s relations −

• T – the temperature of the system

• V – the volume of the system

• P – the pressure of the system

• S – the entropy of the system

If you know about these four variables, you theoretically possess all the knowledge required to start studying Maxwell’s equations. At the same time, we also recommend studying the thermodynamic potentials listed below:

Internal energy (U)

The internal energy of the system, represented by U, is related to the energy that is contained inside it. You can also understand it as the energy required to assemble the system and bring it into the configuration it is currently in.

Note that the kinetic and potential energies do not play a part in the internal energy.

Enthalpy (H)

Consider any thermodynamic system. Let us take a simple cubical box filled with hydrogen. Suppose you start at absolute zero and supply an energy U to create this system. Further, the cubical box occupies some volume, and thus, some work must be done on the surroundings to “make room” for this cubical box. The enthalpy of the system is a measure of these two physical quantities. Thus,

$$\mathrm{H\:=\:U\:+\:pV}$$

Helmholtz Free Energy (F)

This thermodynamic potential is a measure of how much useful work we can obtain from a system at a constant temperature. It is defined as the product of temperature and entropy subtracted from the internal energy of the system. That is,

$$\mathrm{F\:\equiv\:U\:-\:TS}$$

Gibbs Free Energy (G)

Gibb’s free energy is similar to Helmholtz’s free energy. It is a measure of useful work obtainable from the system at a constant temperature as well as pressure. Using the Gibbs free energy, we can also study whether a certain chemical reaction will take place or not under the given conditions.

Mathematically,

$$\mathrm{G\:=U\:+\:pV\:-\:TS\:=\:H\:-\:TS}$$

Maxwell's Relation Derivation

The First Relation

We are now in a position to derive Maxwell’s relations. Let us start with the expression for internal energy given by −

$$\mathrm{dU\:=\:TdS\:-\:PdV}$$

Can you find the expressions for temperature and pressure from this equation? As it turns out, the calculation is quite trivial. We know that if

$$\mathrm{dz\:=\:Mdx\:+\:Ndy}$$

Then, $\mathrm{M\:=\:(\frac{\partial\:z}{\partial\:x})_{y}}\:\mathrm{N\:=\:(\frac{\partial\:z}{\partial\:y})_{x}}$

Therefore, we can see that

Then, $\mathrm{T\:=\:(\frac{\partial\:U}{\partial\:S})_{v}}$ and, $\mathrm{-\:P\:=\:(\frac{\partial\:U}{\partial\:V})_{s}}$

Now, it doesn’t matter what order we differentiate the internal energy in. The outcome is going to be equal. That is,

$$\mathrm{\frac{\partial\:}{\partial\:V}\:(\frac{\partial\:U}{\partial\:V})_{v}\:=\:\frac{\partial\:}{\partial\:S}\:(\frac{\partial\:U}{\partial\:V})_{s}}$$

Entering the values of T and P that we just calculated, we arrive at the first Maxwell relations. That is,

$$\mathrm{(\frac{\partial\:T}{\partial\:V})_{s}\:=\:-\:(\frac{\partial\:P}{\partial\:S})_{v}}$$

The Second Relation

Proceeding just like the first relation, we can arrive at another relation between the derivatives of entropy and temperature. This time, we start with the expression for Helmholtz Free energy,

$$\mathrm{df\:=\:-\:SdT\:-\:PdV}$$

Again, we calculate the values of entropy and pressure using this equation. Simple differentiation leads to −

$$\mathrm{S\:=\:-\:(\frac{\partial\:F}{\partial\:T})_{v}\:P\:=\:\:-\:(\frac{\partial\:F}{\partial\:T})_{v}}$$

We can differentiate these equations, for the volume and temperature, respectively, and equate them since the order of differentiation is irrelevant. This leads us to the following relation −

$$\mathrm{(\frac{\partial\:S}{\partial\:V})_{t}\:=\:(\frac{\partial\:P}{\partial\:T})_{v}}$$

The Third and Fourth Relations

Two more relations can be arrived at in the same manner as above. The complete derivation would be out of the scope of this article. However, we recommend you try to arrive at similar results as above using the following two equations:

$$\mathrm{dH\:=\:TdS\:+\:VdPdG\:=\:VdP\:-\:SdT}$$

Solved Examples

1. Prove that $\mathrm{(\frac{\partial\:U}{\partial\:V})_{t}\:=\:T\:\frac{\alpha}{K_{T}}\:-\:p}$

We start with the expression for the internal energy of the system −

$$\mathrm{dU\:=\:TdS\:-\:pdV}$$

You can see that in the question, there is an isolated pressure term. To arrive at that, we first differentiate this equation for volume. Then

$$\mathrm{(\frac{dU}{dV})_|{t}\:=\:T\:(\frac{dS}{dV})\:-\:p}$$

𝛼 and 𝜅𝑇 are related as follows − $\mathrm{(\frac{dP}{dT})_{v}\:=\:(\frac{\alpha}{K_{T}})}$ And that gives us our result

$$\mathrm{(\frac{\partial\:U}{\partial\:V})_{t}\:=T\:\:(\frac{\alpha}{K_{T}})-\:P}$$

Conclusion

Maxwell’s relations are a set of equations that equate changes in one thermodynamic variable to another. They can be derived using thermodynamic potentials and are universally valid. Just like Maxwell’s equations in electrodynamics, these relations come in particularly helpful in solving various problems and proving various results.

Maxwell’s relations involve four thermodynamic variables, which one must be aware of before diving into their study. These are the temperature, the volume, the pressure, and the entropy of the system. Apart from that, understanding thermodynamic potentials is also a prerequisite for this topic.

Maxwell’s relations are given as follows −

1. $\mathrm{(\frac{\partial\:T}{\partial\:V})_{S}\:=(\frac{\partial\;P}{\partial\:S})_{V}}$

2. $\mathrm{(\frac{\partial\:s}{\partial\:V})_{T}\:=(\frac{\partial\;P}{\partial\:T})_{V}}$

3. $\mathrm{(\frac{\partial\:T}{\partial\:P})_{S}\:=(\frac{\partial\;V}{\partial\:S})_{P}}$

4. $\mathrm{(\frac{\partial\:S}{\partial\:P})_{T}\:=(\frac{\partial\;V}{\partial\:T})_{P}}$

FAQs

1. Are there any relations beyond the four mentioned above?

Most commonly, the above four relations are used. However, t two extra equations can be used in various scenarios −

$$\mathrm{(\frac{\partial\:T}{\partial\:P})_{V}(\frac{\partial\:S}{\partial\:V})_{P}\:-\:(\frac{\partial\;T}{\partial\:V})_{P}\:(\frac{\partial\;S}{\partial\:P})_{V}\:=\:1}$$

And

$$\mathrm{(\frac{\partial\:P}{\partial\:T})_{S}(\frac{\partial\:V}{\partial\:S})_{T}\:-\:(\frac{\partial\;P}{\partial\:S})_{T}\:(\frac{\partial\;V}{\partial\:T})_{S}\:=\:1}$$

2. Is there another way to derive Maxwell’s relations?

One can use the Jacobian approach. That is, from

$$\mathrm{dU\:-\:TdS\:-\:pdV}$$

Take an infinitesimal change. Then, since $\mathrm{d(dU)\:=\:0}$ ,therefore,

$$\mathrm{dPdV\:=\:dTdS}$$

Or, in other words,

$$\mathrm{\frac{\partial\:(T,V)}{\partial\:(P,V)}\:=\:1}$$

From here, one can easily derive the relations.

3. Are the four thermodynamic parameters given above the extent of Maxwell’s relations?

No. The beauty of Maxwell’s relations is that they exist beyond the limits of what thermodynamic parameters are given. Maxwell’s relations are derived analytically. This means that the same process can be followed with any parameter that is given to us. Thus, if our system is defined in terms of something external like an electric field, we can even derive relations for that electric field.

4. Differentiate between intensive and extensive variables?

An extensive variable is a quantity whose value will change depending on how much matter we have at hand. For example, the volume of a given substance will generally increase as we take increasingly larger quantities of it. On the other hand, intensive variables do not depend on the amount. Density is a common example. No matter how much water (or any element) you have, its density does not change.

5. Are Maxwell’s relations applicable to non-reversible processes?

Yes. As we previously mentioned, these relations are analytical and universal.