download the script: Maxwell Relations

At first, we will deal the Internal energy u, Enthalpy h, Gibbs function g and Free energy or Helmholtz function f. All these four are expressed on per unit mass basis:

**Internal Energy u:**

The differential form of 1^{st} law of thermodynamics for a stationary closed system, which contains a compressible substance and undergoes an internally reversible process, can be expressed as:

du=δq_{rev} – p∙dν → δq_{rev} = du + p∙dν

In addition, the entropy is defined as:

T∙ds= δq_{rev}

Then we have:

T∙ds = du + p∙dν → du = T∙ds – p∙dν

and

**Enhalpy h:**

According to the definition of ethalpy, we have: h=u+p∙ν

We then write it in differential form:

dh = du + p∙dν + ν∙dp → dh – ν∙dp = du + p∙dν

Now we can eliminate du by using dh:

dh = T∙ds + ν∙dp

and

**Gibbs Function g:**

Definition of Gibbs Funtion g is: g=h – T∙s

So the Gibbs function g in differential form is:

dg = dh – T∙ds – s∙dT

And knowing from the above equation (dh = T∙ds + ν∙dp), we get:

dg = ν∙dp – s∙dT

and

**Free Energy (or Helmholtz Function) f:**

Definition of Free Energy f is:

f=u – T∙s

The differential form of free energy f is expressed as:

df = du – T∙ds – s∙dT

We can eliminate du by using du – T∙ds =– p∙dν, we obtain:

df =– p∙dν – s∙dT

and

We know from the mathematics that any exact differential the mixed partial derivatives must be equal, which can be expressed as:

dz = M∙dx + N∙dy

where:

Then:

According to this conclusion, we can obtain the following four relations by applying the above partial derivatives of properties pressure p, specific volume ν, temperature T and specific entropy s:

These four equations are called the **Maxwell Relations**.