Maxwell Relations: Definition, Derivation, Examples, And Derivative Forms

The Maxwell relations are derived from Euler’s reciprocity relation. The relations are expressed in partial differential form. The Maxwell relations consists of the characteristic functions: internal energy U, enthalpy H, Helmholtz free energy F, and Gibbs free energy G and thermodynamic parameters: entropy S, pressure P, volume V, and temperature T. Following is the table of Maxwell relations for secondary derivatives:

\begin{array}{l} + (\frac{\partial T}{\partial V})_{S} = - (\frac{\partial P}{\partial S})_{V} = \frac{\partial^{2} U}{\partial S \partial V} \end{array}

\begin{array}{l} + (\frac{\partial T}{\partial P})_{S} = + (\frac{\partial V}{\partial S})_{P} = \frac{\partial^{2} H}{\partial S \partial P} \end{array}

\begin{array}{l} + (\frac{\partial S}{\partial V})_{T} = + (\frac{\partial P}{\partial T})_{V} = \frac{\partial^{2} F}{\partial T \partial V} \end{array}

\begin{array}{l} - (\frac{\partial S}{\partial P})_{T} = + (\frac{\partial V}{\partial T})_{V} = \frac{\partial^{2} G}{\partial T \partial P} \end{array}

What are Maxwell’s relations?

These are the set of thermodynamics equations derived from a symmetry of secondary derivatives and from thermodynamic potentials. These relations are named after James Clerk Maxwell, who was a 19th-century physicist.

Derivation of Maxwell’s relations

Maxwell’s relations can be derived as:

\begin{array}{l} d U = T d S - P d V (differential form of internal energy) \end{array}

\begin{array}{l} d U = (\frac{\partial z}{\partial x})_{y} d x + (\frac{\partial z}{\partial y})_{x} d y (total differential form) \end{array}

\begin{array}{l} d z = M d x + N d y (another way of showing the equation) \end{array}

\begin{array}{l} M = (\frac{\partial z}{\partial x})_{y} and N = (\frac{\partial z}{\partial y})_{x} \end{array}

\begin{array}{l} From d U = T d S - P d V \end{array}

\begin{array}{l} T = (\frac{\partial U}{\partial S})_{V} and - P = (\frac{\partial U}{\partial V})_{S} \end{array}

\begin{array}{l} \frac{\partial}{\partial y} (\frac{\partial z}{\partial x})_{y} = \frac{\partial}{\partial x} (\frac{\partial z}{\partial y})_{x} = \frac{\partial^{2} z}{\partial y \partial x} = \frac{\partial^{2} z}{\partial x \partial y} (symmetry of second derivatives) \end{array}

\begin{array}{l} \frac{\partial}{\partial V} (\frac{\partial U}{\partial S})_{V} = \frac{\partial}{\partial S} (\frac{\partial U}{\partial V})_{S} \end{array}

\begin{array}{l} (\frac{\partial T}{\partial V})_{S} = - (\frac{\partial P}{\partial S})_{V} \end{array}

Common forms of Maxwell’s relations

Function	Differential	Natural variables	Maxwell Relation
U	dU = TdS – PdV	S, V	$\begin{array}{l} (\frac{\partial T}{\partial V})_{S} = - (\frac{\partial P}{\partial S})_{V} \end{array}$
H	dH = TdS + VdP	S, P	$\begin{array}{l} (\frac{\partial T}{\partial P})_{S} = (\frac{\partial V}{\partial S})_{P} \end{array}$
F	dF = -PdV – SdT	V, T	$\begin{array}{l} (\frac{\partial P}{\partial T})_{V} = (\frac{\partial S}{\partial V})_{T} \end{array}$
G	dG = VdP – SdT	P, T	$\begin{array}{l} (\frac{\partial V}{\partial T})_{P} = - (\frac{\partial S}{\partial P})_{T} \end{array}$

Where,

T is the temperature

S is the entropy

P is the pressure

V is the volume

U is the internal energy

H is the entropy

G is the Gibbs free energy

F is the Helmholtz free energy

With respect to pressure and particle number, enthalpy and Maxwell’s relation can be written as:

\begin{array}{l} (\frac{\partial μ}{\partial P})_{S, N} = (\frac{\partial V}{\partial N})_{S, P} = (\frac{\partial^{2} H}{\partial P \partial N}) \end{array}

Solved Examples

Example 1:

Prove that

\begin{array}{l} (\frac{\partial V}{\partial T})_{p} = T \frac{α}{κ_{T}} - p . \end{array}

Solution:

Combining first and second laws:

dU = TdS – pdV

Diving both the sides by dV

\begin{array}{l} \frac{d U}{d V} |_{T} = \frac{T d S}{d V} |_{T} - p \frac{d V}{d V} |_{T} \end{array}

\begin{array}{l} \frac{d U}{d V} |_{T} = (\frac{\partial U}{\partial V})_{T} \end{array}

\begin{array}{l} \frac{T d S}{d V} |_{T} = (\frac{\partial S}{\partial V})_{T} \end{array}

\begin{array}{l} \frac{d V}{d V} |_{T} = 1 \end{array}

\begin{array}{l} (\frac{\partial U}{\partial V})_{T} = T (\frac{\partial S}{\partial V})_{T} - p \end{array}

\begin{array}{l} (\frac{\partial p}{\partial T})_{V} = (\frac{\partial S}{\partial V})_{T} \end{array}

\begin{array}{l} (\frac{\partial U}{\partial V})_{T} = T (\frac{\partial p}{\partial T})_{V} - p \end{array}

\begin{array}{l} (\frac{\partial p}{\partial T})_{V} = \frac{α}{κ_{T}} \end{array}

To know more about Maxwell’s equations and problems on the same, you can visit us BYJU’S

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Frequently Asked Questions – FAQs

Maxwell’s relations are named after which scientist?

It is named after Scientist James Clerk Maxwell.

What are Maxwell’s relations?

These are set of thermodynamics equations derived from a symmetry of secondary derivatives and from thermodynamic potentials.

What is enthalpy?

Enthalpy (H) is the sum of the internal energy (U) and the product of pressure (P) and volume(V).

Define Gibbs free energy.

Gibbs free energy can be defined as the maximum amount of work that can be extracted from a closed system.

Name three thermodynamic parameters.

Thermodynamic parameters are: Pressure P, volume V, and temperature T.

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Maxwell’s Relations

What are Maxwell’s relations?

Derivation of Maxwell’s relations

Common forms of Maxwell’s relations

Solved Examples

Frequently Asked Questions – FAQs

Maxwell’s relations are named after which scientist?

What are Maxwell’s relations?

What is enthalpy?

Define Gibbs free energy.

Name three thermodynamic parameters.

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