 # Maxwell’s Relations

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:

 $$+(\frac{\partial T}{\partial V})_{S}=-(\frac{\partial P}{\partial S})_{V}=\frac{\partial^2 U}{\partial S\partial V}$$ $$+(\frac{\partial T}{\partial P})_{S}=+(\frac{\partial V}{\partial S})_{P}=\frac{\partial^2 H}{\partial S\partial P}$$ $$+(\frac{\partial S}{\partial V})_{T}=+(\frac{\partial P}{\partial T})_{V}=\frac{\partial^2 F}{\partial T\partial V}$$ $$-(\frac{\partial S}{\partial P})_{T}=+(\frac{\partial V}{\partial T})_{V}=\frac{\partial^2 G}{\partial T\partial P}$$

## 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:

$$dU=TdS-PdV$$ (differential form of internal energy)

$$dU=(\frac{\partial z}{\partial x})_{y}dx+(\frac{\partial z}{\partial y})_{x}dy$$ (total differential form)

$$dz=Mdx+Ndy$$ (other way of showing the equation)

$$M=(\frac{\partial z}{\partial x})_{y}$$

And

$$N=(\frac{\partial z}{\partial y})_{x}$$

From $$dU=TdS-PdV$$

T=$$(\frac{\partial U}{\partial S})_{V}$$

And

$$-P=(\frac{\partial U}{\partial V})_{S}$$ $$\frac{\partial }{\partial y}(\frac{\partial z}{\partial x})_{y}=\frac{\partial }{\partial x}(\frac{\partial z}{\partial y})_{x}=\frac{\partial^2 z}{\partial y\partial x}=\frac{\partial^2 z}{\partial x\partial y}$$ (symmetry of second derivatives)

$$\frac{\partial }{\partial V}(\frac{\partial U}{\partial S})_{V}=\frac{\partial }{\partial S}(\frac{\partial U}{\partial V})_{S}$$ $$(\frac{\partial T}{\partial V})_{S}=-(\frac{\partial P}{\partial S})_{V}$$

### Common forms of Maxwell’s relations

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

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:

$$(\frac{\partial \mu }{\partial P})_{S,N} = (\frac{\partial V}{\partial N})_{S,P} = (\frac{\partial^2 H}{\partial P\partial N})$$

### Solved Examples

Example 1:

Prove that $$(\frac{\partial V}{\partial T})_{p}=T\frac{\alpha }{\kappa _{T}}-p$$.

Solution:

Combining first and second laws:

dU = TdS – pdV

Diving both the sides by dV

$$\frac{\mathrm{d} U}{\mathrm{d} V}|_{T}=\frac{T\mathrm{d} S}{\mathrm{d} V}|_{T}-p\frac{\mathrm{d} V}{\mathrm{d} V}|_{T}$$

$$\frac{\mathrm{d} U}{\mathrm{d} V}|_{T}=(\frac{\partial U}{\partial V})_{T}$$

$$\frac{T\mathrm{d} S}{\mathrm{d} V}|_{T}=(\frac{\partial S}{\partial V})_{T}$$

$$\frac{\mathrm{d} V}{\mathrm{d} V}|_{T}=1$$

$$(\frac{\partial U}{\partial V})_{T}=T(\frac{\partial S}{\partial V})_{T}-p$$

$$(\frac{\partial p}{\partial T})_{V}=(\frac{\partial S}{\partial V})_{T}$$

$$(\frac{\partial U}{\partial V})_{T}=T(\frac{\partial p}{\partial T})_{V}-p$$

$$(\frac{\partial p}{\partial T})_{V}=\frac{\alpha }{\kappa _{T}}$$

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