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Chemistry 351

Maxwell Relations

Maxwell Relations

The Maxwell Relations arise from two facts:

  1. Thermodynamic functions such as \(U\), \(H\), \(A\), and \(G\) are state functions.
  2. The differentials of state functions are exact differentials.

For any exact differential of the form

\[ dz = M(x,y)\,dx + N(x,y)\,dy \]

the Euler reciprocity relation requires

\[ \left(\frac{\partial M}{\partial y}\right)_x = \left(\frac{\partial N}{\partial x}\right)_y \]

Applying this rule to the thermodynamic differentials gives four Maxwell Relations.

Maxwell Relation from \(U(S,V)\)

The differential of internal energy is

\[ dU = T\,dS - p\,dV \]

Since the natural variables of \(U\) are \(S\) and \(V\), we may also write

\[ dU = \left(\frac{\partial U}{\partial S}\right)_V dS + \left(\frac{\partial U}{\partial V}\right)_S dV \]

By comparison,

\[ \left(\frac{\partial U}{\partial S}\right)_V = T \qquad \left(\frac{\partial U}{\partial V}\right)_S = -p \]

Because \(dU\) is exact,

\[ \left[ \frac{\partial}{\partial V} \left(\frac{\partial U}{\partial S}\right)_V \right]_S = \left[ \frac{\partial}{\partial S} \left(\frac{\partial U}{\partial V}\right)_S \right]_V \]

Substituting the expressions above gives

\[ \boxed{ \left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial p}{\partial S}\right)_V } \]

Maxwell Relation from \(H(S,p)\)

The differential of enthalpy is

\[ dH = T\,dS + V\,dp \]

Since the natural variables of \(H\) are \(S\) and \(p\), we may write

\[ dH = \left(\frac{\partial H}{\partial S}\right)_p dS + \left(\frac{\partial H}{\partial p}\right)_S dp \]

By comparison,

\[ \left(\frac{\partial H}{\partial S}\right)_p = T \qquad \left(\frac{\partial H}{\partial p}\right)_S = V \]

Since \(dH\) is exact,

\[ \left(\frac{\partial T}{\partial p}\right)_S = \left(\frac{\partial V}{\partial S}\right)_p \]

Therefore,

\[ \boxed{ \left(\frac{\partial T}{\partial p}\right)_S = \left(\frac{\partial V}{\partial S}\right)_p } \]

Maxwell Relation from \(A(V,T)\)

The differential of the Helmholtz function is

\[ dA = -p\,dV - S\,dT \]

Since the natural variables of \(A\) are \(V\) and \(T\), we may write

\[ dA = \left(\frac{\partial A}{\partial V}\right)_T dV + \left(\frac{\partial A}{\partial T}\right)_V dT \]

By comparison,

\[ \left(\frac{\partial A}{\partial V}\right)_T = -p \qquad \left(\frac{\partial A}{\partial T}\right)_V = -S \]

Since \(dA\) is exact,

\[ \left(\frac{\partial (-p)}{\partial T}\right)_V = \left(\frac{\partial (-S)}{\partial V}\right)_T \]

Multiplying both sides by \(-1\) gives

\[ \boxed{ \left(\frac{\partial p}{\partial T}\right)_V = \left(\frac{\partial S}{\partial V}\right)_T } \]

Maxwell Relation from \(G(p,T)\)

The differential of the Gibbs function is

\[ dG = V\,dp - S\,dT \]

Since the natural variables of \(G\) are \(p\) and \(T\), we may write

\[ dG = \left(\frac{\partial G}{\partial p}\right)_T dp + \left(\frac{\partial G}{\partial T}\right)_p dT \]

By comparison,

\[ \left(\frac{\partial G}{\partial p}\right)_T = V \qquad \left(\frac{\partial G}{\partial T}\right)_p = -S \]

Since \(dG\) is exact,

\[ \left(\frac{\partial V}{\partial T}\right)_p = \left(\frac{\partial (-S)}{\partial p}\right)_T \]

Therefore,

\[ \boxed{ \left(\frac{\partial V}{\partial T}\right)_p = -\left(\frac{\partial S}{\partial p}\right)_T } \]

Summary of Maxwell Relations

Function Differential Maxwell Relation
\(U(S,V)\) \(dU = T\,dS - p\,dV\) \(\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial p}{\partial S}\right)_V\)
\(H(S,p)\) \(dH = T\,dS + V\,dp\) \(\left(\frac{\partial T}{\partial p}\right)_S = \left(\frac{\partial V}{\partial S}\right)_p\)
\(A(V,T)\) \(dA = -p\,dV - S\,dT\) \(\left(\frac{\partial p}{\partial T}\right)_V = \left(\frac{\partial S}{\partial V}\right)_T\)
\(G(p,T)\) \(dG = V\,dp - S\,dT\) \(\left(\frac{\partial V}{\partial T}\right)_p = -\left(\frac{\partial S}{\partial p}\right)_T\)

Big picture: Maxwell Relations arise because thermodynamic potentials are state functions with exact differentials. They allow derivatives involving entropy, which is often difficult to measure directly, to be replaced by derivatives involving pressure, volume, and temperature, which are often much easier to measure.

Worked examples

Worked example: Deriving \(\left(\frac{\partial U}{\partial V}\right)_T\)

Derive an expression for \[ \left(\frac{\partial U}{\partial V}\right)_T \] using the Maxwell Relation associated with the Helmholtz function \(A\).

Start with the combined First and Second Laws:

\[ dU = T\,dS - p\,dV \]

Divide both sides by \(dV\) and constrain the change to constant temperature:

\[ \left.\frac{d U}{d V}\right|_T = T \left.\frac{d S}{d V}\right|_T - p \left.\frac{d V}{d V}\right|_T \]

Since

\[ \left.\frac{\partial V}{\partial V}\right|_T = 1 \]

Converting the ratios to partial derivatives, this simplifies to

\[ \left(\frac{\partial U}{\partial V}\right)_T = T \left(\frac{\partial S}{\partial V}\right)_T - p \]

Now use the Maxwell Relation associated with the Helmholtz function:

\[ \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial p}{\partial T}\right)_V \]

Substituting gives

\[ \boxed{ \left(\frac{\partial U}{\partial V}\right)_T = T \left(\frac{\partial p}{\partial T}\right)_V - p } \]

And since \( \left( \frac{\partial p}{\partial T} \right)_V = \frac{\alpha}{\kappa_T} \)

\[ \boxed{\left(\frac{\partial U}{\partial V}\right)_T = T \frac{\alpha}{\kappa_T} - p } \]

Interpretation: This result relates the volume dependence of internal energy to experimentally accessible pressure-volume-temperature behavior. Once an equation of state is known, such as the ideal gas law or the van der Waals equation, the derivative \(\left(\frac{\partial p}{\partial T}\right)_V\) can be evaluated directly.

Worked example: Deriving \(\left(\frac{\partial H}{\partial p}\right)_T\)

Derive an expression for \[ \left(\frac{\partial H}{\partial p}\right)_T \] using the Maxwell Relation associated with the Gibbs function \(G\).

Start with the differential form of enthalpy:

\[ dH = T\,dS + V\,dp \]

Divide both sides by \(dp\) and constrain the change to constant temperature:

\[ \left.\frac{\partial H}{\partial p}\right|_T = T \left.\frac{\partial S}{\partial p}\right|_T + V \left.\frac{\partial p}{\partial p}\right|_T \]

Since

\[ \left.\frac{\partial p}{\partial p}\right|_T = 1 \]

after converting ratios to partial derivatives, this becomes

\[ \left(\frac{\partial H}{\partial p}\right)_T = T \left(\frac{\partial S}{\partial p}\right)_T + V \]

Now use the Maxwell Relation associated with the Gibbs function:

\[ \left(\frac{\partial S}{\partial p}\right)_T = - \left(\frac{\partial V}{\partial T}\right)_p \]

Substituting gives

\[ \left(\frac{\partial H}{\partial p}\right)_T = -T \left(\frac{\partial V}{\partial T}\right)_p + V \]

Rearranging,

\[ \boxed{ \left(\frac{\partial H}{\partial p}\right)_T = V - T \left(\frac{\partial V}{\partial T}\right)_p } \]

Since the isobaric thermal expansivity is defined as

\[ \alpha = \frac{1}{V} \left(\frac{\partial V}{\partial T}\right)_p \]

we can also write

\[ \left(\frac{\partial V}{\partial T}\right)_p = \alpha V \]

Therefore,

\[ \boxed{ \left(\frac{\partial H}{\partial p}\right)_T = V(1-\alpha T) } \]

Interpretation: This expression describes how the enthalpy changes as pressure changes at constant temperature. The result is written in terms of the volume and the thermal expansivity, both of which are experimentally measurable quantities.

Key points (one glance)

Big picture: Maxwell Relations provide a bridge between abstract thermodynamic quantities and experimentally measurable properties. They allow difficult derivatives involving entropy to be replaced by derivatives involving pressure, volume, and temperature, making it possible to derive practical thermodynamic equations of state.