Thermodynamics potentials

3. Thermodynamics potentials#

In this section, principles and the mathematical formalism of classical thermodynamics are reviewed.

3.1. First principle#

For an extensive system with uniform intensive pro

\[d E = T \, d S + \mathbf{F} \cdot d \mathbf{X} \ ,\]

being $E$ the internal energy, $T$ temperature, $S$ entropy, $\mathbf{F}$ generalized force, $\mathbf{X}$ generalized displacement.

Following Gibbs’ formulation, internal energy $E$ can be written as a function of a limited set of independent state variables,

\[E(S, \mathbf{X}) \ .\]

Internal energy $E$, entropy $S$ and the generalized displacement $\mathbf{X}$ are extensive physical quantities, and thus (thus?) the derivatives

\[T := \left.\dfrac{\partial S}{\partial E}\right|_{\mathbf{X}} \quad , \quad \mathbf{F} := \left.\dfrac{\partial E}{\partial \mathbf{X}}\right|_S\]

are intensive quantities. A discussion about the difference between the concept of additivity and extensivity1. Beside being extensive, internal energy in classical thermodynamics is an homogeneous function2 of order 1 of its arguments, namely

\[E(a \, S, a \, \mathbf{X}) = a \, E(S, \mathbf{X}) \ .\]

Euler’s theorem for homogeneous functions holds.

Theorem 3.1 (Euler’s theorem for homogeneous functions)

Let $f(x_i)$ and homogeneous function of order $m$, i.e.

(3.1)#\[f(a \, x_i) = a^m \, f(x_i) \ .\]

It follows that

\[f(x_k) = m \, x_i \, \dfrac{\partial f}{\partial x_i}(x_k) \ .\]

(classical-thermodynamics:potentials:first-principle:examples)

3.1.1. First principle for different systems#

Single-component fluid, $E(S, V, N)$, i.e. internal energy $E$ as a function of the entropy $S$, the volume of the system $V$ and the number of moles $N$ of the substance in the system

\[d E = T \, d S - P \, d V + \mu \, d N\]

As an alternative to the number of moles, it’s possible to the mass of the system $M = N M_m$ as an intependent variable, being $M_m$ the molar mass of the substance.
Multi-component fluid, $E(S, V, N_k)$, i.e. internal energy $E$ as a function of the entropy $S$, the volume of the system $V$ and the number of moles $N_k$ of all the substances in the system

\[d E = T \, dS - P \, d V + \mu_k \, d N_k\]

where the change of number of particles (or moles, it depends on the description - anyway a non-dimensional number) $d N_k$ is governed by the stoichiometric ratios of the reactions occurring in the system. As an alternative to the moles of the susbstances in the system, it’s possible to use the total number of moles $N = \sum_k N_k$ and the molar concentrations $x_k = \frac{N_k}{N}$ as independent variables, with the algebraic constraint $\sum_k x_k = 1$, by the very definition of the molar concentrations; it’s also possible to use the mass of the system and the mass concentration of the substances, $M = \sum_k M_k = \sum_k N_k M_{m,k}$, and $y_k = \frac{M_k}{M}$, with the algebraic constraint $\sum_k y_k = 1$.
Single-component solid, $E(S, \mathbf{X})$

\[d E = T \, dS + \mathbf{F} \cdot d \mathbf{X} \ ,\]

in the limit of small-displacement regime, $d E = T \, d S + V \boldsymbol{\sigma} : d \boldsymbol{\varepsilon}$.

3.1.2. First principle for specific quantities#

First principle and thermodynamics can be writtenin terms of specific quantites, usually either for unit volume or for unit mass.

3.1.2.1. First principle per unit mass#

All the extensive quantites are written as the product of the mass of the system $M$ and its density, namely

\[E = M e \quad , \quad S = M s \quad , \quad \mathbf{X} = M \mathbf{x} \ ,\]

and, using the product rule for differential $d E = d (M e) = dM \, e + M \, de$, first principle can be written as

\[0 = d M \underbrace{\left( e - T s - \mathbf{F} \cdot \mathbf{x} \right)}_{= 0} + M \left( d e - T ds - \mathbf{F} \cdot d \mathbf{x} \right) \ .\]

The first term is identically zero, since it’s the expression of the internal energy divided by $M$. Being $M > 0$, the second term gives the first principle per unit mass

\[d e = T ds + \mathbf{F} \cdot d \mathbf{x} \ .\]

3.1.2.1.1. Different systems#

Single-component fluid;

\[V = M \dfrac{1}{\rho} \quad , \quad N = M \dfrac{1}{m} \ , \]

being $\rho$ the mass density and $m$ the mass of a particle (or mole, it depends on the description) of the medium; $m$ is constant. First principle becomes

\[0 = d M \underbrace{\left( e - T s + \dfrac{P}{\rho} - \dfrac{\mu}{m} \right)}_{= 0} + M \left( d e - T ds - \dfrac{P}{\rho^2} d\rho \right) \ ,\]

and thus

\[de = T ds + \dfrac{P}{\rho^2} d \rho \ .\]
Multi-component fluid

\[V = M \dfrac{1}{\rho} \quad , \quad N_k = M_k \dfrac{1}{m}_k = M \dfrac{M_k}{M} \dfrac{1}{m_k} = M \dfrac{1}{m_k} \dfrac{\rho_k}{\rho} = M \dfrac{1}{m_k} w_k \ , \]

being $\rho$ the mass density and $m$ the mass of a particle (or mole, it depends on the description) of the $k^{th}$ substance; $m_k$ is constant. The first principle becomes

\[0 = d M \underbrace{\left( e - T s + \dfrac{P}{\rho} - \dfrac{\mu_k}{m_k} w_k \dfrac{}{} \right)}_{= 0} + M \left[ d e - T ds - \dfrac{P}{\rho^2} d\rho - \dfrac{\mu_k}{m_k} d w_k \right] \ ,\]

and thus

\[de = T ds + \dfrac{P}{\rho^2} d \rho + \dfrac{\mu_k}{m_k} d w_k\ .\]
Single-component solid

\[\text{\textbf{todo}}\]

3.1.2.2. First principle per unit volume#

All the extensive quantites are written as the product of the mass of the system $M$ and its density, namely

\[E = V \mathcal{E} \quad , \quad S = M \mathcal{S} \quad , \quad \mathbf{X} = M \symbf{\chi} \ ,\]

and, using the product rule for differential $d E = d (V \mathcal{E}) = dV \, \mathcal{E} + V \, d\mathcal{E}$, first principle can be written as

\[0 = d V \underbrace{\left( \mathcal{E} - T \mathcal{S} - \mathbf{F} \cdot \symbf{\chi} \right)}_{= 0} + V \left( d \mathcal{E} - T d \mathcal{S} - \mathbf{F} \cdot d \symbf{\chi} \right) \ .\]

The first term is identically zero, since it’s the expression of the internal energy divided by $V$. Being $V > 0$, the second term gives the first principle per unit volume

\[d \mathcal{E} = T d\mathcal{S} + \mathbf{F} \cdot d \symbf{\chi} \ .\]

3.1.2.2.1. Different systems#

Single-component fluid;

\[V = V \cdot 1 \quad , \quad N = V \dfrac{M}{V} \dfrac{1}{m} = V \dfrac{\rho}{m} \ , \]

being $\rho$ the mass density and $m$ the mass of a particle (or mole, it depends on the description) of the medium; $m$ is constant. First principle becomes

\[0 = d V \underbrace{\left( \mathcal{E} - T \mathcal{S} + P - \mu \dfrac{\rho}{m} \right)}_{= 0} + V \left( d \mathcal{E} - T d\mathcal{S} + 0 - \frac{\mu}{m} \, d \rho \right) \ ,\]

and thus

\[d \mathcal{E} = T d \mathcal{S} + \dfrac{\mu}{m} d \rho \ .\]

This latter formulation is consistent with the principle per unit mass. Volume density can be written as the product of mass density and the mass density of the physical quantity of interest, namely $\mathcal{E} = \rho e$

\[\begin{split}\begin{aligned} 0 & = - d( \rho e ) + T d (\rho s) + \frac{\mu}{m} d \rho = \\ & = d \rho \left( - e + T s + \frac{\mu}{m} \right) - \rho \left( d e + T d s \right) = \\ & = d \rho \underbrace{\left( - e - \dfrac{P}{\rho} + T s + \frac{\mu}{m} \right)}_{\text{def. of $e$}} + \rho \underbrace{\left( - d e + \dfrac{P}{\rho^2} d \rho + T d s \right)}_{1^{st} \text{ pr. per unit mass}} \ . \end{aligned}\end{split}\]
Multi-component fluid

\[\text{\textbf{todo}}\]
Single-component solid

\[\text{\textbf{todo}}\]

3.2. Potentials - specific quantities#

Here mechanical $\mathbf{X}_m$ and non-mechanical $\mathbf{X}_n$ generalized forces and displacements are recognized to define enthaply and Gibbs’ free energy later.

\[e(s, \mathbf{x}) = e \left( \frac{S}{M}, \frac{\mathbf{X}}{M} \right) = \frac{1}{M} E\left( \frac{S}{M}, \frac{\mathbf{X}}{M} \right) \qquad \text{\textbf{todo check}} \]

Internal energy, $e(s, \mathbf{x})$.

\[d e = T ds + \mathbf{F} \cdot d \mathbf{x}\]

Helmholtz free energy, $f(T, \mathbf{x}) = e - T s$.

\[\begin{split}\begin{aligned} d f & = de - T d s - s d T= \\ & =- s d T + \mathbf{F} \cdot d \mathbf{x} \ . \end{aligned}\end{split}\]

Enthaply, $h(T, \mathbf{F}_m, \mathbf{x}_n) = e - \mathbf{F}_m \cdot \mathbf{x}_m$.

\[\begin{split}\begin{aligned} d h & = de - d \mathbf{F}_m \cdot \mathbf{x}_m - \mathbf{F}_m \cdot d \mathbf{x}_m = \\ & = T d s - \mathbf{x}_m \cdot d \mathbf{F}_m + \mathbf{F}_n \cdot d \mathbf{x}_n \ . \end{aligned}\end{split}\]

Gibbs’ free energy, $g(T, \mathbf{F}_m, \mathbf{x}_n) = h - T s = e - \mathbf{F}_m \cdot \mathbf{x}_m - T s = f - \mathbf{F}_m \cdot \mathbf{x}_m = \mathbf{F}_n \cdot \mathbf{x}_n$.

\[\begin{split}\begin{aligned} d g & = dh - T ds - s dT = \\ & = - s dT - \mathbf{x}_m \cdot d \mathbf{F}_m + \mathbf{F}_n \cdot d \mathbf{x}_n = \\ & = \mathbf{x}_n \cdot d \mathbf{F}_n + \mathbf{F}_n \cdot d \mathbf{x}_n \ . \end{aligned}\end{split}\]

Partial derivatives of potentials.

\[\begin{split}\begin{aligned} T & = \quad \left.\frac{\partial e}{\partial s}\right|_{\mathbf{x}} && = \quad \left.\frac{\partial h}{\partial s}\right|_{\mathbf{F}_m,\mathbf{x}_n} \\ s & = - \left.\frac{\partial f}{\partial T}\right|_{\mathbf{x}} && = - \left.\frac{\partial g}{\partial T}\right|_{\mathbf{F}_m,\mathbf{x}_n} \\ \mathbf{F}_m & = \quad \left.\frac{\partial e}{\partial \mathbf{x}_m}\right|_{s,\mathbf{x}_n} && = \quad \left.\frac{\partial f}{\partial \mathbf{x}_m}\right|_{T,\mathbf{x}_n} \\ \mathbf{F}_n & = \quad \left.\frac{\partial e}{\partial \mathbf{x}_n}\right|_{s,\mathbf{x}_m} && = \quad \left.\frac{\partial f}{\partial \mathbf{x}_n}\right|_{T,\mathbf{x}_m} && = \quad \left.\frac{\partial h}{\partial \mathbf{x}_n}\right|_{s,\mathbf{F}_m} && = \quad \left.\frac{\partial g}{\partial \mathbf{x}_n}\right|_{T,\mathbf{F}_m} \\ \mathbf{x}_m & = -\left.\frac{\partial h}{\partial \mathbf{F}_m}\right|_{s,\mathbf{x}_n} && =- \left.\frac{\partial g}{\partial \mathbf{F}_m}\right|_{T,\mathbf{x}_n}\\ \end{aligned}\end{split}\]

Maxwell’s relations.

\[\text{\textbf{todo} Uncomment}\]

1: H.Touchette, When is a quantity additive, and when is it extensive?
2: https://physics.stackexchange.com/q/677855