Week 1 Thermodynamics Revision, with an eye to Statistical Mechanics

Week 1 Thermodynamics Revision, with an eye to Statistical Mechanics#

Objective of these notes:#

To guide revision of the principles of Thermodynamics to: olecu

  • Revise the first and second law of thermodynamics

  • Define the concept of thermodynamic equilibrium

  • Define state functions and thermodynamic potentials

First Law of Thermodynamics.#

The first law of thermodynamics is about internal Energy. It is based on two postulates:

  1. Internal Energy is an extensive property.

  2. Internal Energy is conserved.

This leads to the formulation of the first law of thermodynamics:

\[ dU=\mathbf{d}Q+\mathbf{d}W \]

where \(Q\) and \(W\) represent heat and work, and \(\mathbf{d}\) indicate inexact differentials.

\(\mathbf{d}W\) is the differential work done by manipulating mechanical constraints. It can be written generally as:

\[ dW=\mathbf{f}{d}\mathbf{X} \]

where \(\mathbf{f}\) represents a vector of applied forces and \(\mathbf{X}\) a vector of their corresponding mechanical extensive variables.

In its simplest formulation, \(\mathbf{X}\) has a single element, the system’s volume \(V\). The corresponding thermodynamic force, corresponding to volume \(V\), is \(-P\), where \(P\) is pressure.

It should be noted that heat and work only represent forms of energy transfer, and once the transfer has taken place, there is no \(Q\) or \(W\) quantity that remains associated with a given state. This is why \(\mathbf{d}Q\) and \(\mathbf{d}W\) are inexact differentials.

Second Law of Thermodynamics#

The second law of thermodynamics concerns the reversibility of transformations and emerges from assumptions on the nature of equilibrium states. Following Chandler’s approach, we start with a postulate of the second law:

  • There is an extensive function of state, \(S(U, X)\), which is a monotonically increasing function of U, and if state B is adiabatically accessible from state A, then \(S_B\geq{S}_A\).

This implies that the change \(\Delta{S} = S_B - S_A\) is zero for a reversible adiabatic (\(dQ=0\)) process, and otherwise \(\Delta{S}\) is positive for any natural irreversible adiabatic process.

The extensive function of the state, \(S(U, X)\), is the system’s entropy. Hence, the entropy change for a reversible adiabatic process is null. Note also that entropy is a function of state. That means it is defined for those states characterized by \(U\) and \(\mathbf{X}\).

Definition of Temperature:#

Since \(S\) is a monotonically increasing function of \(U\), \(\rightarrow\) \(\left(\frac{\partial{S}}{\partial{U}}\right)_X>{0}\), or \(\left(\frac{\partial{U}}{\partial{S}}\right)_X\geq{0}\). This derivative defines the absolute temperature \(T\):

\[ \left(\frac{\partial{U}}{\partial{S}}\right)_X\equiv{T} \]

Mathematical formulation of the Second Law: the fundamental thermodynamic equation.#

Let’s start by writing the differential of the Entropy function introduced above:

\[ dS(U,X)=\left(\frac{\partial{S}}{\partial{U}}\right)_XdU + \left(\frac{\partial{S}}{\partial{X}}\right)_UdX \]

This indicates a variety of entropy associated with any variation of internal energy \(dU\) and of the vector of extensive mechanical properties \(X\). For any reversible transition \(dU=\mathbf{d}Q_{rev}+f\mathbf{d}X\), and \(dS=0\), hence:

\[ 0=\left(\frac{\partial{S}}{\partial{U}}\right)_X(\mathbf{d}Q_{rev}+f\mathbf{d}X)+\left(\frac{\partial{S}}{\partial{X}}\right)_UdX \]

This equation is true for any reversible transition, including the adiabatic ones. Hence it must be that:

\[ 0=\left[\left(\frac{\partial{S}}{\partial{U}}\right)_Xf+\left(\frac{\partial{S}}{\partial{X}}\right)_U\right]\mathbf{d}X \]

and

\[ \left(\frac{\partial{S}}{\partial{X}}\right)_U=-\left(\frac{\partial{S}}{\partial{U}}\right)_Xf \]

By introducing the definition of absolute temperature gives:

\[ -\frac{f}{T}=\left(\frac{\partial{S}}{\partial{X}}\right)_U \]

This expression enables to rewrite the differential of entropy \(dS\) as:

\[ dS=\frac{1}{T}dU - \frac{f}{T}dX \]

which, for a single component, systems unable to exchange matter with the environment (\(dn=0\)) leads to the fundamental expression:

\[ dU=TdS-PdV \]

N.B. This equation establishes the triplet of state variables \(U\), \(S\) and \(V\) as a so-called fundamental grouping. Different choices are possible, leading to the definition of different fundamental groupings with their thermodynamic potential.

Changing the fundamental grouping from V, S, U: Legendre transform and Thermodynamic potentials#

So far we have discussed of Energy, Entropy and Volume - and learned that the equilibrium conditions for a system as a function of S and V correspond to minimum E. If one needs to characterise equilibrium as a function of a different set of variables than entropy and volume, can a therodynamic function that plays the same role of Energy be identified?

The answer is, yes. Such function, hereafter indicated as thermodynamic potential, can be obtained by swapping variables by performing a Legendre Transform.

The simplest way of demonstrating the process of operating a Legendre transform to obtain a function of a new set of variables is to start from original function, and subtract the product of the variables that need to be swapped. Differentiating such new function allows to derive a new fundamental relationship defining a thermodynamic potential for the new grouping.

Swapping T and S, from Energy to Helmholtz Free Energy.#

If we define a new grouping in which the entropy and temperature are swapped, i.e. we want to define a new thermodynamic potential able to characterise equilibrium for a system at constant T and V we should start by defining a new function as follows:

\[ A=U-TS \]

where \(A\) is the new thermodynamic potential, the Helmholtz free energy, \(T\) is the temperature, and \(S\) is the entropy. Computing the differential for A gives us:

\[ dA=dU-SdT-TdS=TdS-PdV-SdT-TdS=-SdT-PdV \]

which is the new thermodynamic relationship that allows the characterization of equilibrium as a function of \(V\) and \(T\) rather than \(V\) and \(S\).

Swapping P and V from Energy to Enthalpy.#

Let’s start by Legendre transforming the \(U\) to swap \(V\) with its intensive conjugate variable \(P\), this yields H, or enthalpy:

\[ H=U+PV \]

the fundamental equation, in a system at constant P and S, then becomes:

\[ dH=dU+VdP+PdV=TdS-PdV+VdP+PdV=TdS+VdP \]

Gibbs Free Energy#

Gibbs Free Energy is obtained by swapping both T/S and P/V. It is defined as:

\[ G=U-TS+PV \]

the expression of the Gibbs free energy differential \(dG\) is thus:

\[ dG=dH-SdT-TdS=TdS+VdP-SdT-TdS=-SdT+VdP \]

The Phase Equilibrium Problem#

References:#

  • Chapters 1, 2 Prausnitz

  • Chapters 1, 2 Chandler

The problem:#

Relate quantitatively the variables that describe the state of equilibrium between two or more homogeneous phases that are free to exchange energy and matter.

Definitions:#

  • Homogeneous phase at equilibrium: Region of space where the intensive properties are constant.

  • Intensive properties: properties that are independent of mass, size, or shape of the phase (i.e. T, P, r, x, y, etc.)

Equilirbium in a multiphase, multicomponent, system:#

Consider a system where:

  • Two or more phases are present

  • Each phase can be considered an open system within the over all closed system

How do we know when such a system is at equilibrium?

The energy, \(U\), is an extensive property of the system. In an heterogeneous multipase system containing at least two phases \(\alpha\) and \(\beta\) the total energy of the system is \(U=U^\alpha+U^\beta\). The same can be written for the other extensive variables defining the state of the multiphase system such as entropy:

\[ S=S^\alpha+S^\beta \]

volume

\[ V=V^\alpha+V^\beta \]

and the number of moles of each i\(^{th}\) component in the different phases:

\[ n_i=n_i^\alpha+n_i^\beta \]

Consider now to define the equilibrium conditions for a system where the total \(S\), \(V\) and \(n_i\) are constant, but these quantities can be redistributed between phases \(\alpha\) and \(\beta\). This requires that:

\[ \delta{S}^{\alpha} = - \delta{S}^{\beta} \]
\[ \delta{V}^{\alpha} = - \delta{V}^{\beta} \]
\[ \delta{n_i}^{\alpha} = - \delta{n_i}^{\beta} \]

The equilibrium can be investigated with a variational principle, i.e. by introducing a perturbation \(\delta{U}\) from equilibrium. If the system is in equilibrium the perturbation can only be associated to and increase in U:

\[ \delta{U}_{S,V,n_i}\geq 0 = (T^\alpha-T^\beta)\delta{S^{\alpha}}-(p^{\alpha}-p^{\beta})\delta{V^{\alpha}}+(\mu_i^{\alpha}-\mu_i^{\beta})\delta{n_i^{\alpha}} \]

Since the variations in \(n\), \(S\), and \(V\) are all decoupled, and can be both positive or negative, the only valid solution for the equilibrium condition stated above is that the temperature, pressure, and chemical potential of component i remain constant across the phase boundary between \(\alpha\) and \(\beta\), namely:

\[ {T}^{\alpha} = {T}^{\beta} \]
\[ {p}^{\alpha} = {p}^{\beta} \]
\[ {\mu_i}^{\alpha} = {\mu_i}^{\beta} \]

Degrees of freedom of a single multicomponent phase.#

Given what we have seen so far, can we determine how many degrees of freedom are associated to a single phase?

In a single homogeneous multicomponent phase at constant \(T\) and \(P\), containing \(m\) component the number of degrees of freedom is:

\[ DOF=m+2-c \]

where \(c\) is the number of constraints. For a single phase there is only one constrain that applies, the Gibbs Duhem equation. Hence \(DOF=m+1\).

The Gibbs Duhem Equation#

The fundamental equation emerging from the first and second laws of Thermodynamics for a multicomponent system reads:

\[ dU=TdS-pdV+\sum_i \mu_idn_i \]

Integrating this equation to obtain \(U\) along from a state with no molecules to a state where a final number \(n_i\) of molecules is present, along an isothermal, isobaric transformation at constant compositionleads to the following expression for \(U\):

\[ U=TS-pV+\sum_i\mu_in_i \]

Differentiating \(U\) leads to:

\[ dU=TdS+SdT-pdV-VdP+\sum_i\mu_idn_i+\sum_id\mu_in_i \]

Equating the right hand sides of these two expressions leads to:

\[ TdS-pdV+\sum_i \mu_idn_i=TdS+SdT-pdV-VdP+\sum_i\mu_idn_i+\sum_id\mu_in_i \]

which implies that the following equality is always satisfied for an homogenous multicomponent phase:

\[ SdT-Vdp+\sum_{i}n_{i}d\mu_i=0 \]

This equality is the Gibbs-Duhem equation, which allows express the chemical potential as a function of the temperature and pressure.

Degrees of fredom in a multiphase, multicomponent system: The Gibbs phase rule#

Let’s consider a system comprising \(\nu\) homogenous phases at equilibrium, each containing \(m\) components. The number of degrees of freedom associated to each phase is:

\[ DOF=\nu(m+1) \]

However at equlibrium we have that for every couple of phases (\(\nu-1\)), \(T\), \(P\) and all \(m\) chemical potentials needs to be constant across phase boundaries. This means that we have a number of constraints equal to:

\[ c=(\nu-1)(m+2) \]

This leads to a number of degrees of freedom equal to:

\[ DOF=\nu(m+1)-(\nu-1)(m+2)=m+2-\nu \]

Contributions#

Ivan Boguslavskyi, 1.10.2024

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