Another set of characteristic functions for a single-substance system can be defined by performing the Legendre transformations on the entropy \(S(U, V)\) rather than on the internal energy \(U(V, S)\). The thermodynamic potentials turn out to be particularly useful in statistical mechanics and the theory of irreversible thermodynamics, in contrast to equilibrium thermodynamics presented in this book. (a) Show that Legendre transformation of \(S(U, V)\) that produces the characteristic function \(J(1 / T, V)\), known as the Massieu function, is given by the transform $$ J=-\frac{U}{T}+S=-\frac{A}{T}, $$ and $$ d J=\frac{U}{T^{2}} d T+\frac{P}{T} d V $$ (b) Show that Legendre transformation of \(J(1 / T, V)\) that produces the thermodynamic potential \(Y(1 / T, P / T)\), known as the Planck function, is defined by the transform and $$ \begin{aligned} &Y=-\frac{H}{T}+S=-\frac{G}{T} \\ &d Y=\frac{H}{T^{2}} d T-\frac{V}{T} d P \end{aligned} $$

Step by step solution

01

Understanding the Problem

The problem involves performing Legendre transformations on entropy functions and deriving the expressions for the Massieu and Planck functions along with their differentials. Legendre transformations are used to switch between different thermodynamic potentials by changing their natural variables.

02

Legendre Transformation for Massieu Function

Start with the entropy function, given by: \[ S(U, V) \] Perform the Legendre transformation with respect to the internal energy and temperature: \[ J = S - \frac{U}{T} \] By rearranging, this becomes: \[ J = -\frac{U}{T} + S \] Given that the Helmholtz free energy is defined as \( A = U - TS \), we can rewrite it in terms of Massieu function as: \[ J = -\frac{A}{T} \]

03

Deriving the Differential for Massieu Function

Take the differential of the Massieu function \(J\): \[ dJ = -d\left(\frac{U}{T}\right) + dS \] First, calculate \(d\left(\frac{U}{T}\right)\): \[ d\left(\frac{U}{T}\right) = \frac{dU}{T} - \frac{U}{T^2} dT \] Using the first law of thermodynamics, \( dU = TdS - PdV \), substitute for \( dU \): \[ d\left(\frac{U}{T}\right) = \frac{TdS - PdV}{T} - \frac{U}{T^2} dT = dS - \frac{P}{T} dV - \frac{U}{T^2} dT \] So the differential of \(J\) is: \[ dJ = -\frac{U}{T^2} dT + \frac{P}{T} dV \]

04

Legendre Transformation for Planck Function

Next, perform the Legendre transformation on \(J\) for temperature and pressure variables:\[ Y = J - \frac{P}{T}V \] Substitute for \(J\) from the previous transformation: \[ Y = -\frac{U}{T} + S - \frac{P}{T}V \] Using the enthalpy definition \(H = U + PV\), we get: \[ Y = -\frac{H}{T} + S \] We also know that \(G = H - TS\), so we can express \(Y\) as: \[ Y = -\frac{G}{T} \]

05

Deriving the Differential for Planck Function

Take the differential of the Planck function \(Y\): \[ dY = -d\left(\frac{H}{T}\right) + dS \] Calculate \(d\left(\frac{H}{T}\right)\): \[ d\left(\frac{H}{T}\right) = \frac{dH}{T} - \frac{H}{T^2} dT \] Using the definition \( dH = TdS + VdP \), substitute for \( dH \): \[ d\left(\frac{H}{T}\right) = \frac{TdS + VdP}{T} - \frac{H}{T^2} dT = dS + \frac{V}{T} dP - \frac{H}{T^2} dT \] So the differential of \(Y\) is: \[ dY = \frac{H}{T^2} dT - \frac{V}{T} dP \]

06

Summary of the Transformations

For the Massieu function: \[ J = -\frac{U}{T} + S = -\frac{A}{T} \]\[ dJ = -\frac{U}{T^2} dT + \frac{P}{T} dV \]For the Planck function: \[ Y = -\frac{H}{T} + S = -\frac{G}{T} \]\[ dY = \frac{H}{T^2} dT - \frac{V}{T} dP \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Massieu function

The Massieu function is a thermodynamic potential obtained through a Legendre transformation of entropy with respect to temperature. This transformation requires starting with the entropy function, expressed as:
\ S(U, V)
To perform the transformation, we use the internal energy \( U \) and temperature \( T \), resulting in the Massieu function \( J \):
\ J = S - \frac{U}{T}
By rearranging this relationship, we can also express it as:
\ J = -\frac{U}{T} + S
Since the Helmholtz free energy \( A \) is given by \( A = U - TS \), we can rewrite the Massieu function in terms of \( A \):
\ J = -\frac{A}{T}
The differential form of the Massieu function is also useful:
\ dJ = -d\left(\frac{U}{T}\right) + dS
Through the first law of thermodynamics and differentiation, it simplifies to:
\ dJ = -\frac{U}{T^2} dT + \frac{P}{T} dV
This transformation highlights the connection between entropy, internal energy, and temperature in thermodynamic systems.

Planck function

The Planck function, another thermodynamic potential, is derived from the Massieu function via a further Legendre transformation. Starting from the Massieu function \( J \), we perform the transformation with pressure and temperature variables:
\ Y = J - \frac{P}{T}V
By substituting our earlier result for \( J \), we get:
\ Y = -\frac{U}{T} + S - \frac{P}{T}V
Utilizing the definition of enthalpy \( H = U + PV \), we can express this as:
\ Y = -\frac{H}{T} + S
Knowing that Gibbs free energy \( G \) equals \( H - TS \), we find:
\ Y = -\frac{G}{T}
To find the differential form, consider:
\ dY = -d\left(\frac{H}{T}\right) + dS
Through thermodynamic laws and differentiation, we derive:
\ dY = \frac{H}{T^2} dT - \frac{V}{T} dP
The Planck function provides valuable insights into systems where temperature and pressure are the primary variables.

Thermodynamic potentials

Thermodynamic potentials are functions that help us describe the state of a thermodynamic system. They provide different perspectives depending on the natural variables used. The key thermodynamic potentials include:

• Internal Energy \( U \)
• Helmholtz Free Energy \( A = U - TS \)
• Enthalpy \( H = U + PV \)
• Gibbs Free Energy \( G = H - TS \)
• Massieu Function \( J = -\frac{A}{T} \)
• Planck Function \( Y = -\frac{G}{T} \)
• Entropy \( S \)

Each potential is useful in different thermodynamic scenarios. For example, Helmholtz Free Energy is used when volume and temperature are constant, while Gibbs Free Energy is relevant under constant pressure and temperature. The Massieu and Planck functions are particularly helpful in the fields of statistical mechanics and irreversible thermodynamics.

Irreversible thermodynamics

Irreversible thermodynamics studies systems that are not in equilibrium. Unlike equilibrium thermodynamics, which deals with static states, irreversible thermodynamics examines dynamic processes:

• Diffusion
• Heat transfer
• Viscous flow

These processes are governed by the principles of nonequilibrium entropy and thermodynamic fluxes. The Massieu and Planck functions, for example, help in describing such systems. They offer a perspective on how energy and entropy distribution changes over time during irreversible processes. This is crucial for understanding real-life applications where gradients in temperature, pressure, and chemical potentials drive the systems away from equilibrium.

Statistical mechanics

Statistical mechanics connects the microscopic properties of individual particles with macroscopic thermodynamic properties. By using probability theory and statistics, it explains the behavior of systems composed of a large number of particles:

• Microstates: The different configurations a system can be in.
• Macrostates: The observable states characterized by macroscopic quantities like temperature and pressure.
• Partition function: A function that sums up the statistical weights of all possible microstates to predict the thermodynamic properties.

The Massieu and Planck functions relate to statistical mechanics by providing alternate, convenient ways to express the potentials and partition functions. This makes it easier to derive properties and work with systems that are not in equilibrium. Understanding these concepts is essential for fields ranging from condensed matter physics to chemical thermodynamics.