A rather specific thermodynamic potential

In summary: These potentials can be related through a Legendre transformation, which allows for a different perspective on the system and its variables. The physical interpretations of these potentials lie in the different types of energies and contributions that make up the overall energy and free energy of the system.
  • #1
johnmadsen88
2
0
Today, during class, our professor went through a simple example about Piezoelectricity. We have a cylinder with a trapped ideal gas and a piston which is part of a capacitor together with the bottom of the cylinder. The voltage across the capacitor is [tex]\Phi[/tex], and the charge is [tex]Q[/tex]. The distance between the piston and the bottom of the cylinder is [tex]l[/tex]. A potential for such a system could be

[tex]\mathcal{A}(T,l,\Phi )=-\frac{\epsilon A}{2l}\Phi ^2-Nk_BT\ln{Al}+\frac{3}{2}Nk_BT-\frac{3}{2}Nk_BT\ln{T}[/tex]​

with [tex]k_B[/tex] and [tex]N[/tex] denoting the Boltzman constant and the number of molecules, respectively. [tex]\epsilon[/tex] is the dielectric constant. He then said that one can (I assume, by Legendre transformation) express this using another potential [tex]\mathcal{B}[/tex] which is relevant for variables [tex](T,Q,l)[/tex]:

[tex]\mathcal{B}(T,Q,l)=-\frac{1}{2\epsilon A}Q^2-Nk_BT\ln{Al}+\frac{3}{2}Nk_BT-\frac{3}{2}Nk_BT\ln{T}[/tex]​

I was wondering if someone could give this a more physical twist. In other words, I'd like it if someone explained where these contributions come from and what characterizes them. If I have been too unclear, please tell me, so I can reword my writing. Thanks.

PS: I cannot currently see my TeX code properly when previewing (either my computer or the site is having problems), but I hope I haven't made errors.
 
Physics news on Phys.org
  • #2
The potential \mathcal{A} is related to the energy of the system, which includes the mechanical and electrical energies. The first term, -\frac{\epsilon A}{2l}\Phi ^2, is the electrical energy of the capacitor due to the voltage \Phi across it, where A is the area of the capacitor plates and \epsilon is the dielectric constant of the material between the plates. The second term, -Nk_BT\ln{Al}, is the thermal energy associated with the ideal gas, where T is the temperature of the gas and N is the number of molecules in the cylinder. The third term, \frac{3}{2}Nk_BT, is the kinetic energy of the molecules, and the fourth term, -\frac{3}{2}Nk_BT\ln{T}, is the entropy contribution due to the thermal fluctuations of the molecules. The potential \mathcal{B} is related to the free energy of the system, which is a combination of the energy and entropy contributions. The first term, -\frac{1}{2\epsilon A}Q^2, is the electrical energy of the capacitor due to the charge Q stored on it. The second and third terms are the same as in \mathcal{A}, while the fourth term is the entropy contribution associated with the thermal fluctuations of the molecules.
 

Related to A rather specific thermodynamic potential

1. What is a rather specific thermodynamic potential?

A rather specific thermodynamic potential is a mathematical concept used in thermodynamics to describe the state of a system. It is a function that depends on the state variables of the system, such as temperature, pressure, and volume, and can be used to calculate the properties of the system.

2. How is a rather specific thermodynamic potential different from other thermodynamic potentials?

A rather specific thermodynamic potential is different from other thermodynamic potentials in that it is tailored to a specific system or process. It takes into account specific constraints or conditions that may be present in the system, making it more accurate and useful for certain applications.

3. What are some examples of rather specific thermodynamic potentials?

Some examples of rather specific thermodynamic potentials include the Helmholtz free energy, Gibbs free energy, and enthalpy. These potentials are commonly used in different thermodynamic processes and systems, such as chemical reactions, phase transitions, and heat transfer.

4. How is a rather specific thermodynamic potential calculated?

A rather specific thermodynamic potential is calculated by taking the partial derivatives of the general thermodynamic potential with respect to the state variables of the system. These derivatives are then used to solve for the specific potential, which can then be used to calculate the properties of the system.

5. Why is a rather specific thermodynamic potential important in thermodynamics?

A rather specific thermodynamic potential is important in thermodynamics because it allows for a more accurate and precise description of a system. It takes into account specific conditions and constraints that may affect the behavior of the system, making it a valuable tool in understanding and predicting the properties of different processes and systems.

Similar threads

Replies
1
Views
682
Replies
4
Views
901
  • Introductory Physics Homework Help
Replies
2
Views
841
  • Introductory Physics Homework Help
Replies
15
Views
338
  • Introductory Physics Homework Help
Replies
1
Views
178
  • Introductory Physics Homework Help
2
Replies
64
Views
2K
Replies
25
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
425
  • Classical Physics
Replies
0
Views
243
Replies
1
Views
1K
Back
Top