0

In this question Does the volume of a thermodynamic system always have to change for it to do work? it is concluded that a change in pressure is not considered as work as work is per definition force times path.

Beside for an ideal gas the inner (thermal) energy does not depend on pressure so the first law of thermodynamics holds $dU = dQ + dW$.

However what would happen for example if we consider the piezo electric effect where a change in pressure indeed causes a change in inner energy. Would we than consider this as work?

Community
  • 1
bodokaiser
  • 543

1 Answers1

1

The internal energy is always a function of the extensive variables of the system under consideration $U(S,V)$ for example when the work being done involves changes in volume $PdV$, or $U(S,\varepsilon)$ when we want to consider general strains $\sigma_{ij} d\varepsilon_{ij}$, or even $U(S,D)$ if the material has a polarization expressed as an electric displacement (polarization) in an electric field $EdD$.

For a piezoelectric, the mechanical properties (strain) and electrical properties (polarization) are in fact coupled. If the polarization arises due to mechanical stress as for the direct piezoelectric effect then $$D_i = d_{ijk} \sigma_{jk}$$ where $d_{ijk}$ is the piezoelectric strain tensor. When an electric field results in a material strain as in the converse piezoelectric effect we have $$\varepsilon_{ij}=d_{ijk}E_i$$

We can write the general form of the internal energy as $$dU(S,D,\varepsilon) = TdS + EdD + \sigma d \varepsilon.$$

With this potential, or other characteristic potentials that depend on intensive variables that can be controlled in the lab, you can find more coupling equations by investigating the equations of state and the Maxwell relations.

compmatsci
  • 441