1
$\begingroup$

Let the external intensive and extensive "mechanical" variables be denoted by $Y_k, X_k$. These variables are well defined irrespective of the system is in equilibrium or not. For an adiabatic process then we always have $dU=\sum_k Y_kdX_k$. Assume that the thermal and caloric equations of the system are given.

Now let us start from an arbitrary equilibrium state, that is both the external and the system's internal parameters are equal $Y_k^0, X_k^0$ with internal energy and entropy denoted by $U^0, S^0$. Now let the external control parameters vary according to some prescribed time function, i.e., $Y_k(t), X_k(t)$ ending in the equilibrium state $Y_k^1, X_k^1$ and $U^1$, a pure work process, such that the initial entropy is not the same as the final one. Since we assumed that the process is adiabatic the system entropy must increase $S^1 > S^0$ and the process is irreversible.

My question has two parts:

  1. given the start and end states, $\mathscr P^0= \{Y_k^0, X_k^0, U^0, S^0\}$ and $\mathscr P^1=\{Y_k^1, X_k^1, U^1, S^1\}$, are the control functions $Y_k(t), X_k(t)$ unique? In other words, can there be more than one irreversible adiabatic processes that start and end in the same equilibrium starting and ending states?
  2. if the control functions are not unique then how "large" is that set of function and how to characterize them analytically?
Improve this question
$\endgroup$
8
  • $\begingroup$ Are you asking whether there can be more than 1 adiabatic irreversible process between the same two end states? $\endgroup$
    – Chet Miller
    Commented Jan 17, 2022 at 10:37
  • $\begingroup$ @ChetMiller the words "in finite time" were missing in the title, my bad... $\endgroup$
    – hyportnex
    Commented Jan 17, 2022 at 13:50
  • $\begingroup$ There is some ambiguity in what is called adiabatic process - in the context of Carnot cycle one usually implies a reversible process. However, more generally the word is taken to mean absence of heat exchange. $\endgroup$
    – Roger V.
    Commented Jan 17, 2022 at 14:16
  • $\begingroup$ @RogerVadim I assume no heat exchange. Can there be two adiabatic processes starting in the same equilibrium state and end in the same equilibrium state both in finite time assuming that there is at least one such? $\endgroup$
    – hyportnex
    Commented Jan 17, 2022 at 14:55
  • $\begingroup$ Yes, for some adiabatic reversible processes. For example, shear deformation of a viscous incompressible fluid between two infinite adiabatic parallel plates, the final state depends only on the amount of work done. Also, stirring of a viscous fluid adiabatically. $\endgroup$
    – Chet Miller
    Commented Jan 17, 2022 at 15:00

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.