0
$\begingroup$

In a cyclic process, the system returns to the same state i.e. the same $P, V, T$. If this is an irreversible cycle, how can the surrounding not return to it's original state? Wouldn't this imply that the surrounding is no longer at it's own $P, V, T$ hence the system and surroundings would not be back in equilibrium with each other, resulting in some change again?

Does this mean that the surroundings cannot be described by state variables or it would just need more state variables than this? Yes the entropy must go up but what does this correspond to macroscopically? An increase/decrease in temperature or pressure or volume?

Improve this question
$\endgroup$
2
  • $\begingroup$ In thermodynamics it's usually assumed that the surroundings are so big that it is not affected by anything that happens in the system - so e.g. if the system loses energy, it is absorbed by the surroundings without the surroundings' temperature increasing. $\endgroup$
    – Allure
    Commented Mar 14, 2021 at 22:40
  • $\begingroup$ So something happens to the surroundings "far away" but it has no affect on the system? $\endgroup$
    – cheekylittleduck
    Commented Mar 14, 2021 at 22:51

2 Answers 2

Reset to default
1
$\begingroup$

When we build thermodynamic systems to analyze, we typically construct the environment to be simpler to calculate, at the expense of some precision. For example, if we have a simple 2nd order model which is 0.0000000000000000000001% less accurate than a 3718147289214-order model, we'll use the 2nd order model.

As Allure points out, it is common to assume things like "the surroundings are very large and have high thermal conductivity, so the system we are exploring cannot meaningfully change it."

In theory, the surroundings could contain something which does return a system to its original state. For example, in a 4 stroke engine cycle, the action of the piston is irreversible, but the outside system happens to squirt a little gas in at the right time, which is sufficient to make the system containing only the piston reversible in that environment.

Improve this answer
$\endgroup$
1
$\begingroup$

In a cyclic process, the system is returned to its original state each cycle, but not the surroundings. The internal energy and entropy of each reservoir comprising the surroundings change each cycle. In the case of a reversible cycle, the combined entropy changes of the reservoirs is zero, but in an irreversible cycle, the combined entropy changes of the reservoirs increase. In both irreversible and reversible cycles, the combined internal energy of the reservoirs changes.

If the cycle is carried out reversibly, both the system and the surroundings can be returned to their original states by running in reverse. But, if the cycle is carried out irreversibly, it will not be possible to return both the system and the surroundings to their original states.

Improve this answer
$\endgroup$
4
  • $\begingroup$ This is the part that bothers me. What is the "state" of the surroundings then, is it just the total energy and entropy of the surroundings? What does it mean for certain energy exchanges to be irreversible. Does that mean that energy is no longer useable? $\endgroup$
    – cheekylittleduck
    Commented Mar 15, 2021 at 0:49
  • $\begingroup$ In thermodynamics, the surroundings are assumed to consist thermally of ideal infinite isothermal reservoirs (of infinite heat capacity). As such, the temperature of a reservoir can change only infinitesimally, but this is accompanied by finite changes in internal energy and entropy. In an irreversible cycle, all the entropy generation is assumed to occur exclusively within the system, but some or all of this entropy is transferred to the surroundings vis heat exchange, such that the entropy change of the system in the cycle is zero. $\endgroup$
    – Chet Miller
    Commented Mar 15, 2021 at 1:31
  • $\begingroup$ Actually, in an irreversible cycle, all the entropy generated is transferred to the surroundings via heat exchange. $\endgroup$
    – Chet Miller
    Commented Mar 15, 2021 at 12:26
  • 1
    $\begingroup$ @ChetMiller I'm reminded of the case we discussed involving the irreversible transfer of heat in an isochoric heat addition process. As I recall, the entropy generated in the system was not transferred to the surroundings and instead less entropy was transferred from the surroundings to the system. The end result is the same, an increase in the entropy of the surroundings. $\endgroup$
    – Bob D
    Commented Mar 15, 2021 at 14:03

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.