4
$\begingroup$

I have been reading up (on this website) about the difference between reversible and irreversible thermodynamic process and it defines them saying that reversible processes are ideal processes while irreversible processes are the natural processes. Are these definitions correct?

If not how are they defined?

|improve this question|||||
$\endgroup$
4
$\begingroup$

The simplest answer in thermodynamic terms is "a reversible process is one in which the entropy generation term is zero."

Yes, I understood that but can you explain what difference is there in the change in enthalpy, internal energy, heat, work, etc. if the process is reversible and irreversible.

For any process, all of these would be the same whether it was reversible or not. The only one that would differ would be change in entropy.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Sorry to dig up something this old but this is wrong. The heat transferred and work done will typically be different in a reversible and an irreversible process. U, H, S are state functions, so as long as the initial and final states are kept constant, $\Delta U, \Delta H$ and $\Delta S$ are independent of the reversibility of the process. $\endgroup$ – orthocresol♦ Jul 5 '15 at 2:21
  • $\begingroup$ @orthocresol - you are right in that heat and work will probably be different for an irreversible process compared to a reversible one. You are also right that the $\Delta$'s of the state functions will be the same if the starting and ending states are the same. My point was that the only difference we know for sure is that the entropy generation term will be zero - that's different from $\Delta S$. I was assuming that op is talking about terms in a thermodynamic equation, and you wouldn't change anything other than $dS_{gen}$ if the process is reversible. $\endgroup$ – thomij Jul 5 '15 at 14:33
0
$\begingroup$

See my recent article on this very subject at the following link:

https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

The conclusion is that, mechanically at least, the difference between reversible and irreversible processes is the result of viscous dissipation of mechanical energy in the irreversible case.

Chet

|improve this answer|||||
$\endgroup$
-1
$\begingroup$

These are not definitions of reversible and irreversible processes, but it is a description of one of the differences between the processes.

In real world situations, however, for some processes, the system undergoing the changes responds much faster than the applied change, the deviation from reversibility may be negligible. In a reversible cycle, the system and its surroundings will be exactly the same after each cycle. (see here)

A (...) definition of a reversible process is a process that, after it has taken place, can be reversed and causes no change in either the system or its surroundings. In thermodynamic terms, a process "taking place" would refer to its transition from its initial state to its final state. again, wikipedia

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Yes, I understood that but can you explain what difference is there in the change in enthalpy, internal energy, heat, work, etc. if the process is reversible and irreversible. $\endgroup$ – Ris97 Dec 7 '13 at 2:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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