1
$\begingroup$

The principle of maximum work states that for any process between two states, the work done by the system is maximised for a reversible process (and heat transfer is minimised), and that the work done by any reversible process between these two states is equal.

I don't see how that can be compatible with going between 2 states with equal entropy by different paths. For example:

-An adiabatic expansion between 2 states with the same entropy.

-An isovolumetric pressurisation followed by an isobaric expansion with the same initial and final states as the adiabatic expansion.

The area below of the second path in a P-V diagram is clearly higher, so we have a reversible process between two states doing less work than another process between those states. How can this apparent contradiction be resolved?

$\endgroup$
  • $\begingroup$ Your 2nd example is not an adiabatic process, and that the final states are the same as in the 1st example means just that and no more. The states maybe the same but both heat transfer and working are process dependent though their sum is the difference between the corresponding internal energies. $\endgroup$ – hyportnex Jul 11 '16 at 19:25
  • $\begingroup$ Does this mean that I should take the principle of maximum work to mean "If a process between two states is not reversible, then a process that delivers more work exists, and that process is reversible"? That seems reasonable, but the statement from Herbert Callen's textbook "the delivery of work is identical for every reversible process" would contradict that. $\endgroup$ – Senrade Jul 11 '16 at 19:42
  • $\begingroup$ I assume you mean chapter 4-5 in Callen. There Callen constrains not only the internal energy change but also the entropy change and in that case the reversible process will always deliver more work than the irreversible one because in the latter the "sub-system" will also have to shed excess heat (due to internal irreversibilities, such as friction) thus the sum can only be the same if the work is less. $\endgroup$ – hyportnex Jul 11 '16 at 19:59
  • $\begingroup$ But my second process has the same entropy change as the adiabatic expansion, as they have the same initial and final states. The example I gave is in fact problem 4.5-6 from that section, so if I've phrased it poorly it's also there. $\endgroup$ – Senrade Jul 11 '16 at 20:06
  • $\begingroup$ Ok, so it seems to be a homework problem. Hint: check Eq. 4-13, what is $T_{res}$ in your problem? $\endgroup$ – hyportnex Jul 11 '16 at 21:27
1
$\begingroup$

It doesn't seem possible for the original premise to be correct. If you start at a certain state, and carry out a reversible Carnot cycle ending up at the same original state, you can design a great big Carnot cycle and you can design a tiny little Carnot cycle. Certainly the reversible work for these two cycles will not be the same.

$\endgroup$
  • $\begingroup$ I realise now that my two examples don't qualify for "any reversible process between the two states", as I was only considering the primary subsystem's state. The second process involves the transfer of more heat from a reservoir in order to make up for the extra work. Considering the state of the reservoirs clearly means that the initial and final states of the entire system are not identical. I think how the reconciliation takes place. $\endgroup$ – Senrade Jul 11 '16 at 23:47

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.