1
$\begingroup$

I need a sanity check. I am comparing two expansion processes with the same volume change. In both processes, there is a load on a piston that needs to be lifted a specified height. The piston cylinder is filled with a gas.

For the first process, the gas is heated reversibly so that the internal pressure is constantly equal to the external pressure (atmosphere + load). The amount of work done by the gas in this reversible process is $ W=p \Delta V $.

For the second process, the piston is pinned at its initial height, and the gas is heated at constant volume. Once the gas is heated sufficiently, the pins are removed and the gas is allowed to expand irreversibly and adiabatically against the constant external pressure (atmosphere + load). For this set of processes, the work done by the gas is still $W = p \Delta V $.

The initial and final states of the two processes are the same, so $ \Delta U $ for the two processes must also be equal. Because $\Delta U$ and $W$ are equal, the heat transfer into the gas must also be equal.

So my question is: Am I correct that the heat and work involved in these two processes, one which is a single-step reversible process and another which is a two-step irreversible process, are equal? Am I missing something in this analysis? This seemed surprising at first glance, though after I considered what these processes might look like on a $p-V$ diagram they seemed to be roughly the same. For the two-step irreversible process, the gas pressure isn't well-defined, but one could imagine that once the pins are removed, the local pressure next to the piston head would be roughly equal to the external pressure throughout the process, and so the fact that the two work values are the same perhaps makes sense.

Improve this question
$\endgroup$
2
  • $\begingroup$ Have you analyzed the two processes in detail? In the 2nd irreversible process, do you allow the gas to equilibrate with the piston and outside atmosphere, or do you force the piston to stop and V2? Is the heat added to the gas the same in both cases? $\endgroup$
    – Chet Miller
    Feb 21 at 18:34
  • $\begingroup$ The net change in volume is chosen to be the same in both processes. In the second process, the adiabatic expansion occurs rapidly as soon as the pins are released, so it is certainly far from mechanical equilibrium through most of the process. Because it is adiabatic, the gas is also not equilibrating with the external temperature, but instead the temperature will be determined by the equation of state of the gas based on the pressure, volume and temperature after the constant-volume heating process. $\endgroup$
    – scmartin
    Feb 21 at 20:11

1 Answer 1

Reset to default
0
$\begingroup$

Very interesting question.

I've analyzed both these processes as you have described them, and they do indeed have the same final state, heat Q, work W, internal energy change $\Delta U$, and entropy change $\Delta S$. So what is different? Well the 2nd process has a smaller $\int{(dQ/T)}$ than the first process. So, for the 2nd process, $\Delta S>\int{(dQ/T)}$

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you! I'm actually planning to compare these to a third process that has the same initial and final states but where one of the steps is a reversible adiabatic expansion. I'm putting this together as an exercise for my students, but wanted to make sure my ducks were in a row before we get started. I'm glad you reminded me of the difference in the total entropy production of the two processes. I need to give more thought to the $dQ/T$ integrals for the irreversible processes. $\endgroup$
    – scmartin
    Feb 21 at 21:08

Your Answer

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

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