Problem statement: 2 kg of air at 500 kPa expands adiabatically in a closed system until its volume (V) is doubled and it's temperature become equal to that of surrounding which is at 100 kPa, 5°C. For the process determine

  1. maximum work
  2. change in availability
  3. irreversibilty

For air take $c_v= 0.718\text{ kJ/kg K}$, $u = c_v T$ and $PV=mRT$ where $R= 0.287\text{ kJ/kg K}$

In the solution, maximum work is solved using state equation including entropy which I have no doubt about. But my doubt is:

  1. why can't we use integration $p\partial V$ for an adiabatic process to find maximum work here?
  2. as per the question, final temperature is same as that of surrounding and final volume is 2 times of initial volume also it is an adiabatic process. Final pressure using $PV^{1.4}=\text{constant}$ and $PV=mRT$ are giving different result for the final pressure of the system.
  1. If you simply integrate the internal energy change from pressure-volume changes you are assuming that going from state 1 -> 2 is fully reversible (isentropic). The way the question has been created makes it irreversible, so there is an increase in entropy in your system. If the system changed reversibly your thoughts would be correct.

  2. For the same reason as above. the relationship where you are using the ratio of specific heats (Cp/Cv) in the exponent is only valid for an isentropic process.

That dang entropy gets us every time ...

  • $\begingroup$ Exactly. After spending a good amount of time on the question I came to following conclusion: If the process was to be isentropic it would not go through a state which would give double the initial volume and at the same time have the same temperature as that of surrounding which suggest that there is irreversibilty because I got 2 different values for final pressure using PV^1.4 and PV=mRT. And maximum work would be in reversible process so I after finding the final temperature of the process in isentropic process and using that temperature to find the work gave me the correct answer. $\endgroup$ – huministic Apr 7 '20 at 3:40
  • $\begingroup$ But i guess one should go with considering it as an irreversible process from the beginning itself and use property equation like TdS = dU + PdV to work through the problem. $\endgroup$ – huministic Apr 7 '20 at 3:43

1) The maximum mechanical work of a process is not the same as the actual work done on/by the system. The former is calculated using the assumptions:

  • reversible
  • ideal gas

In the reversible case,

$$ \Delta U = 0 \rightarrow w = -q\ \ \mathrm{IUPAC\ form,\ for\ Clausius\ form\ } w = q $$

Next, the starting equation for work is $w = -\int p_{ext}\ dV$. Only when we say that the process is reversible can we state that $w = -\int p\ dV$.

2) For a reversible adiabatic processes with ideal gases, we can substitute into the mechanical work using $pV^\gamma = $constant. An adiabatic reversible process is also always isentropic. An isentropic process is not always a reversible adiabatic one. An isentropic process also does not always mean that we are using an ideal gas. In summary, we cannot say that we can use $pV^\gamma = $constant for the work of any generic isentropic process. This is a shortcut that ignores the three founding rules for the substitution (an adiabatic, reversible process on an ideal gas).

Finally, regardless of whether the real process is reversible or irreversible, we can always write

$$\Delta U = \int T\ dS - \int p\ dV $$

All terms in the above are state functions, and the change $\Delta U$ does not care about the path.

  • $\begingroup$ So bottom line is unless otherwise stated that the process is reversible and other ideal conditions we should approach a problem as irreversible and real conditions.. also there's an edit I would like to suggest. An adiabatic process is not necessarily isentropic process $\endgroup$ – huministic Apr 9 '20 at 17:06
  • $\begingroup$ No real process is truly reversible. Reversibility must always be stated as an explicit assumption, not presumed implicitly. Adiabatic reversible processes are isentropic. $\endgroup$ – Jeffrey J Weimer Apr 9 '20 at 17:23

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.