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We know that for constant pressure thermodynamic processes, $dH=dQ_p$. My question is, does it implies that only reversible work is possible in this processes so that $dw=0$ because $dv$ is zero? In addition, does $Q_v$ necessarily be reversible heat transfer in this case? What if the heat transfer is irreversible?

Similar question for $dU=dQ_v$, does the process need to be reversible?

Another question is, do the above relations have anything to do with whether or not the system is an ideal gas? I have heard from a lecture that $du≠dQ_v$ in Joule's free expansion for non-ideal gas. Consider reversibility and whether it's ideal gas there are four combinations of situations (ideal gas reversible,ideal gas irreversible, non-ideal gas reversible, non-ideal gas irreversible). I get confused with how these factors affect the thermodynamic relations.

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  • $\begingroup$ "We know that for constant volume thermodynamic processes, $dH=dQ_p$." What is $Q_p$? Don't you mean constant pressure thermodynamic processes? $\endgroup$
    – Ján Lalinský
    Commented Jun 10, 2014 at 6:18
  • $\begingroup$ yes, dQp is constant pressure heat transfer. $\endgroup$
    – Kelvin S
    Commented Jun 10, 2014 at 6:40
  • $\begingroup$ If volume is constant then there is no volume work done. There may be other work, for example electric work or work of capillary forces. These are not necessarily reversible. Whether pressure is constant or changes not does not matter. $\endgroup$
    – Ján Lalinský
    Commented Jun 10, 2014 at 8:08
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    $\begingroup$ @KelvinS Reversibility should be characterised by whether entropy increases or keeps the same. $\endgroup$
    – M. Zeng
    Commented Aug 6, 2015 at 12:42

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In general $dU=\delta Q + \delta W$ and if the only work is pressure work then $dU=TdS-pdV$, but one always has $$TdS \ge \delta Q $$ and $$\delta W \ge -pdV$$, where equality only in reversible process. Then $dH=d(U+pV)=\delta Q + \delta W +pdV + Vdp$, therefore $$dH \ge \delta Q +Vdp$$ and if the process is isobaric $dp=0$, then $$dH \ge \delta Q_p$$ with equality only in a reversible process. The same argument shows that $dU \ge \delta Q_v$. None of this is specific for an ideal gas, rather, it is true for any material whose state can be described with a single pair of mechanical parameters, here pressure and volume, so that there is a thermal and a caloric equation of state $T=f_1(p,V)$ and $U=f_2(p,V)$

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  • $\begingroup$ you have mentioned δW≥−pdV, where equality only in reversible process. Does it mean that reversible work can only be expansion or compression? $\endgroup$
    – Kelvin S
    Commented Jun 11, 2014 at 4:10
  • $\begingroup$ Is it true that reversible process can only occur in the form of infinitesimal expansion/compression? $\endgroup$
    – Kelvin S
    Commented Jun 11, 2014 at 4:17
  • $\begingroup$ I did not mean that at all but in your question you implied with the notation $\delta Q_v $ that you are interested in the simplest case when only volume work is done. In general, the work $\delta W$ is a linear 1st degree differential form of the system variables if the process is reversible and is larger than that if if it is irreversible: $\delta W \ge y_1dx_1+ y_2dx_2+, y_3dx_3+...$ where $y_k$ are the "forces" (pressure, tension, stress, electric field, magnetic field,..., and $dx_k$ are "displacements" (volume, dilation, strain, electric dipole moment, magnetic dipole moment). $\endgroup$
    – hyportnex
    Commented Jun 11, 2014 at 12:44
  • $\begingroup$ Thanks. Besides the work do we also need to consider the reversibility of δQp and δQv in order to determine whether dU=δQv and dH=δQp ? $\endgroup$
    – Kelvin S
    Commented Jun 12, 2014 at 1:13
  • $\begingroup$ In addition, in a process is it possible that the work is irreversible but the heat transfer is reversible? Or the work is reversible but the heat transfer is irreversible? $\endgroup$
    – Kelvin S
    Commented Jun 12, 2014 at 1:15

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