I am slightly confused about graphs of irreversible processes. Firstly, they are difficult to find and whenever I find them the form is something similar to:-

enter image description here

I have read about the explanation given behind these graphs and a brief summary of what I understand is:-

  1. In both a reversible and an irreversible process, we plot Pext and not Pgas. In a reversible process, Pgas happens to be equal to Pext and so we can consider the pressure to be either Pext or Pgas but in general, Pext is what we plot.

  2. In the reversible graph Pext=Pgas=(nRT/V) and we can use this expression to plot the graphs of various processes(in the above image it is an isothermal process).

  3. In the irreversible curve, Pext is not equal to Pgas and it rises(or drops) instantaneously(which Pgas does not) and then the process is done under constant pressure.

Now, I understand the above points but my question is that if Pext=constant in an irreversible process(after the instantaneous drop/rise) then shouldn't all irreversible processes be isobaric? Suppose we have an irreversible isochoric/isothermal/adiabatic process(which is not isobaric simultaneously), how do plot its graph? How does Pext vary in these processes?

Please note: Providing graphs of irreversible isothermal, isochoric and adiabatic processes will be helpful.

  • $\begingroup$ "Why aren't all irreversible processes isobaric?" Why would you think that all irreversible processes are isobaric? I intend to to post an answer, but first I need to know why you would think that. $\endgroup$
    – Bob D
    Commented Jun 28, 2022 at 21:29
  • $\begingroup$ @BobD Because Pext is constant in all irreversible processes(afaik) which is the condition to be an isobaric process. $\endgroup$
    – Boson
    Commented Jun 29, 2022 at 2:25
  • 1
    $\begingroup$ @Boson This is incorrect. That a process is irreversible says nothing about $P_\text{ext}$, it only tells you than the created entropy term is strictly positive. There's no relation between the two. $\endgroup$
    – Miyase
    Commented Jun 29, 2022 at 6:55
  • $\begingroup$ @Miyase Ok. so is the above graph only for isobaric irreversible process? $\endgroup$
    – Boson
    Commented Jun 29, 2022 at 7:00
  • 1
    $\begingroup$ I know this sort of graph, but it's technically wrong unless there's a quasi-static assumption. Without this assumption, pressure $P$ is undefined during the process, so the graph is meaningless. With this assumption, $P$ is defined. Almost automatically, you can add a mechanic equilibrium assumption, so $P=P_\text{ext}$. If your graph is in this case, then yes $P$ is constant, so $P_\text{ext}$ is constant too. But it's completely unrelated to the fact that it's irreversible. $\endgroup$
    – Miyase
    Commented Jun 29, 2022 at 7:19

1 Answer 1


If it were not for the word "irreversible" on your graph, it would technically not be possible to determine whether the constant volume and constant pressure paths are reversible or irreversible based solely on the graph, although it is generally assumed that the processes are quasi-static (and friction free) and therefore reversible, unless told otherwise.

For example, the irreversible path can theoretically consist of any of the following combinations of processes:

  1. A reversible isochoric heat addition followed by an irreversible isobaric compression.

  2. An irreversible isochoric heat addition followed by a reversible isobaric compression.

  3. An irreversible isochoric heat addition followed by an irreversible isobaric compression.

  4. An irreversible sudden external pressure increase followed by an irreversible isobaric compression.

Note that the external pressure has no bearing on the irreversible isochoric heat addition processes.

Please note: Providing graphs of irreversible isothermal, isochoric and adiabatic processes will be helpful.

Irreversible Isothermal Path:

Presumably, the dotted line curve is intended to represent a reversible isothermal process between states 1 and 2 governed (in the case of an ideal gas) by the equation $PV$ = constant. The two equilibrium states are therefore related by $P_{2}V_{2}=P_{1}V_{1}$.

An irreversible isothermal path is any alternative (to $PV$ = constant) path between equilibrium states 1 and 2 that is carried out in a constant temperature environment. Thus the term "isothermal" in this case refers to the constant temperature at the boundary between the system and surroundings, and not the interior temperature of the gas (which is not constant due to temperature gradients). Irreversible path (4) above, provided we are told is carried out in a constant temperature environment of $T_1$, could constitute an irreversible isothermal compression path going from state 2 to state 1.

To visualize how the path occurs, consider a vertically oriented cylinder of gas fitted with a movable piston of area A. The initial temperature of the gas and the environment is $T_1$ and initial pressure of the gas and environment is $P_2$, say 1 atmosphere.

An object of mass $m$ is abruptly placed on top the the piston. This results in a sudden increase in external pressure of $mg/A$ so that $P_{1}= mg/A + 1$ atm. Because the external pressure change happens so quickly, there is no time for the piston to move or for any heat transfer to occur. Now, since the external is greater than the gas pressure, the gas begins to get compressed. This raises the temperature of the gas above the temperature of the environment ($T_1$) so that heat transfers irreversibly from the gas to the environment. This continues until the temperature and pressure of the gas re-equilibriate with the environment at state 1.

Irreversible Isochoric Path:

There is virtually no difference between the paths of a reversible and irreversible isochoric transfer of heat between the same two equilibrium states. You would need to be told that the isochoric heat transfer process is irreversible. The reason is as follows:

From the first law, $\Delta U=Q-W$. For an isochoric process, $W=0$ and thus $\Delta U=Q$. Since $U$ is a state function, $\Delta U$ between two equilibrium states is the same regardless of whether the process is reversible or irreversible.

For an ideal gas, any process, $\Delta U=C_{v}\Delta T$ where $C_v$ is the heat capacity of the gas a constant volume. So the heat transfer for a reversible and irreversible isochoric process between the same two states will be the same, $Q=C_{v}\Delta T$.

The difference between the reversible and irreversible isochoric processes is how the heat is transferred. For the reversible isochoric heat addition, the gas obtains heat from an infinite series of thermal reservoirs with temperatures ranging between the initial and final equilibrium temperature, each reservoir temperature being infinitesimally greater than the gas temperature. For the irreversible heat addition the gas obtains heat from single thermal reservoir of temperature equal to the final equilibrium temperature of the gas. The increase in entropy of the gas is the same for the reversible and irreversible path (entropy being a state function like internal energy). For the reversible process the decrease in entropy of the environment will equal the increase in entropy of the gas for a total entropy change of zero. For the irreversible process, the decrease in entropy of the environment will be less than the increase of the gas, for an overall entropy change (gas plus environment) of $\Delta S_{tot}\gt 0$.

Irreversible Adiabatic Path:

The adiabatic process is unique in that it is not possible to connect the same two equilibrium states with a reversible and irreversible adiabatic path. That is to say, any irreversible path connecting the same two equilibrium states as a reversible adiabatic path, cannot be adiabatic. For a discussion of this see Non Quasi-static expansion in a piston cylinder

Hope this helps.


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