Thermodynamics of Irreversible process

I'm a high school student with introductory knowledge about thermodynamics. I'm confused regarding irreversible processes. I was told that in irreversible processes we cannot define intermediate states of the undergoing process. For example if the initial state is $(P_1, V_1, T_1)$ and the final is $(P_2, V_2, T_2)$, I can't say what's the pressure when it's going through that irreversible expansion. Then how can we plot this curve for irreversible process?

EDIT :- The graph represents an isothermal expansion processes the first being irreversible and the second one being reversible. In my opinion I just can't plot the irreversible curve because I don't know the internal pressure of the gas at the instants it's undergoing the process.What I interpret out of the first graph ( if I'm not given the description that it's an irreversible isothermal process) is that the pressure is slowly decreased from P1 to P2 and then the volume is expanded slowly from V1 to V2 (by slow I mean slow enough to reach the reversible limit). What I mean to say is that I think that only equilibrium states can be plotted on the graph. Thanks to all for extending a helping hand.

• The graph in one of the many graphs that take the same system from the initial state to the final state. The presented graph can tell you the desired detail but you cannot say anything about a thermodynamic property for a change in general. – Mitchell Sep 4 '17 at 15:49

It is not true that we cannot define a path on a thermodynamic state space for an irreversible process. As long as the process is quasi-static it will be a succession of equilibrium states and it will have a definite path. On the other hand it can be quasi-static and irreversible (Is there a quasistatic process that is not reversible?). For example, consider an idealized thermal engine working between two sources and following Diesel or Otto cycle. Although those cycles are well defined they are irreversible since there are heat being exchanged through finite temperature difference (therefore they have less efficiency than Carnot cycle).

• If the process is irreversible, it will not consist of a succession of equilibrium states, and the path will not be determinate. It is possible for a process to be quasi-static yet irreversible if the process involves finite temperature gradients within the system. This will occur if the reservoir temperature is very different from the initial temperature of the system. The OPs question seemed to focus on the issue of "how can an irreversible process be isobaric if the pressures of the system in the initial and final equilibrium states are different?" This answer doesn't address that. – Chet Miller Sep 4 '17 at 22:37
• @ChesterMiller It seems that your first and second sentences are inconsistent with each other since a quasi-static process has a determinate path. Moreover, my answer is consistent with your second and third sentences. To me it is clear that OP has some misunderstanding about reversibility and that is what I tried to address. If its premisse (an irreversible process cannot have a definite path) is false, then there is no problem with his first diagram. On the other hand that first diagram may intend to show a discontinuity at $(P_1,V_1)$. It lacks some of context to directly address. – Diracology Sep 4 '17 at 23:49
• I guess our interpretations of what the OP was asking are different. – Chet Miller Sep 5 '17 at 0:22
• Thanks for your answer... But I really don't know the difference between quasi-static and irreversible processes.. I've never come across this term. – user150098 Sep 5 '17 at 13:24
• @gowreeshmago To understand what quasi-static and irreversible processes are you please read this answer. It is quite clear and straightforward. – Diracology Sep 5 '17 at 16:56

I think what you are really asking is that, if the initial gas pressure is $P_1$ and the final gas pressure is $P_2$ in an "irreversible isobaric" process, how can we call the process "isobaric?" Well, if, in an irreversible process, we suddenly change the externally applied pressure on the (massless, frictionless) piston from $P_1$ to $P_2$ and then hold it at that value until the system re-equilibrates, we define this as an irreversible isobaric process. Under such circumstances, the work done by the gas on its surroundings is $P_2(V_2-V_1)$. If we wanted to illustrate this on a P-V diagram, what we would do would be to show a discontinuous change in pressure from $P_1$ to $P_2$ at the initial volume $V_1$, after which the pressure would be plotted as constant at $P_2$. That's what your first diagram implies.

So, in calculating the work, you use the externally imposed pressure acting on the boundary of the system and multiply by the volume change. Even if the pressure of the gas within the system is not uniform, at its boundary (such as a massless frictionless piston), the gas pressure matches the externally imposed pressure.

On a thermodynamic diagram you can plot only equilibrium states.

I was told that in irreversible processes we cannot define intermediate states of the undergoing process.

The meaning is that you cannot plot all the intermediate states, simply because not all the intermediate states are equilibrium states. It is possible that the system, in going from initial to final state, hops between intermediate equilibrium states. These intermediate equilibrium states can be plotted on the thermodynamic diagram. But a string of intermediate equilibrium states no matter how numerous does not necessarily result in a path, because a thermodynamic path is understood to be continuous curve on the thermodynamic diagram with no gaps in between. You obtain a reversible path only in the limit as the number of intermediate equilibrium states goes to infinity in such a way that you obtain a continuous curve on the thermodynamic diagram.

• Thanks for your answer. But if I can't plot non equilibrium states then how come I plot the aforementioned graph? What I interpret out of the first graph ( if I'm not given the description that it's an irreversible isothermal process) is that the pressure is slowly decreased from P1 to P2 and then the volume is expanded slowly from V1 to V2. What I mean to say is that I think that only equilibrium states can be plotted on the graph. – user150098 Sep 6 '17 at 10:14
• @gowreeshmago If you have plotted it then it is not irreversible. To say that you have plotted an irreversible process on a thermodynamic diagram is a contradiction. It is possible that part of the process was reversible, which you have then plotted. – Deep Sep 7 '17 at 4:50

Here is what Moran et al, Fundamentals of Engineering Thermodynamics (incidentally, a wonderful thermodynamics text in my opinion) has to say about this issue:

"There is no requirement that a system undergoing a process be in equilibrium during the process. Some or all of the intervening states may be non equilibrium states. For many such processes, we are limited to knowing the state before the process occurs and the state after the process is completed.

Typically, at a non equilibrium state intensive properties vary with position at a given time. Also, at a specified position intensive properties may vary with time, sometimes chaotically. In certain cases, spatial and temporal variations in properties such as temperature, pressure, and velocity can be measured or obtained by solving appropriate governing equations, which are generally differential equations."

I would add that a system that is not passing through a continuous sequence of equilibrium states during a process is experiencing an irreversible process.

protected by Qmechanic♦Sep 6 '17 at 10:51

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