# Work done in Irreversible Isothermal Expansion

This is the graph for the Isothermal Expansion of an Ideal Gas. Sorry for the extremely basic doubt, but I just could not understand how the expansion was Isothermal if the graph was a straight line. The pressure is changed once suddenly and then remains constant.

Why don't we consider this as an Isobaric expansion instead of Isothermal if the graph is not even an isotherm? If a gas is expanding but the pressure has to be constant then why doesn't the temperature increase?

There is some confusion about terminology. The figure shows two processes. The curved path represents a reversible expansion in which there is pressure balance between system and surroundings throughout, so that the applied pressure can be described by the ideal gas law (or Boyle's law $$p=\frac{p_1V_1}{V}$$). This process is clearly isothermal, since $$pV=nRT$$.

The second process shown as two straight paths consists of a pressure drop and and an expansion against constant external pressure and is also isothermal (in thermodynamic lingo). The fact that the initial and final points are at equal T makes it isothermal. The temperature (and other properties except presumably composition) of the system during the process is on the other hand not well-defined. This process is irreversible.

• I disagree. Isothermal means constant-T, just as isobaric means constant-p. A process in which p_initial and p_final are the same, but in which p was varying all over the place during the process, would not be isobaric. Same thing with isothermal. What you are describing as "thermodynamic lingo" is not -- it runs counter to standard usage in physical chemistry textbooks. Jul 17, 2020 at 9:27
• For instance, from Castellan 3rd Edition: "In the isothermal cycle described above, the net work produced in the irreversible cycle was negative, that is, net work was destroyed. This is a fundamental characteristic of every irreversible and therefore every real isothermal cyclic transformation. If any system is kept at a constant temperature and subjected to a cyclic transformation by irreversible processes (real processes), a net amount of work is destroyed in the sur­ roundings. " Jul 17, 2020 at 9:27
• And from Engel and Reid, 3rd Edition: "The simplest processes are those in which one of P, V, or T remains constant. A constant temperature process is referred to as isothermal, and the corresponding terms for constants P and V are isobaric and isochoric, respectively." Jul 17, 2020 at 9:28
• @theorist What does the irreversible step in Castellan's isothermal cycle look like? Jul 17, 2020 at 11:12
• @theorist I realize it's not an ideal source, but the Wikipedia says: "Once obtained, these formulas can be applied to an irreversible process, such as the free expansion of an ideal gas. Such an expansion is also isothermal and may have the same initial and final states as in the reversible expansion." The temperature of the system is ill-defined during a free expansion. Jul 17, 2020 at 11:17

In the parlance of thermodynamics, an irreversible isothermal process is defined as one in which the system and constant temperature reservoir start out at the same temperature and the system is held in contact with the constant temperature reservoir throughout the process, so that, in the end, they re-equilibrate at the same temperature.

In an irreversible process, one cannot use the ideal gas law to describe the gas behavior, because the ideal gas law is valid only at thermodynamic equilibrium. A reversible process is characterized by a continuous sequence of thermodynamic equilibrium states, so the ideal gas law can be used for such a case.

In an irreversible process, in addition to the pressure being a function of the gas volume, it is also a function of the rate of change of volume. Fluid Mechanics tells us that this is caused by viscous stresses which contribute to the force per unit area exerted by the gas on the piston. In an irreversible expansion, these stresses lower the pressure that the gas exerts on the piston, so that, for a massless frictionless piston, the lower pressure exactly matches the externally applied reduced constant pressure. To provide the required match, the gas has to be expanding rapidly and at the right rate so that the viscous stresses are just right.

During the irreversible expansion, the gas temperature is not spatially uniform. The part of the gas in contact with the constant temperature reservoir is, of course, at the reservoir temperature, but the interior of the gas becomes cooler. However, by the end of the expansion, when thermodynamic equilibrium has been re-established, the gas temperature is again uniform and equal to the reservoir temperature.

This is all a one-stage process, and, by controlling the pressure on the piston so that it suddenly drops initially and then stays constant throughout the deformation, this can be considered a constant-pressure expansion.

In short, the answer given by @theorist isn't even close to being correct. @Buck Thorn's answer is much closer to the target.

I agree: It's not an isothermal process. It merely starts and ends at the same temperature.

It consists of two steps: an isochoric (constant-V) process, in which the pressure is decreased by cooling the gas at constant V, followed by an isobaric process, in which the volume is increased by heating the gas at constant p, returning it to its original temperature (but not its original pressure or volume).

As to what Chet Miller writes:

In short, the answer given by @theorist isn't even close to being correct.

"Isn't even close to being correct" is actually a good description of his own comment.

I stand by what I said. The OP posted a graph that showed how the pressure and volume of an ideal gas varied during a process. I can only answer based on what was posted. Logically, for the first leg, if p is going down at constant V, the system must be cooling. And for the second leg, if V is going up at constant p, the system must be heating. [We're assuming the system is closed.] It is easy to accomplish this: Simply put the system in contact with a lower-T bath while keeping V constant, and then put the system in contact with a bath at the original T, keeping p constant.

For instance, we could start at 300 K, 1 L, and 2 atm. Then we could gradually cool at constant V to 150 K, 1 L, 1 atm. Then we could gradually heat at constant p to 300 K, 2 L, 1 atm.

As for his niggle that T and p can never be defined during an irreversible process, that's an unfortunate and absolutist position. In practice, it certainly is possible to define and measure T and p during irreversible processes, particularly if we make the change slow enough that the system can relax close enough to an equilibrium state such that our measuring devices can't tell the difference—or that the difference is less than the precision with which we care to report our measurements. [This is not to be confused with a true reversible process, which is an idealization that can't exist is the real world.]

And the diagram, by showing the pressure of the system throughout the process, is telling us just that: That the system is sufficiently relaxing during the process to have a measurable p. I.e., yes, normally the default assumption during an irreversible process is that the intensive variables are undefined (this is sometimes indicated using a dashed line). But here the diagram, by using an unbroken solid line to describe p throughout the process, is telling us p is measurable.

Consider the converse: If we hold to such an absolutist view, that systems must be at equilibrium for us to be able to describe their temperature and pressure, then we can never measure or discuss the temperature or pressure of any real-world system, since no real-world system is ever truly at equilibrium. So essentially all the temperature and pressure measurements in all the world's scientific literature are invalid.

Note that this would also prevent us from talking about the temperature or pressure change in a Joule-Thomson expansion, since such an expansion is a steady-state process so it is, by design, never at equilibrium.

I see from Chet's profile that he's a retired chemical engineer. To be logically consistent with the view he presents here, anytime his colleagues or bosses asked him about the temperature or pressure of a system, he'd always have to say: "No real-world system is ever at equilibrium, so they have no definable temperature or pressure." Do you think that's what he always said?

The process Chet is ascribing to the question is different from what's pictured. He's taking the diagram to represent the V of the system, but the p of the surroundings. Based on that assumption, his description is correct. But, while that may be the intent of the question, there's nothing in the presented diagram to indicate that's the case. He is, as the lawyers say, "assuming facts not in evidence."

As to his other comment:

Really???? So, in stage 1 you start out with two entries, the gas and the reservoir, at the same temperaturer. Then, spontaneously, heat flows from the gas (at constant volume) to the reservoir so that, at the end of stage 1, they are at different temperatures (with the gas colder). Do you really think that this is consistent with the 2nd law of thermodynamics?

I applaud his restraint in limiting himself to four question marks. Had he used five, I would be concerned.

He's attributing to me a bizzare process that bears no relationship to anything I said, nor to what I had in mind, nor to what's pictured by the diagram. I have no idea how or why he came up with this. It's so strange it took me a while to figure out what he was trying to describe, and almost seems like a straw man. To adopt Chet's parlance: Really??? [I'm not as well-trained as Chet in hyperbole, so I can only go to three question marks, max.]

• Comments are not for extended discussion; this conversation has been moved to chat. Jul 18, 2020 at 19:32
• Here is a detailed analysis of irreversible isothermal expansion and compression in a thread from PhysicsForums.com: physicsforums.com/threads/…. Please feel free to join and weight in. Jul 18, 2020 at 20:33
• I stand by what I said about an irreversible expansion. For a gradual reversible expansion, we are totally in agreement. But, to me, the diagram indicates that the expansion takes place irreversibly. The implication of the diagram is that the (external) pressure is suddenly reduced at time zero, and the gas is allowed to expand at this constant externally applied pressure to the final volume (in contact with a constant temperature reservoir at the original gas temperature). To me, no other interpretation is possible. Sep 22, 2021 at 14:37
• If you disagree with my interpretation of the process diagram, then I guess we will just have to agree to disagree. Sep 22, 2021 at 14:40
• In the process you describe, it you put the gas at 300 K in contact with a constant temperature reservoir at 150 K and cool it at constant volume, it will not cool gradually and and it will not cool homogeneously. Then if you put the gas at 150 K in contact with a constant temperature reservoir at 300 K and heat it at constant pressure, it won't heat gradually and it won't heat homogeneously. However, the actual process you described is irreversible, and an alternate irreversible path to the one I described. For either path, the integral of dq/T will be less than the entropy change. Sep 22, 2021 at 16:54

As a followup to the answer I gave previously and to my discussions with @Theorist, I have analyzed in detail the alternate irreversible process paths (1) specified by Theorist in his answer and (2) specified in the problem statement (and referred to in my answer as well as that of @Buck Thorn). To provide a common basis for comparison, I have derived the integral of $$dQ/T_R$$ for each of the two alternate irreversible process paths, where $$T_R$$ is the reservoir temperature exchanging heat with the gas (this is the temperature that should be used in conjunction with the Clausius inequality, as pointed out in Thermodynamics by Fermi).

For the process path specified by Theorist, it is found that $$\int{\frac{dQ}{T_R}}=nR\left(1-\frac{V_1}{V_2}\right)-nC_v\left(\frac{V_2}{V_1}+\frac{V_1}{V_2}-2\right)$$ For the process path specified in the problem statement (and by Buck Thorn and myself), it is found that $$\int{\frac{dQ}{T_R}}=nR\left(1-\frac{V_1}{V_2}\right)$$ For any reversible path, the integral is equal to the entropy change, which, of course, is the same for all three paths:$$\Delta S=nR\ln{(V_2/V_1)}$$For an irreversible path, the amount of entropy generated in the process is given by $$\sigma$$: $$\sigma=\Delta S-\int{\frac{dQ}{T_R}}$$So, for the path specified by Theorist, $$\sigma_{Theorist}=nR\ln{(V_2/V_1)}-nR\left(1-\frac{V_1}{V_2}\right)+nC_v\left(\frac{V_2}{V_1}+\frac{V_1}{V_2}-2\right)$$and, for the path specified in the problem statement, $$\sigma_{problem\ statement}=nR\ln{(V_2/V_1)}-nR\left(1-\frac{V_1}{V_2}\right)$$Note that the first two terms on the RHS of these entropy generated expressions are identical; however, the third term in expression for Theorist's path is positive for all values of the volume ratio. This indicates that, in this sense, Theorist's path is more irreversible since it involves more entropy generation. This is primarily the result of the first step in his process which involves irreversible cooling of the gas at constant volume by use of a lower temperature reservoir. This is not a criticism of the irreversible path he chose to specify (although it is not consistent with the problem statement). It is just a feature of that particular irreversible path.