Why is the final temperature of irreversible adiabatic processes higher than that of reversible adiabatic processes?

Suppose an irreversible adiabatic expansion process and a reversible adiabatic expansion process are starting from the same initial state, say, P1V1. Now, let both of these processes have equal pressure in their final state. My teacher said that the magnitude of the work done in the reversible process will be more than that done in the irreversible process. The reason behind this, my teacher said, was that the final temperature of irreversible process will always be higher than that of reversible process. But we were not taught why this happens. Is there a formula based proof (not graph based) that uses equations related to the first law to explain why this happens?

*I specifically say first law because we haven't been taught the other laws yet.

• Did your teacher teach you about the second law? Commented Dec 3, 2022 at 15:15
• @BobD not yet but we will be learning about it on monday Commented Dec 3, 2022 at 15:23
• Now that I have reread what I wrote I missed one main point although I alluded to it in the comment, namely, what your teacher told you is essentially equivalent to almost all formulation of the so-called "2nd law", and this is what you find in Pau-Chang Lu article doi.org/10.1119/1.13048 very well described. In other words, you have already been taught the 2nd law not necessarily in its most convenient form for all applications but one equivalent formulation of it! Commented Dec 3, 2022 at 19:07

2 Answers

It is better to think of it the other way around and understand/derive the "2nd law" from the observation that in an adiabatic process the work done is maximum if it is reversible. When I say "derive" the 2nd law I mean making equivalent statements to that for this is at the heart of it. Take that statement as being an axiom derived from observation and base the rest on that.

In general, you cannot avoid using concepts/statements that are observational and use them as basis for the rest of physics. You can even phrase the observational content of your question by stating that no adiabatic irreversible cycle exists, all adiabatic cycles are reversible. At this level "reversibility" is almost a basic undefined concept in the sense of a point or line in Euclidean geometry so that axioms/postulates "define" the concept.

• So basically you are saying that there is no reason and that it is just an observation? Commented Dec 3, 2022 at 13:57
• Kind of like gravity? Like we observe that it happens and form laws based on it? Commented Dec 3, 2022 at 14:00
• yes, it is a basic observation and accept it as the foundation of the theory. Of course, you can start from other similar or (almost) equivalent observations (such Kelvin-Planck or Clausius) and derive this as a consequence. See Chapter 4 in archive.org/details/thermodynamicsan0000call/page/n9/mode/2up but for your original question read this doi.org/10.1119/1.13048 Commented Dec 3, 2022 at 14:19
• For a less abstract view how the temperature or internal energy increase follows from irreversibility based on other observations here is Finck's internal (hidden) variables explanation babel.hathitrust.org/cgi/…. Commented Dec 3, 2022 at 14:22
• The beauty of the Sears-Kestin axiom (Pau-Chang Lu) is that everybody knows from experience that "friction" reduces the work that can be extracted from a system and the energy that is dissipated is warming it up. In that sense it is less abstract than Kelvin or Clausius or Caratheodory but it is essentially equivalent to them. Finck attempts to explain it why is that the case when there are internal variables not accessible from the outside. Commented Dec 3, 2022 at 14:29

Although at the time of your post your teacher hadn't yet taught you the second law, you can, based on the first law alone, show why the final temperature of the gas is higher for the irreversible process in the case of an ideal gas and the assumption that more work is done for the reversible process for the same final pressure.

From the first law, any process:

$$\Delta U=Q-W\tag{1}$$

Where $$W$$ is positive when work is done by the system (expansion) and negative if work is done on the system (compression).

For both the reversible adiabatic process and the irreversible adiabatic process there is no heat transfer, so $$Q=0$$ for both processes and we have

$$\Delta U=-W\tag{2}$$

For an ideal gas, any process, the change in internal energy depends only on temperature change, or, for one mole of gas

$$\Delta U=C_{v}\Delta T\tag{3}$$

Where $$C_{v}$$ is the molar specific heat at constant volume, assumed to be constant.

Equating the right side of equation (2) with the right side of equation (3)

$$\Delta T=\frac{-W}{C_v}\tag{4}$$

Since $$W$$ is less for the irreversible process at the same final pressure, the decrease in internal energy and thus decease in temperature is less, meaning the final temperature is higher for the same final pressure.

When you are taught the second law you will find out why the work done is less for the irreversible process.

Hope this helps.

• 'When you are taught the second law you will find out why the work done is less for the irreversible process.'--------------> Is this similar to what hyportnex said in his answer? Like is the second law based on this phenomenon? Or is the second law based on something else but is also explaining why the work done for reversible is less? Commented Dec 3, 2022 at 15:59
• @Obinna I think it is way too premature to go into the reason less work is done based on the second law until you are taught the second law. Commented Dec 3, 2022 at 16:07
• ok, then. I will revisit this question after I have been taught the second law :) Commented Dec 3, 2022 at 16:09
• Thank you for the help Commented Dec 3, 2022 at 16:10
• @Obinna OK. But can you now see why the final temperature of an ideal gas is greater for the irreversible process based only on the first law given (without proof) that less work is done for the irreversible process? Commented Dec 3, 2022 at 16:17