# Adiabatic paths

Can two adiabatic process paths intersect on a plot? This question is for 2 cases- 1.) Can two irreversible adiabatic curves intersect. 2.) Can a reversible & an irreversible adiabatic curve intersect?

There are some answers for this question on Quora, but they mostly address the intersection of two reversible adiabatic processes.

Also if the answer to either of the above 2 questions is 'yes', then I would like to know that if we do the proof given in the first answer of Quora link below with a reversible & an irreversible adiabatic process (or even two irreversible one) instead of two reversible adiabatic processes, it appears to be in violation of Kelvin Planck's statement. That is if we take an initial state and expand the system reversible adiabatically to a temperature 'T', now we take same system with same initial state but this time expand it irreversible adiabatically to same temperature 'T' (but with different P & V ofcourse) & join these two final states by an isothermal curve, we get a cycle with 3 processes (diagram in Quora link below, first answer). Now considering this cycle, we get a net positive work since it's area on a P-V diagram is positive but the system interact with surrounding only in 1 of the process, i.e the system recieves/rejects heat in only one of the processes, i.e the isothermal one. So it appears that we have a device which produces work continuously in a cycle just by interacting with a single thermal reservoir thus violating Kelvin Planck's statement of the second law of thermodynamics. how is that possible?

Also I want a physical interpretation of the processes that if they don't intersect then why don't and what it would be physically for a system.

Here are answers on Quora- https://www.quora.com/Why-do-two-adiabatic-curves-never-cut-each-other

• I would advice to try again on physics stackexchange, this question clearly belongs there. Oh, and for the rest of the people, if you are down-voting a question from a new contributor, please give a reason. Commented Sep 6, 2018 at 19:11
• @user190081 questions get downvoted based on the quality or clarity of the question, not the rep of the user... Commented Sep 6, 2018 at 20:03
• @SolarMike I'm not arguing that, I'm asking for them to provide a the reason for the downvote so the user can make better questions in the future. Getting downvotes for no apparent reason may be frustrating for new users. Commented Sep 6, 2018 at 21:10
• @user190081 it's frustrating for ALL users, but it's the system that we have. Perhaps it should only be possible to downvote with constructive criticism but we can all dream, until then we have to use and live with the system that we have. Commented Sep 6, 2018 at 21:13
• I recommend the question is appropriate for engineering. Both mechanical and chemical engineers work with adiabatic processes. Physicists tend to get lost in theory, what they appreciate about thermodynamics in practice is poor by comparison, and the poster wants a "physical" interpretation (i.e. fundamentals grounded to applications). As for the "it is the system we have" ... The "rule" absolutely does not disallow comments to why downvotes are given. A note as to the reason for a down vote would have taken less time, and a response akin to "duly noted" would have been less impetuous. Commented Sep 7, 2018 at 13:18

## 2 Answers

Intuition should tell you that it is possible because irreversible paths can result in an infinite array of end points sharing the same volume (or pressure or temperature).

Consider this: You have 2 identical systems. For the first system, reversibly compress the system to $\frac{1}{2}$ its volume. For the second system, irreversibly compress the system to $\frac{1}{2}$ its volume. Intuitively it should made sense that you have imparted more energy into the system in the irreversible case, and thus, it should be at a greater temperature and/or pressure than the second system.

Mathematically, we can show this is true as well. I will use $$PV_m=RT$$ to represent the ideal gas system, where $V_m$ is the molar volume. $c_v = 3/2R$ for a monatomic gas (a simply explanation of this is Degrees of Freedom of a monatomic gas; see equipartition theorem for more information). Also, for simplicity, I will assume $n=1$, so I will use $V$ in place of $V_m$ for the rest of the derivation.

Let us consider the following system: You 2 identical systems with $1$ $mol$ of a monatomic ideal gas with volume $1 m^3$ and pressure of $1$ $Pa$. It follows that the starting temperature of this system is $\frac{1}{R}K$. I will consider the compression to $\frac{1}{2}m^3$.

Reversible: Let's begin with the more restricted reversible case. Since $q=0$ we have $\Delta U=w$. We know pressure volume work is given by $w=-\int_{1}^{1/2} P dV$, but $P = \frac{RT}{V}$ and we can flip the integral bounds by multiplying by $-1$, giving $w=\int_{1/2}^{1} \frac{RT}{V} dV$. We also know (can easily derive) $dU = c_v dT$ for an ideal gas. This implies $\Delta U = \int_{T_i}^{T_f} c_v dT$. We then have $\int_{T_i}^{T_f} c_v dT = \int_{1/2}^{1} \frac{RT}{V} dV$. We can then "carefully" rearrange the equation to $\int_{T_i}^{T_f} \frac{c_v}{T} dT = \int_{1/2}^{1} \frac{R}{V} dV$, which gives $c_v \ln(\frac{T_f}{T_i}) = R \ln(\frac{1}{\frac{1}{2}})$. (I say "carefully" because such a rearrangement of dividing within integrals isn't always allowed, though I've never come across a case in a science where it is not allowed to be done. The theorem regarding it's use, however, is not trivial at all. To finish the problem up, we rearrange to obtain $\ln (\frac{T_f}{T_i}) = \ln (2^{\frac{R}{c_v}})$, raise both sides to the power of $e$, and solve $T_f = T_i \cdot 2^{\frac{R}{c_v}} = \frac{1}{R} \cdot 2^{\frac{2}{3}}$. Thus $T_f \approx 0.191K$. From the ideal gas law, we get $P \approx 3.175$ $Pa$.

Irreversible: For the irreversible case, we will undergo a 1-step compression by applying a constant pressure on the system. Since the system is still adiabatic, $q=0$ and $\Delta U=w$ still hold true. Expanding $\Delta U$ in the same way as above, we get $\int_{T_i}^{T_f} c_v dT = -\int_{1}^{1/2} P dV$; however since $P$ is now constant, we can easily simplify the equation down to $c_v \cdot (T_f-T_i) = P \cdot (1-\frac{1}{2})$. We don't know the $T_f$ of this path, so we replace $T_f$ with $\frac{PV_f}{R}$ by the ideal gas law. After distributing, we see that $P \cdot \frac{1}{2} = \frac{Pc_vV_f}{R} - c_vT_i$. We can then factor $P$ out and rearrange to get $P = \frac{c_vT_i}{\frac{c_vV_f}{R}-\frac{1}{2}}$. Plugging in numbers, we see that $P = \frac{\frac{3}{2}}{\frac{3}{2} \cdot \frac{1}{2} - \frac{1}{2}} = 6$ $Pa$. It follows that $T_f = 3 \cdot Ti \approx 0.361K$

As you can see, although these two systems started at the same place, they clearly diverged from each other, ending at a different pressure and temperature for the same volume. You can show this is true for expansion in the same manner. It follows naturally then, that you can have two systems cross in the middle of a path. Please check my math, as I may have made an error trying to format my thoughts into mathjax. Regardless, only reversible paths (of the same process) cannot intersect with each other. Irreversible paths can intersect with other irreversible paths (you can show this in a similar manner as with reversible paths) or with reversible paths. I hope this clears up the "impossibility" of system paths crossing each other.

Edit: The Kelvin-Planck statement is essentially a consequence (possibly an equivalent statement) of the 2$^{nd}$ Law of Thermo. It essentially says that the energy you put into a system cannot (as a consequence of probability) be entirely converted to [macroscopic] work. In other words, if you heat a system $100J$, the system cannot perform $100J$ of work. Why? Because this would mean that $\Delta S_{univ}$ does not increase. Why? In a cycle, the system will return to its original state ($\Delta S = 0$), and the entropy lost by the heat reservoir is $\frac{q}{T_{res}}$. This means the 2$^{nd}$ external body (the one you want to do work on) must have $\Delta S \geq \frac{q}{T_res}$. If all the heat gained from the reservoir were converted to energy, however, $\Delta S \ngeq \frac{q}{T_res}$, as such a system would have q = 0. Thus, a functioning heat engine cycle would need a second heat sink (a lower temperature heat reservoir) to restore the cycle to its initial state. (Think about why a heat “sink” of the same temperature would cause the system to perform no work.)

What is the significance of the Kelvin-Planck statement?

It tells us that if the energy imparted into a cycle is derived from heat (such as from an external heat source reservoir), a second heat sink/reservoir is needed to restore the initial state of the cycle. What this does not say, however, is that you cannot have only one heat reservoir in a cycle.

It is possible for a cycle to have only one heat reservoir in its path – it just must be a heat sink (absorbs heat from the cycle). This heat lost by the system to the heat sink must then be provided to the system from an isentropic* source (such as an adiabatic expansion/compression). In other words, the logic argued on Quora is flawed, or incomplete at best. (I find quora and Chegg are flawed all the time… Take everything you read/hear/etc. with a grain of salt. Think about whether or not what someone says makes sense with what you think.) In fact, if you complete the scenario I presented as a system such that you have 3 steps: 1. Irreversible adiabatic compression ($1m^3$ to $\frac{1}{2}m^3$; 2. Constant volume cooling ($.361K$ to $.191K$); 3. Reversible adiabatic expansion ($\frac{1}{2}m^3$ to $1m^3$), you can see that it indeed does not violate the 2$^{nd}$ Law of Thermo (the Kelvin-Planck statement). In fact, for one full cycle, $\Delta S_{univ} \approx 0+7.938+3.16 = 11.098$ $J/K$, a clearly positive value (assuming I didn't mess up my math).

Why is the Kelvin-Planck statement often stated and misinterpreted as “a cycle can’t must have more than one heat reservoir”? This is because there’s little reason in applying work to a system to extract less work from the system (while this might be needed occasionally, there are better ways of going about this than a heat engine). Say you have a piston and you put in 200kJ of energy to compress the gas, the cylinder is then used to spin a turbine with 150kJ of energy. Why not just turn the turbine directly? (Obviously, real world scenarios are not this simple, but I think this gets the point across.)

*I say isentropic, but I really mean any process that would decrease the entropy its source less than the entropy gained by the the sink. It is safe to ignore the system since you are considering a cycle. Note that while an irreversible adiabatic compression is not an isentropic process, the entropy of the source does not change.

• Intuitively it does make sense that a reversible & an irreversible Adiabatic system starting at same state will end up differently. So they can intersect. But if you look at the proof in given Quora link, it appears as though the condition of a process being an adiabatic one is enough for two processes to not intersect. Because we can have an initial state & two different, 1 reversible & other irreversible adiabatic process that end up at same temperature (by doing different work), & so it appears that Kelvin Planck's statement is indeed violated by this cycle ?? Commented Sep 8, 2018 at 17:33
• You are adding comments to answers that subsequently change the bounds of your starting question. Either amend your question appropriately to document better what you really want to be answered or accept that answers given so far have addressed the core aspect of your (otherwise somewhat broad and poorly developed) initial question. Commented Sep 8, 2018 at 18:09
• Which answer are you referring to on the quora link? None of them provide a "proof" of why such an intersection cannot occur between processes that are merely adiabatic? Also, what do you mean by "end up at same temperature (by doing different work)"? Commented Sep 8, 2018 at 19:13
• I've added information based on the comments you've made. Irreversible compression imparts more energy onto the system, increasing the internal energy, which will manifest, at least partially, as heat (completely as heat for an ideal gas). Commented Sep 8, 2018 at 21:06
• Thank you @SmallFish and JeffreyJWeimer for being patient & helping me out. :-) Commented Sep 9, 2018 at 3:40

A simple approach considers only the first law for a system.

$$dU = \delta q + \delta w$$

Internal energy change $$dU$$ is caused by heat flow $$\delta q$$ and work flow $$\delta w$$ across a boundary between the system and surroundings. Heat leaving the system (an exothermic process) decreases internal energy. Work done by the system leaves the system (and is negative), thereby also decreasing the internal energy. Recall that $$U$$ is a state function. For a closed system, it depends only on $$T, V$$.

A reversible process allows that any step that increases $$dU$$ can be identically and exactly reversed to decrease $$dU$$ just by changing the signs of $$\delta q$$ and $$\delta w$$. So, when heat leaves and work enters to increase $$dU$$, the reversible step is to have exactly the same heat enter as left and exactly the same work leave as entered. These heat and work flows are always across a boundary between the system and surroundings.

An irreversible process has an additional heat flow $$\delta q_{irr}$$ that always leaves the system. That heat flow can be due to friction or internal interactions in the material in the system. That heat flow is never returned during the reverse path. It also does NOT flow (leave or enter) through the boundary between the system and surroundings. It is either directly absorbed in the system or surroundings (as a consequence of having internal interactions of the fundamental particles -- i.e. as a consequence of using a "real" substance rather than an ideal gas) or it is absorbed at the boundary (i.e. as friction). Because we loose f $$q_{irr}$$, to return to the same internal energy $$U$$ as the starting point after taking an irreversible step, we must supply more work during the irreversible path than during the reversible path.

An adiabatic process has no reversible heat flow. A reversible adiabatic process has no heat flow. An irreversible has irreversible heat flow $$\delta q_{irr}$$.

The work done by the system during a compression or expansion process is $$\delta w = -p_{ext} dV$$. We can allow that $$p_{int} = p_{ext}$$ during the process regardless that it is reversible or irreversible (reversible processes must have $$p_{int} = p_{ext}$$ by definition).

On a graph of $$p$$ versus $$V$$, work is the area under the curve. During a reversible, adiabatic process, that area will be $$dU$$ exactly. During an irreversible, adiabatic process, that area will be $$dU \pm \delta q_{irr}$$. When the two curves are drawn on a graph, they can intersect either at one common starting point or at one common ending point.

In summary, the total work done during a reversible, adiabatic process is never the same as the total work done during an irreversible, adiabatic process that either starts or ends at the same set of $$T, V$$ conditions (i.e. starts or ends at the same $$U$$). Therefore, adiabatic curves for the two different processes, reversible and irreversible, will diverge from a common starting point or intersect at a common end point.

Any concerns about whether this violates the Kelvin-Planck form of the second law cannot be addressed here. This discussion is about the intersection of individual paths. The Kelvin-Planck statement is for cycles.

• Same question here, please comment on the proof given in the link while taking 1 adiabatic process as reversible and other as irreversible in given cycle as I mentioned in comment above. Also if the system is Adiabatic then where will the heat generated by irreversibilities inside the system go? Commented Sep 8, 2018 at 17:38
• So, if the heat is absorbed by the system then how will it manifest itself(will it increase internal energy? Or some other forms of energy related to system?)? Also I know we cannot have 2 different reversible path for same type of process & initial state. What I am asking is that if we do the proof given in link with a reversible & an irreversible adiabatic instead of 2 reversible, even then it seems to work out the same way it did with 2 reversible & hence seem to be violating the Kelvin Planck's statement. How? Commented Sep 8, 2018 at 18:12
• The Kelvin-Planck statement is for a CYCLIC process, not for a single path. Commented Apr 14, 2019 at 13:19