# Entropy, (ir)reversibility & Otto

I've been trying to grasp it very thoroughly (excuse me for my english, I'm belgian) as I have a course of thermodynamics now in my first year of Physics. There is one last thing bothering me about entropy:

Take the Otto-cycle (two adiabats, two ischores/isometrics). We know the cycle is even theoretically irreversible, because its efficiency (which we can calculate) is smaller than Carnot's and all reversible processes have the same efficiency. But now I want to calculate the change of entropy in the universe after one cycle to SEE its irreversibility.

I assume in some way it HAS to be possible to calculate it, because you already know the change of entropy is greater than zero (cf. last paragraph). Take an isochore (I will continue with the isochore from cold to hot [2 -> 3 in bottom image]). My teacher said, after inquiry, that you can calculate the entropy change in the universe (of this isochoric process) by assuming the process is equivalent to placing an infinite reservoir at Th in contact with the gas in a solid container, initially at Tc, until it's warmed up to Th. In this case (not really important how) it can be shown the total entropy increase of the universe is Cv(ln(Th/Tc) + Tc/Th - 1) which can be proven to be greater than zero.

BUT this process is not quasi-static and since you've drawn the cycle in a PV-diagram (indicating at each moment its T,P,V is well-defined), it has to be quasi-static (and ANY non-quasistatic process is irreversible, so it's kind of a lame and non-interesting choice in this case). So a process that would fit would be one where you first place against the gas an infinite reservoir at the temperature of Tc + delta, then Tc + 2delta, ... Th. But in this case the entropy change of the system (which is the same as the earlier case is) is Cvln(Th/Tc) and that of the reservoir/exterior/... is Cvln(Tc/Th) and thus total entropy change zero...

But the Otto cyclus is irreversible

What is the deal?

Thank you tremendously for any help, I've been breaking my head on it,
Ruben

PS: if my explenation was hard to follow, feel free to ignore it, basically my question is: how can I calculate the change in the universal entropy (i.e.: show it is irreversible) after going through one isochoric process in an Otto-cycle

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All cycle that can be drawn on a p-V diagram ARE reversible, and so is Otto cycle.
Otto cycle is just a model;it is far from what really happens in the car engine.

All cycle that can be drawn on a p-V diagram ARE reversible, and so is Otto cycle.
Otto cycle is just a model;it is far from what really happens in the car engine.

That is NOT true. You can calculate the Otto's efficiency from the PV-diagram and see that it is less than the reversible carnot engine and it's a (proveable) theorem that all theoretically reversible processes have the same efficiency.

Andrew Mason
Homework Helper
BUT this process is not quasi-static and since you've drawn the cycle in a PV-diagram (indicating at each moment its T,P,V is well-defined), it has to be quasi-static (and ANY non-quasistatic process is irreversible, so it's kind of a lame and non-interesting choice in this case). So a process that would fit would be one where you first place against the gas an infinite reservoir at the temperature of Tc + delta, then Tc + 2delta, ... Th. But in this case the entropy change of the system (which is the same as the earlier case is) is Cvln(Th/Tc) and that of the reservoir/exterior/... is ln(Tc/Th) and thus total entropy change zero...

But the Otto cyclus is irreversible

What is the deal?
The entropy change of the gas is 0 in one complete cycle. This is because entropy is a state function and the gas returns to the initial state after one complete cycle. However, the change in entropy of the surroundings is greater than 0.

To calculate the change in entropy of the surroundings during the isochoric processes use:

$$\Delta S_h = \int_{P_i}^{P_f} dQ/T_h = \int_{P_i}^{P_f} nC_vdT/T_h = \int_{P_i}^{P_f} nC_vV_1dP/nRT_h = nC_vV_1/nRT_h \int_{P_i}^{P_f} dP = C_vV_1\Delta P/RT_h$$

Similarly:

$$\Delta S_c = C_vV_2\Delta P/RT_c$$

All cycle that can be drawn on a p-V diagram ARE reversible, and so is Otto cycle.
Otto cycle is just a model;it is far from what really happens in the car engine.
The Otto cycle is not reversible. It cannot be reversible because the system is not in equilibrium with the surroundings (ie the hot and cold reservoirs) while heat flows (ie during the isochoric processes). In the Carnot cycle, the system is quasi-static while heat flows (during the isothermal expansion/compression).

AM

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DrDu
I also don't see why the Otto cycle cannot be run (as allways that means almost) reversible. All that has to be done is to make sure that the heat is provided by a heat reservoir which has always the same temperature as the working medium at each instant of the isochorus process. As the temperature of the reservoir changes, there won't be a unique reversible Carnot process to compare with.

I must say I did the same calculation as I said in my original post but in it you assume the temperature of the environment (relative to the gas) is constant. But is this process quasi-static? As I tried to say in my first post: wouldn't you first have to attach a reservoir at Tc + delta against it and once the gas and that reservoir is in equilibrium, change the reservoir with one of temperature Tc + 2delta etc? Cause if indeed your way of doing it (immediately holding a reservoir of Th against the gas) is not quasi-static, that would mean you can't even draw it on a pv-diagram because then the variables aren't well defined (because there would be a temperature gradient in your gas and thus not an overall "T")

But if you do the calculations with the delta reservoirs, the otto cycle SEEMS to be reversible...

Thanks again for your time, I hope I'll finally get this issue resolved

The entropy change of the gas is 0 in one complete cycle. This is because entropy is a state function and the gas returns to the initial state after one complete cycle. However, the change in entropy of the surroundings is greater than 0.

To calculate the change in entropy of the surroundings during the isochoric processes use:

$$\Delta S_h = \int_{P_i}^{P_f} dQ/T_h = \int_{P_i}^{P_f} nC_vdt/T_h = \int_{P_i}^{P_f} nC_vV_1dP/nRT_h = nC_vV_1/nRT_h \int_{P_i}^{P_f} dP = C_vV_1\Delta P/RT_h$$

Similarly:

$$\Delta S_c = C_vV_2\Delta P/RT_c$$

The Otto cycle is not reversible. It cannot be reversible because the system is not in equilibrium with the surroundings (ie the hot and cold reservoirs) while heat flows (ie during the isochoric processes). In the Carnot cycle, the system is quasi-static while heat flows (during the isothermal expansion/compression).

AM

Th and Tc are NOT constants so you cannot extract them out of the integration sign!!!

Otto cycle is definetely reversible
you should notice that Carnot engine operates between TWO reservoirs whose temperature are held contant respectively,but Otto cycle operates between(or among?) many reservoirs

That is NOT true. You can calculate the Otto's efficiency from the PV-diagram and see that it is less than the reversible carnot engine and it's a (proveable) theorem that all theoretically reversible processes have the same efficiency.

The Carnot theorem just states that any reversible process between a heat reservoir T1 and a "cold" reservoir T2 has the same efficiency$$$\eta = 1 - \frac{{T_2 }}{{T_1 }}$$$
BUT you should notice that Carnot engine operates between TWO reservoirs whose temperature are held contant respectively,but Otto cycle operates between(or among?) many reservoirs.

If you are comparing Otto cycle to a standard Carnot cycle,tell me what the T1 and T2 in Otto cyle are

Hm, very interesting!

So are you two claiming the Otto cycle IS reversible? Our professor quite literally and resolutely claimed its irreversibility due to some unspecified universal entropy change in the isochores/isometrics of the PV-diagram. (this isn't meant as proof that you're wrong, it's a genuine question explaining my confusion)

Now you can claim there are two Otto cycles: one with the infinite reservoirs that make it reversible, and one with two reservoirs which is irreversible, but that last one can't even be drawn on a pv-diagram so it can't be applied here, right?

Hm, very interesting!

So are you two claiming the Otto cycle IS reversible? Our professor quite literally and resolutely claimed its irreversibility due to some unspecified universal entropy change in the isochores/isometrics of the PV-diagram. (this isn't meant as proof that you're wrong, it's a genuine question explaining my confusion)

Now you can claim there are two Otto cycles: one with the infinite reservoirs that make it reversible, and one with two reservoirs which is irreversible, but that last one can't even be drawn on a pv-diagram so it can't be applied here, right?

It's true, if the latter one can be taken as "Otto cycle"

Conclusion: my professor (who was talking about the drawable otto cycle, hence the quasi-static one, hence the infinite reservoir one, hence the reversible one) was wrong?

Discomforting...

Thanks a lot for the input, I needed it.

Any extra comments always welcome (anybody in disagreement perhaps?)

Andrew Mason
Homework Helper
Th and Tc are NOT constants so you cannot extract them out of the integration sign!!!
If they are not constant, then there is heat flowing into the hot reservoir. In that case you have to define another reservoir external to the hot reservoir from which heat flows to the first hot reservoir. Ultimately, the surroundings all reduce to a single hot and a single cold reservoir whose temperature does not change during the process.

Otto cycle is definetely reversible
you should notice that Carnot engine operates between TWO reservoirs whose temperature are held contant respectively,but Otto cycle operates between(or among?) many reservoirs
No. An engine cannot be reversible unless all processes are quasi-static.

AM

AM, if the otto cycle is not quasi-static, then how can you even draw it on a pv-diagram?

Andrew Mason
Homework Helper
AM, if the otto cycle is not quasi-static, then how can you even draw it on a pv-diagram?
You can draw a PV diagram for virtually any cycle at all. Pressure and volume are measurable quantities in non-quasi-static processes. What is the problem?

AM

You can draw a PV diagram for virtually any cycle at all. Pressure and volume are measurable quantities in non-quasi-static processes. What is the problem?

AM

You are absolutely wrong
In process that are not quasi-static,pressure and volume are NOT well-defined.

Andrew Mason
Homework Helper
You are absolutely wrong
In process that are not quasi-static,pressure and volume are NOT well-defined.
They may not be precisely defined for a very brief time, but one usually ignores that in drawing the PV diagram. For example, if you take a fuel/air mixture in a closed cylinder and ignite it, the heat flows into the cylinder contents and pressure builds up extremely quickly - a neglible time compared to the cycle period. In drawing the PV diagram one does not bother with that very brief time when P and T may not be uniform throughout the cylinder contents.

Here is a PV diagram of a http://en.wikipedia.org/wiki/File:DieselCycle_PV.svg" [Broken]. Are you saying that this is a reversible process?

AM

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"They may not be precisely defined for a very brief time, but one usually ignores that in drawing the PV diagram." -A.M.

Well then it becomes pretty unscientific in my eyes -- you can read off equilibrium states that never were close to reality. Then the irreversibility follows somewhat directly by how you describe it. You could just as well draw a PV-diagram and add a note "...but it must be done irreversibly".

I do get what you're saying and it's acceptable in a practical point of view, but if you want to deal with it in a purely theoretical way (to gain insight), it does seem to be more interesting to not let the PV diagram be an approximation and calculate things as if it were really done quasi-statically. That brings us back to the infinite reservoirs/two reservoirs-point. I must say I didn't quite get what you meant with (paraphrasing) "but that would need a second reservoir in combination with the first [...] it comes down to one huge reservoir of one Th". Could you enlighten me, as it does sound interesting?

Andrew Mason
Homework Helper
You could just as well draw a PV-diagram and add a note "...but it must be done irreversibly".

I must say I didn't quite get what you meant with (paraphrasing) "but that would need a second reservoir in combination with the first [...] it comes down to one huge reservoir of one Th". Could you enlighten me, as it does sound interesting?

The only way you can have a reversible flow of heat is to do it infinitessimally slowly where the heat source is at almost the same temperature (ie. an infinitessimal amount hotter) as the body receiving the heat. The Otto cycle does not contemplate heat flow of that kind. It contemplates heat flowing into or out of the gas at constant volume. The means that the temperature of the gas is constantly changing as heat flows.

You cannot have such an engine operating between an infinite capacity hot reservoir and an infinite capacity cold reservoir in a reversible fashion. These reservoirs represent the entire surroundings (ie. the rest of the universe). There is no flow of heat into or out of these reservoirs to/from anywhere else in the universe other than the engine.

You want to imagine a hot reservoir whose temperature is constantly increasing, and a cold reservoir whose temperature is decreasing, in step with the Otto cycle. But the problem is that this does not represent the surroundings. Such reservoirs could only exist if heat was flowing into or out of them from somewhere else. To determine the total change in entropy, you would have to add in the change in entropy of the other reservoirs.

AM

You want to imagine a hot reservoir whose temperature is constantly increasing, and a cold reservoir whose temperature is decreasing, in step with the Otto cycle. But the problem is that this does not represent the surroundings. Such reservoirs could only exist if heat was flowing into or out of them from somewhere else. To determine the total change in entropy, you would have to add in the change in entropy of the other reservoirs.

AM

Thank you for going deeper into your explanation. I get your point and it is a very fair one, in the case of a reservoir whose temperature changes. But I proposed an infinite set of infinite reservoirs, each infinitesimally hotter than the other, all the way from Tc to Th. They'd each be placed after each other in contact with the gas-container. Do you still argue, even in this case, it is irreversible? You might object it is not a two-reveservoir heat engine anymore, but I do not see why that should be a prerequisite.

They may not be precisely defined for a very brief time, but one usually ignores that in drawing the PV diagram. For example, if you take a fuel/air mixture in a closed cylinder and ignite it, the heat flows into the cylinder contents and pressure builds up extremely quickly - a neglible time compared to the cycle period. In drawing the PV diagram one does not bother with that very brief time when P and T may not be uniform throughout the cylinder contents.

Here is a PV diagram of a http://en.wikipedia.org/wiki/File:DieselCycle_PV.svg" [Broken]. Are you saying that this is a reversible process?

AM

Read the Dissel Cycle thoroughly and you will find that Dissel Cycle is just a model for Dissel engine ,not what really happens in Dissel engine,as Otto for inner combustion engine.The real Dissel engine is irreversible,but the Dissel Cycle is reversible.

And I don't want to debate with you on this question any longer.It is just like debating with someone who claims the earth is rectangular.

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Andrew Mason
Homework Helper
Thank you for going deeper into your explanation. I get your point and it is a very fair one, in the case of a reservoir whose temperature changes. But I proposed an infinite set of infinite reservoirs, each infinitesimally hotter than the other, all the way from Tc to Th. They'd each be placed after each other in contact with the gas-container. Do you still argue, even in this case, it is irreversible? You might object it is not a two-reveservoir heat engine anymore, but I do not see why that should be a prerequisite.
Try to work out the entropy change of the hot reservoirs and cold reservoirs after one complete cycle:

$$\Delta S_h = - \sum_{j=1}^\infty dQ_j/T_{j} = - \int_{T_{hi}}^{T_{hf}} dQ/T = - \int_{T_{hi}}^{T_{hf}} nC_vdT/T = - nC_v\ln{\frac{T_{hf}}{T_{hi}}$$

Determine the condition for the sum of the entropy changes of the hot and cold reservoirs to equal 0 (after one cycle, there is no entropy change in the gas). You will see that it cannot be made to equal 0.

AM

Andrew Mason
Homework Helper
Read the Dissel Cycle thoroughly and you will find that Dissel Cycle is just a model for Dissel engine ,not what really happens in Dissel engine,as Otto for inner combustion engine.The real Dissel engine is irreversible,but the Dissel Cycle is reversible.

And I don't want to debate with you on this question any longer.It is just like debating with someone who claims the earth is rectangular.
;You don't have to debate it with me. Just show me a thermodynamics text that claims the diesel cycle is reversible.

AM

Try to work out the entropy change of the hot reservoirs and cold reservoirs after one complete cycle:

$$\Delta S_h = - \sum_{j=1}^\infty dQ_j/T_{j} = - \int_{T_{hi}}^{T_{hf}} dQ/T = - \int_{T_{hi}}^{T_{hf}} nC_vdT/T = - nC_v\ln{\frac{T_{hf}}{T_{hi}}$$

Determine the condition for the sum of the entropy changes of the hot and cold reservoirs to equal 0 (after one cycle, there is no entropy change in the gas). You will see that it cannot be made to equal 0.

AM

That seems like the right formula. Okay let's try it then for one isometric (if the one isometric is reversible, so is the other one and so are the two adiabats and so the whole cycle).

$$\Delta S_{environment} = - nC_v\ln{\frac{T_{hf}}{T_{hi}}$$

And seeing as $$T_{hi}$$ is infinitesimally hotter than $$T_{c}$$, it is equal to this. And $$T_{hf} = T_{h}$$ (I am using $$T_{c}$$ and $$T_{h}$$ as (resp.) the beginning and end temperature of the gas for the isometric).

So:

$$\Delta S_{environment} = - nC_v\ln{\frac{T_{h}}{T_{c}}$$

Now we also know:

$$\Delta S_{gas} = \int_{T_{c}}^{T_{h}} dQ/T = nC_v\ln{\frac{T_{h}}{T_{c}}$$ (NB: if you argue that I cannot use dS = dQ/T for the gas (cause you might say it already implies it can be reversible, cause in an irreversible case dS > dQ/T): I also calculated the same entropy change (because it is a variable of state) of the gas by going from start to end through an adiabat and an isotherm and got the same value)

As you can see, for the isometric:

$$\Delta S_{universe} = \Delta S_{environment} + \Delta S_{system} = 0$$

Andrew Mason
Homework Helper
That seems like the right formula. Okay let's try it then for one isometric (if the one isometric is reversible, so is the other one and so are the two adiabats and so the whole cycle).
I think you mean isochoric - constant volume.

$$\Delta S_{environment} = - nC_v\ln{\frac{T_{hf}}{T_{hi}}$$

And seeing as $$T_{hi}$$ is infinitesimally hotter than $$T_{c}$$
How do you get that? This is actually NOT the case. The initial hot reservoir temperature is determined by the adiabatic compression from the final cold reservoir temperature. That relationship is given by the adiabatic condition:

$$TV^{\gamma-1} = K$$ so:

(1) $$T_{ci} = T_{hf}\left(\frac{V_h}{V_c}\right)^{\gamma-1}}$$

(2) $$T_{cf} = T_{hi}\left(\frac{V_h}{V_c}\right)^{\gamma-1}}$$

So: $$\Delta S_{environment} = - nC_v\ln{\frac{T_{h}}{T_{c}}$$
No. The change in entropy of the environment (surroundings) is the sum of the change in entropy of the hot reservoir(s) and the change in entropy of the cold reservoir(s):

$$\Delta S_{env.} = \Delta S_h + \Delta S_c$$

Now we also know:

$$\Delta S_{gas} = \int_{T_{c}}^{T_{h}} dQ/T = nC_v\ln{\frac{T_{h}}{T_{c}}$$ (NB: if you argue that I cannot use dS = dQ/T for the gas (cause you might say it already implies it can be reversible, cause in an irreversible case dS > dQ/T): I also calculated the same entropy change (because it is a variable of state) of the gas by going from start to end through an adiabat and an isotherm and got the same value)

As you can see, for the isometric:

$$\Delta S_{universe} = \Delta S_{environment} + \Delta S_{system} = 0$$

This is not correct. After one complete cycle, there is no change in the entropy of the gas (engine) since it returns to its initial state. Entropy is a state function. So the change in entropy of the hot and cold reservoirs represents the change in entropy of the universe after completion of one cycle.

AM

Please note that I had stated I wasn't looking at the whole cycle in my calculations, but only ONE process, namely the isochoric one (I thought isometric was a synonym, but indeed, I meant ischoric). In this case obviously the entropy change of the gas is not zero and you'll find that also the rest of my calculations make sense if you bear in mind I wasn't looking at a full cycle. My argument is that once I've shown the reversibility of ONE process of the cycle (which I claim to have done in my previous post that you misinterpreted), all the processes are reversible (in the same fashion) and thus the whole cycle.

Curious as to what you think.