# Are textbooks sloppy with the entropy change of an irreversible process?

by Confusus
Tags: entropy, irreversible, thermodynamics
 P: 10 Trying hard to understand a basic textbook model meant to illustrate that entropy (of the universe) increases for irreversible processes. Help me out please? I get this part: A gas is compressed isothermally (constant T) and reversibly, getting worked on and expelling heat. To calculate ΔS for the system, you can use the formula ΔS=q/T since it is a reversible process. The same formula can be used for the entropy change in the surroundings, and of course, the entropy changes are equal and opposite since qsys=-qsurr. Total entropy change of the universe is zero. Now for the irreversible compression, with same initial and final system states. Since entropy is a state function, ΔS for the system is exactly the same value as above. However when you calculate the heat transfer q, that has a higher value now (more work was required for the irreversible compression, so more heat was expelled). To get entropy change of the surroundings, textbooks use ΔSsurr=qsurr/T=-qsys/T, which is higher than ΔSsys. But isn't it illegal to use that formula for this irreversible process? I'm totally confused. All of the textbooks brush right over this point, but it seems like an obvious objection. Thank you for your help!
 Sci Advisor P: 8,373 Let's say the process is quasi-static but irreversible due to work done against friction. Work done against friction is irreversible since when you reverse the work and heating on the system, the work against friction produces heat that does not reverse sign. The same system and environment initial and final states can also be produced by a quasi-static process in which the work done is frictionless so that the process is reversible and the heat from friction is explicitly accounted for as a source of heat. In this equivalent reversible process, the heat from friction is supplied to the environment.
HW Helper
P: 6,654
 Quote by Confusus Now for the irreversible compression, with same initial and final system states.
If they had the same initial and final states there would be no difference in the entropy calculation. So there has to be a difference between the final state of the irreversible process and the final state of the reversible process.

As an example, consider a quasi-static reversible adiabatic expansion of an ideal gas from Vi to Vf and an irreversible adiabatic expansion (let's say a free expansion - no work done - to make it simple) from Vi to Vf. The free expansion results in no change in internal energy/temperature, since no work is done. The reversible expansion does work so the internal energy/temperature decreases.

To calculate the change in entropy of the irreversible free expansion, you have to find the integral of dQ/T over the reversible path between the initial and final states. That would be an isothermal reversible expansion in which there is heat flow into the gas (dQ>0), so the integral of dQ/T over that path is > 0.

AM

P: 10
Are textbooks sloppy with the entropy change of an irreversible process?

 Quote by Andrew Mason If they had the same initial and final states there would be no difference in the entropy calculation. So there has to be a difference between the final state of the irreversible process and the final state of the reversible process.
The textbook examples I am citing ALL go against what you say here, which is why I'm confused. Quite explicitly, the initial and final states of the system are identical comparing the reversible and irreversible processes, so they can say the ΔSsystem values are the same. BUT the irreversible process dumps more heat into the surroundings, so how can the final state of the surroundings be the same as in the reversible process? Since ΔS is different, the final state can't be the same, yet the process is isothermal, the volume should be the same, the pressure should be the same... ? Where is the state change manifested?
P: 10
Thanks for the reply, atyy. I understand the qualitative difference between the processes. However I want to understand the textbook justification for using ΔS=q/T to calculate entropy change of the SURROUNDINGS for an irreversible process. It seems that if the system is in the same final state, so to must be the surroundings, at the end of the irreversible process. Yet ΔS is indeed higher, so it can't be at the same final state (S is a state function!) So where's the flaw in the reasoning?

 Quote by atyy Let's say the process is quasi-static but irreversible due to work done against friction. Work done against friction is irreversible since when you reverse the work and heating on the system, the work against friction produces heat that does not reverse sign. The same system and environment initial and final states can also be produced by a quasi-static process in which the work done is frictionless so that the process is reversible and the heat from friction is explicitly accounted for as a source of heat. In this equivalent reversible process, the heat from friction is supplied to the environment.
P: 33
 I get this part: A gas is compressed isothermally (constant T) and reversibly, getting worked on and expelling heat
Sorry, but isn't an adiabatic process the reversible one?
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P: 6,654
 Quote by Confusus The textbook examples I am citing ALL go against what you say here, which is why I'm confused. Quite explicitly, the initial and final states of the system are identical comparing the reversible and irreversible processes, so they can say the ΔSsystem values are the same. BUT the irreversible process dumps more heat into the surroundings, so how can the final state of the surroundings be the same as in the reversible process?
Read your text again. Better still, give us the quote that you think says that the initial and final states of the system and surroundings are identical comparing the reversible and irreversible process. They cannot possibly be the same. The free expansion example I gave you shows why.

To calculate the change in entropy of any process you have to calculate the integral of dQ/T over the reversible path between the initial and final states of the system and then do the same for the surroundings. The theoretical reversible path between the beginning and end states for the system in an actual irreversible process is not the same as the theoretical reversible path between the beginning and end states for the system in an actual reversible process.

Irreversible processes do not necessarily dump more heat into the surroundings than reversible processes. An irreversible adiabatic compression uses more work, which means that the internal energy of the system will be higher than in a reversible adiabatic compression between the same initial and final volumes (ie. higher T than in a reversible adiabatic compression). There is no change in state of the surroundings (no heatflow), so there is no change in entropy of, the surroundings. But there is an increase in entropy of the system in the irreversible process. This is because the reversible path between the initial and final states is not adiabatic - it requires heatflow into the system.

AM
P: 8,373
 Quote by Confusus Thanks for the reply, atyy. I understand the qualitative difference between the processes. However I want to understand the textbook justification for using ΔS=q/T to calculate entropy change of the SURROUNDINGS for an irreversible process. It seems that if the system is in the same final state, so to must be the surroundings, at the end of the irreversible process. Yet ΔS is indeed higher, so it can't be at the same final state (S is a state function!) So where's the flaw in the reasoning?
The final state of the surroundings for irreversible quasi-static isothermal compression is not the same as when the process is reversible quasi-static isothermal compression. What I meant is that the surroundings are in the same final state as after a quasi-static process consisting of reversible isothermal compression plus heating of the surroundings. Then the question is whether the heating reversible? Yes, because heating during a quasi-static process is reversible. An irreversible quasi-static process is irreversible not because of heating, but because of work against friction - the equivalent reversible process in this case for calculating the entropy change is "frictionless work+heating". Basically you can think of the friction as a source of heat that does not reverse sign when the action is reversed.

In this case, I have imagined that the irreversible compression was quasi-static because you described it as isothermal. There are presumably other ways of achieving the same system final state which are not even quasi-static, in which case the entropy change of the surroundings cannot be calculated by this method.

 Sci Advisor HW Helper P: 6,654 Let's use your example of an isothermal compression. If we are talking about an isothermal compression of an ideal gas we are talking about a compression in which there is no change in the internal energy of the gas (initial and final temperatures of the system are the same). The surroundings have done work on the system but none of that work has resulted in a change of U of the system, so there must be heat flow out of the system to the surroundings. The only reversible process for this ($\Delta S_{sys} + \Delta S_{surr} = 0$), is one in which heatflow occurs reversibly. This must be one in which the system and surroundings are arbitrarily close to the same temperature. An irreversible process for this is one that occurs with a finite temperature difference between the system and surroundings. So the temperatures of the system and surroundings for an isothermal irreversible compression cannot be the same whereas for an isothermal reversible compression they must be the same. Here is how you determine the change in entropy: 1. Determine the reversible path between the initial and final states for the system. 2. Calculate $\Delta S_{sys} = \int dQ/T$ for that path 3. Determine the reversible path between the initial and final states for the surroundings. 4. Calculate $\Delta S_{surr} = \int dQ/T$ for that path 5. Add to get the total change in entropy: $\Delta S = \Delta S_{sys} + \Delta S_{surr}$ So let's do it for the irreversible process. System is at constant temperature T1: Surroundings are at temperature T2. T1>T2. 1. The reversible path betweein the initial and final states (Pi,Vi,T1) and PiVi/Vf, ViPi/Pf, T1) is a quasistatic isothermal compression in which the work done on the system creates heatflow out of the gas such that Q = -W where W is the work done ON the gas. T is constant = T1. 2. $\Delta S_{sys} = \int dQ/T = -W/T1 < 0$ 3. The reversible path is one in which heat flows into the surroundings in an amount equal to the work done by the surroundings on the gas. T is constant = T2 0[/itex] 5. $\Delta S = \Delta S_{sys} + \Delta S_{surr} = W/T2 - W/T1 = W(1/T2 - 1/T1) > 0$ since T2
 P: 10 Andrew, your immediately prior comment was excellent. (For the record, you have a typo in your Step 3, change "Determine the reversible path..." to "...irreversible path...".) However, isn't there a flaw when you assume the work done is the same in the reversible and irreversible processes (you factor it out in step 5). Work done on the system should be lower for the irreversible process. The ΔSsys is unaffected of course, but when you bring down the path-function w down to the step 4, that is not correct. You would need to calculate for a particular case, but both wirrev
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P: 6,654
 Quote by Confusus Andrew, your immediately prior comment was excellent. (For the record, you have a typo in your Step 3, change "Determine the reversible path..." to "...irreversible path...".)
Not a typo. Entropy is always calculated along the reversible path between the initial and final states. The actual path determines what the initial and final states are.
 However, isn't there a flaw when you assume the work done is the same in the reversible and irreversible processes (you factor it out in step 5). Work done on the system should be lower for the irreversible process. The ΔSsys is unaffected of course, but when you bring down the path-function w down to the step 4, that is not correct.
In THIS CASE the work done on the gas is the same in the reversible and irreversible process. That is because it is an isothermal compression. The work done on the gas is $-\int P_{int}dV = -nRT\int dV/V = -nRT\ln(Vf/Vi) = nRT\ln(Vi/Vf)$

The only difference between the irreversible and reversible isothermal processes is that in the irreversible process, there is a finite temperature difference between the system and the surroundings. How do you think the irreversible process occurs here?

 You would need to calculate for a particular case, but both wirrev
You need to study this very carefully. It is a confusing area. What I wrote in my previous post is correct.

 For the record, my textbook source is Engel & Reid's Physical Chemistry book, though many physical chemistry books have a very similar example to demonstrate entropy change being positive for irreversible processes. What confused me was they details. They assume T is the same in surroundings as system, which as you carefully noted, is nonsense since there cannot be an irreversible heat flow in that case. Thank you for that.
It depends on the example. Why not give us a particular example that your text uses that you think is wrong and we will analyse it.

AM
P: 192
 Quote by Andrew Mason How do you think the irreversible process occurs here?
Your example does not fit the problem Confusus describes. In the problem he describes, the gas and the surroundings are at the same temperature T throughout. The difference between the reversible and irreversible process is presumably due to irreversibilities within the gas, generated during the compression process. These might be, for example, due to vortices generated by compression of a piston at a finite rate.

Warning: I use the "engineering" sign convention, where positive heat transfer is into the system, and positive workflow is out of the system, as Mr. Watt intended, and dU=dQ-dW. In the case of compression that is reversible or nearly so, the work and heat transfers are both negative. Note that the conditions on irreversible heat and work with this sign convention is that dQ_irrev < T*dS and dW_irrev < P*dV.

The effect of the irreversibilities is that the heat transfer and work transfer are both smaller than the reversible values. Since these quantities are negative, this means that the magnitude of the heat and work transfer for the irreversible process are larger than the magnitudes for the reversible process. That is, |W_irrev| > |W_rev| and |Q_irrev| > |Q_rev|. Because the initial and final states are stated to be the same in both cases, the entropy change of the gas is the same in both cases. Now, you mention the need to integrate dQ/T for a reversible process connecting the initial and final states. However, in this case, because the temperature is uniform, we can hypothesize that all the irreversibilities are inside the gas. So we can assume that the heat transfer is reversible! The entropy change in the surroundings is the integral of dQ/T, which for constant T is always Q/T. But we know |Q_irrev| > |Q_rev|, so we see that the entropy change for the surroundings is larger for the irreversible case than for the reversible case. Thus, the total entropy change is zero for the reversible case and some positive number for the irreversible case.

We can also note how the problem is slightly different for an expansion. Again we have dQ_irrev < T*dS and dW_irrev < P*dV. But with an expansion, both heat and work transfers are positive with respect to the gas. So |Q_irrev| < |Q_rev|, opposite from before. Again since the initial and final gas states are deemed the same for the irreversible and reversible cases, the entropy change for the gas is the same. But for the surroundings, heat is removed, so the entropy change for the surroundings is some negative number -Q/T for the reversible case, and some numerically larger negative number for the irreversible case. Thus the total entropy change is again zero for the reversible case and some positive number for the irreversible case.

BBB
P: 192
 Quote by Cipherflak Sorry, but isn't an adiabatic process the reversible one?
In the paradigms of thermodynamics, "reversible" is an idealization that is closely modeled in systems with slow changes and negligible gradients of pressure, temperature, etc. Since heat transfer can be made arbitrarily slow, you can have a process with heat transfer that is effectively reversible. So you do not have to have adiabaticity to have reversibility.
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P: 6,654
 Quote by bbbeard Your example does not fit the problem Confusus describes. In the problem he describes, the gas and the surroundings are at the same temperature T throughout. The difference between the reversible and irreversible process is presumably due to irreversibilities within the gas, generated during the compression process. These might be, for example, due to vortices generated by compression of a piston at a finite rate.
I disagree. There is only one way to have an isothermal compression of a gas where the temperature of the surroundings and the gas are the same throughout: it must be reversible. If you compress the gas too quickly, creating dynamic flows within the gas, when the vortices settle down the temperature of the gas will increase and this is not permitted. This is just the first law: dQ = dU + dW. If you do work on the gas, you either increase the internal energy or you increase heat flow. Neither are permitted to occur if both the gas and the surroundings are maintained at the same temperature UNLESS the work is done infinitely slowly.

 Note that the conditions on irreversible heat and work with this sign convention is that dQ_irrev < T*dS and dW_irrev < P*dV.
In general that is true. The problem is that I don't see how you can have an irreversible isothermal compression of gas with the surroundings at the same temperature as the gas. Perhaps you could explain that.

AM
P: 192
 Quote by Andrew Mason I disagree. There is only one way to have an isothermal compression of a gas where the temperature of the surroundings and the gas are the same throughout: it must be reversible. If you compress the gas too quickly, creating dynamic flows within the gas, when the vortices settle down the temperature of the gas will increase and this is not permitted. This is just the first law: dQ = dU + dW. If you do work on the gas, you either increase the internal energy or you increase heat flow. Neither are permitted to occur if both the gas and the surroundings are maintained at the same temperature UNLESS the work is done infinitely slowly.
I would first note that as the OP stated the problem, the irreversible process has the same initial and final states as the reversible compression. But the problem does not state that the irreversible process is isothermal, just that the endpoints have the same temperature. [I misspoke when I wrote "In the problem he describes, the gas and the surroundings are at the same temperature T throughout."] This is appropriate, since for real irreversible processes there is usually not one single temperature that can be seen in the system -- that's why, on PV or TS diagrams, textbooks usually denote irreversible processes with a dotted or fuzzy line. Looking back on the thread, this seems to be a confusion that you injected.
P: 192
 Quote by Confusus Now for the irreversible compression, with same initial and final system states. Since entropy is a state function, ΔS for the system is exactly the same value as above. However when you calculate the heat transfer q, that has a higher value now (more work was required for the irreversible compression, so more heat was expelled). To get entropy change of the surroundings, textbooks use ΔSsurr=qsurr/T=-qsys/T, which is higher than ΔSsys. But isn't it illegal to use that formula for this irreversible process? I'm totally confused. All of the textbooks brush right over this point, but it seems like an obvious objection.
I suspect that your textbook is assuming the irreversibilities are generated inside the gas volume, but that the external heat transfer can be modeled as reversible. This makes it simple to calculate the entropy change in the surroundings.
P: 192
 Quote by Andrew Mason The only reversible process for this ($\Delta S_{sys} + \Delta S_{surr} = 0$), is one in which heatflow occurs reversibly. This must be one in which the system and surroundings are arbitrarily close to the same temperature.
You've used the phrase "the reversible process" several times in this thread, as if there were a unique reversible process joining two states. I feel obligated to clarify that in general there are an infinite number of reversible processes joining any two states, generally with different amounts of heat and work transfer (with ΔQ-ΔW = ΔU fixed by the endpoints).

BBB
P: 192
 Quote by Confusus Thanks for the reply, atyy. I understand the qualitative difference between the processes. However I want to understand the textbook justification for using ΔS=q/T to calculate entropy change of the SURROUNDINGS for an irreversible process. It seems that if the system is in the same final state, so to must be the surroundings, at the end of the irreversible process. Yet ΔS is indeed higher, so it can't be at the same final state (S is a state function!) So where's the flaw in the reasoning?
You always have to use some caution when dealing with idealizations about the surroundings. A "heat bath", or the atmosphere, is ostensibly infinite compared to the finite "system". That means that the actual state does not budge -- it's a fixed point on a PV or TS diagram. So you can dump varying amounts of heat, or extract varying amounts of work, and the state of the atmosphere does not change. To balance the books, you can assume that the state of the heat bath changes by an infinitesimal amount, e.g. MatmΔsatm + mgasΔsgas = 0 for reversible processes, but as Matm→∞, Δsatm becomes infinitesimal. For different amounts of heat transfer, the infinitesimal is different, but it's still infinitesimal.

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