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

1. Oct 3, 2011

### Confusus

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.

2. Oct 4, 2011

### 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.

3. Oct 4, 2011

### 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.

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

4. Oct 4, 2011

### 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? 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?

Last edited: Oct 4, 2011
5. Oct 4, 2011

### 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?

6. Oct 4, 2011

### Cipherflak

Sorry, but isn't an adiabatic process the reversible one?

7. Oct 4, 2011

### Andrew Mason

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

Last edited: Oct 4, 2011
8. Oct 4, 2011

### atyy

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.

Last edited: Oct 4, 2011
9. Oct 4, 2011

### Andrew Mason

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<T1 Since heat flow is into the surroundings, Q is positive.

4. $\Delta S_{surr} = \int dQ/T = W/T2 > 0$

5. $\Delta S = \Delta S_{sys} + \Delta S_{surr} = W/T2 - W/T1 = W(1/T2 - 1/T1) > 0$ since T2<T1.

For the reversible process, do the same steps. The only difference is that T2 = T1. In that case $\Delta S = W(1/T2 - 1/T1) = 0$

AM

10. Oct 4, 2011

### 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...".)

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<wrev and Tsurr<Tsys, so the comparison of the w/T ratios is indeterminate as yet.

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.

11. Oct 4, 2011

### Andrew Mason

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.
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 need to study this very carefully. It is a confusing area. What I wrote in my previous post is correct.

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

12. Oct 5, 2011

### 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.

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

13. Oct 5, 2011

### bbbeard

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.

14. Oct 5, 2011

### 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.

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

15. Oct 5, 2011

### bbbeard

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.

Last edited: Oct 5, 2011
16. Oct 5, 2011

### bbbeard

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.

17. Oct 5, 2011

### bbbeard

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

18. Oct 5, 2011

### bbbeard

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.

19. Oct 5, 2011

### bbbeard

Yes, but if the system is surrounded by a heat bath, after the reversible adiabatic expansion, the surroundings will dump heat into the system until it reaches thermal equilibrium, i.e. will have the same final temperature as the initial temperature (which equals the post-free expansion temperature). This is assuming the initial temperature in both cases was the heat bath temperature. This heat transfer from surroundings to system can be accomplished reversibly -- just hook up a Carnot engine and extract reversible work to go along with the reversible heat transfer.

Or, you can add a segment of irreversible heat transfer to the free-expansion process, and bring it down to the final temperature of the reversible adiabatic process. In either case, we have a reversible and an irreversible process with the same initial and final states.

20. Oct 5, 2011

### atyy

Hmmm, as an example of an irreversible isothermal process, why can't we take the reversible process to be a container of gas with a frictionless piston in a thermal bath at T, and have the gas compressed quasi-statically to a particular final volume by placing weights on the piston. An isothermal irreversible process would be the same set up, but with friction on the piston (I think the friction is similar to bbbeard's vortices). This is irreversible because removing and adding an infinitesimal amount of weight to the piston both produce positive heat from friction. Then assuming everything is quasi-static, the irreversible frictiony work in the forward direction can be treated for accounting purposes as reversible frictionless work with reversible heating of the environment.

Last edited: Oct 5, 2011
21. Oct 5, 2011

### Andrew Mason

If the gas is not at the same temperature throughout it is not a problem. It is not an isothermal compression. In that case, you are correct that there will be more work done and more heat generated in the gas and, therefore, transferred to the surroundings.

Let's follow the steps then to calculate change in entropy:

The system is an ideal gas. There is no difference between the initial and final thermal energy content of the system: $\Delta U = 0$. Thus, the reversible path between the beginning and end states for the gas is an isothermal compression (there is only one reversible path between these two states). For such an isothermal compression:

$$\Delta S_{sys} = \Delta Q_{sys-rev}/T = -W_{sys-rev}/T = nR\ln\left(\frac{V_i}{V_f}\right)$$ where W is the work done ON the gas (dW=-PdV) in the reversible process.

The reversible path for the surroundings between its beginning and end states is also an isothermal path since the surroundings have an arbitrarily high capacity to absorb heat without an increase in temperature.

[Note: Since we are finding a reversible path for the surroundings only, we don't have to worry about how the system generates that heat flow into the surroundings. But we could imagine the extra heat flow coming from a reversible isothermal compression of the same gas to a smaller final volume.].

Unlike the system, the thermal energy state of the surroundings has increased. The surroundings have actually retained the heat generated in the actual process: ($\Delta Q_{surr} = W_{actual}$). The thermal energy content of the surroundings has increased by the amount of actual work done on the system, (which is greater than the work done in a reversible process). We have to account for that heat flow in determining the entropy change of the surroundings:

$$\Delta S_{surr} = \Delta Q_{surr-rev}/T = W_{actual}/T > W_{sys-rev}/T = - nR\ln\left(\frac{V_i}{V_f}\right)$$, since $W_{actual} > W_{sys-rev}$

As a result, $\Delta S = \Delta S_{sys} + \Delta S_{surr} = W_{actual}/T - W_{rev}/T > 0$.

Actually I was addressing atyy's #8 post (my #9 post) in which he referred to an irreversible isothermal compression. It is not possible to have an isothermal irreversible compression if the surroundings and system are at the same temperature.

AM

Last edited: Oct 5, 2011
22. Oct 5, 2011

### Andrew Mason

All reversible processes joining two states have to be equivalent and must involve the same net heat flows and net work.

For example, the reversible process between the beginning and final end states of the gas in question could be a reversible isothermal compression from Vi to .5Vf and then a reversible isothermal expansion from .5Vf to Vf. But that process is equivalent to a reversible isothermal compression from Vi to Vf.

Perhaps you could give an example for a reversible compression of an ideal gas from Vi to Vf in which the initial and final temperatures are the same that is not equivalent to an isothermal compression from Vi to Vf.

AM

23. Oct 5, 2011

### bbbeard

I think we're on the same page now. I would go one further and say that if the system is undergoing an irreversible process and there are gradients of concentration, pressure, temperature, etc. then not only is the process not isothermal, but that the system temperature itself is poorly defined during the process. At the beginning and end of the process, before the gradients arise and after they dissipate, it is reasonable to describe the system with a single temperature and stipulate that the irreversible process has the same endpoints as some hypothetical reversible process.

24. Oct 5, 2011

### bbbeard

I think in a moment you will realize that your statement is untenable.

Consider a http://en.wikipedia.org/wiki/Carnot_cycle" [Broken]. For the benefit of the uninitiated reading this, the Carnot cycle comprises four processes that connect four equilibrium states. It describes the operation of a ideal heat engine of maximum efficiency.

Read more at the Wiki page. But note that by definition each process is reversible.

So Andrew, consider the processes 1-2 and 2-3 that connect states 1 and 3. Isn't it obvious the processes 1-4 and 4-3 (the reversal of the direction of the processes in the figure) also connect states 1 and 3? And clearly 1-2-3 and 1-4-3 have different heat transfers, and different work transfers? (But ΔQ-ΔW = ΔU = U3-U1 is the same.)

Exercise: show that even though Q123 and Q143 are different, the integral of dQ/T is the same for both processes.

Oh, and Andrew, the specific answer to your question is that process 1-2 is reversible and isothermal. Process 1-4-3-2 is reversible and not isothermal and has a different work and heat transfer from 1-2.
BBB

Last edited by a moderator: May 5, 2017
25. Oct 5, 2011

### Andrew Mason

/

Well, that's true. I see your point.

An adiabatic expansion followed from 1-4 by an isothermal expansion from 4-3 will get you to the same place as an isothermal expansion followed by an adiabatic expansion. That results in less work being done and smaller heat flow into the gas but the final state is the same.

And I suppose you could break the adiabatic part into two parts and have a path with a shorter adiabatic expansion to a temperature higher than 3 followed by an isothermal expansion to a volume less than at 3 followed by a second adiabatic expansion to 3. Or you could have a longer adiabatic expansion followed by a longer isothermal expansion followed by an adiabatic compression to 3. &c.

Ok. I agree. So you are right, there are an infinite number of possible reversible paths with different dWs and dQs so long as all heat flow occurs isothermally and all changes in temperature occur adiabatically. Although the dQs are different for the different paths, $\int dQ/T$ is the same. Good point.

AM