Are textbooks sloppy with the entropy change of an irreversible process?by Confusus Tags: entropy, irreversible, thermodynamics 

#1
Oct311, 10:16 PM

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 q_{sys}=q_{surr}. 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 ΔS_{surr}=q_{surr}/T=q_{sys}/T, which is higher than ΔS_{sys}. 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! 



#2
Oct411, 12:30 AM

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P: 8,004

Let's say the process is quasistatic 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 quasistatic 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
Oct411, 12:44 AM

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As an example, consider a quasistatic 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
Oct411, 08:08 AM

P: 10

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



#5
Oct411, 08:20 AM

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?




#6
Oct411, 08:36 AM

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#7
Oct411, 08:40 AM

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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 



#8
Oct411, 10:49 AM

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P: 8,004

In this case, I have imagined that the irreversible compression was quasistatic because you described it as isothermal. There are presumably other ways of achieving the same system final state which are not even quasistatic, in which case the entropy change of the surroundings cannot be calculated by this method. A useful reference is sections 2.14 and 2.15 of http://books.google.com/books?id=sTu...gbs_navlinks_s 



#9
Oct411, 12:16 PM

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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 ([itex]\Delta S_{sys} + \Delta S_{surr} = 0[/itex]), 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 [itex]\Delta S_{sys} = \int dQ/T[/itex] for that path 3. Determine the reversible path between the initial and final states for the surroundings. 4. Calculate [itex]\Delta S_{surr} = \int dQ/T[/itex] for that path 5. Add to get the total change in entropy: [itex]\Delta S = \Delta S_{sys} + \Delta S_{surr}[/itex] 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. [itex]\Delta S_{sys} = \int dQ/T = W/T1 < 0[/itex] 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. [itex]\Delta S_{surr} = \int dQ/T = W/T2 > 0[/itex] 5. [itex]\Delta S = \Delta S_{sys} + \Delta S_{surr} = W/T2  W/T1 = W(1/T2  1/T1) > 0[/itex] since T2<T1. For the reversible process, do the same steps. The only difference is that T2 = T1. In that case [itex]\Delta S = W(1/T2  1/T1) = 0[/itex] AM 



#10
Oct411, 09:49 PM

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 ΔS_{sys} is unaffected of course, but when you bring down the pathfunction w down to the step 4, that is not correct. You would need to calculate for a particular case, but both w_{irrev}<w_{rev} and T_{surr}<T_{sys}, 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
Oct411, 11:19 PM

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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? AM 



#12
Oct511, 12:22 AM

P: 192

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=dQdW. 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
Oct511, 12:27 AM

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#14
Oct511, 08:04 AM

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AM 



#15
Oct511, 08:59 AM

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#16
Oct511, 09:18 AM

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#17
Oct511, 09:23 AM

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BBB 



#18
Oct511, 09:38 AM

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