- #31
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Helping other members is my pleasure, especially those as motivated as you are.Juanda said:
Undergrad Focus Problem for Entropy Change in Irreversible Adiabatic Process
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SUMMARY
The discussion centers on the analysis of an irreversible adiabatic process involving an ideal gas in a cylinder with a massless, frictionless piston. The initial state is defined by conditions To, Po, and Vo, and the process is initiated by dropping the external pressure to P1. The participants derive the new volume (V1) and temperature (T1) using the first law of thermodynamics and the ideal gas law, concluding that T1 and V1 can be expressed in terms of the initial conditions and P1. They also explore various reversible processes to achieve the same final state, emphasizing the importance of entropy changes in both irreversible and reversible processes.
PREREQUISITES- Understanding of the First Law of Thermodynamics
- Familiarity with the Ideal Gas Law
- Knowledge of entropy and its implications in thermodynamic processes
- Basic concepts of reversible and irreversible processes in thermodynamics
- Study the derivation of the First Law of Thermodynamics in detail
- Learn about the Ideal Gas Law and its applications in thermodynamic systems
- Investigate the concept of entropy and how it applies to reversible and irreversible processes
- Explore various thermodynamic cycles and their efficiency, focusing on reversible processes
Students and professionals in thermodynamics, mechanical engineers, physicists, and anyone interested in understanding the principles of gas behavior under adiabatic conditions and the implications of entropy in thermodynamic processes.
- #32
guhag
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There was a typo in the sign (sigh!) in calculations for process A. Apologies for the delay, I have started to use Freeoffice and not excel, so things took time to figure out.Chestermiller said:
I have attached my working and the stacked graphs. Thanks and have a good weekend.
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- #33
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The graphs are definitely wrong. They should all show the same value of W=0 for Po/P1 =1. The data in the table also looks incorrect. Dimensionless W for C is ln(V1/Vo). Dimensionless W for D is ##\frac{T_1}{T_0}\ln(V_1/V_0)##. V1/Vo is not equal to Po/P1. I haven't checked your data for A. The data for A and Irreversible is correct.guhag said:
- #34
guhag
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PROCESS CChestermiller said:
1. Isothermal expansion at temperature To from Vo to V1
2. Isochoric cooling at volume V1 from To to T1
I got this for W(total)= RT0ln(V1/V0). For the overall process, V1/V0 = (Po/P1) * (T1/T0). For the 1st step,V1/V0 = (Po/P1) . Since work for step 2 =0, W(total) = W1, hence I used V1/V0 = Po/P1 from the 1st step. Why is that not correct?
Similar case with process D.
- #35
guhag
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I got all the graphs starting at zero. Not sure but something to do with the software, maybe? I will check this data tomorrow in my lab using excelChestermiller said:
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- #36
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- #37
guhag
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Thanks, this is what I have, BUT the calculations for processes C and D would change depending on your clarification for the question I had asked earlier.Chestermiller said:
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- #38
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l don't quite understand. Are you saying that my results for C and D in post #36 are incorrect, or are you saying that you results for C and D in this post need updating?guhag said:
The results in these figures for part (b) show that the reversible work between the initial and final states of our system in this focal problem depends on the reversible path we select. So far we've seen that, for all the reversible paths we've looked at, the reversible work exceeds that actual irreversible work. Do you think that this holds in general, or is it possible that there are reversible paths that give less work than the irreversible path? If so, can you propose any?
- #39
guhag
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Chet, I meant I would have to correct *my* work once my doubt in post 34 is clarified and that affects processes C and D. Thanks.Chestermiller said:
- #40
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At the end of step 1, the pressure is not P1. This is because there is a pressure- and temperature change during step 2. The volume ratio for step 1 is the same as the volume ratio for the whole process, since volume does not change in step 2.guhag said:
- #41
guhag
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I got it, thanks ! Here is the corrected graph for all the processes, finally it seems to match with yours !Chestermiller said:
- #42
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I’m
Excellent. Next, please consider the questions I posed in post 39.guhag said:
- #43
guhag
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I can't prove that this holds in general but I think that reversible paths produce more work than irreversible as in the reversible case, the entire area under the path connecting initial and final states is used. I need to think more about it, though !Chestermiller said:
- #44
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Consider this 3-step path E:guhag said:
1. Isochoric cooling from To to To/4
2. Constant temperature expansion at To/4 from Vo to V1
3. Isochoric heating from To/4 to T1
- #45
guhag
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Dimensionless work, W/nRT0 = 0.25* ln(V1/V0) = W( for path C) /4. This is less than W(irrev).Chestermiller said:
How did you came up with this example, I wonder? I don't mean to disrupt your flow of thought (so you can check the question below, when you see fit):
Another question: I tried to chart this process out on a P-V diagram to see if I could deduce this graphically, for instance BUT was not successful.
As an exercise within an exercise ad infinitum :), the state variables for each of these steps I calculated are (to plot them in the P-V diagram):
1. (P0,T0,V0) to (Pa, T0/4, Vo) where Pa = P0/4
2. (Pa, T0/4, Vo) to (Pb,T0/4, V1) where Pb* V1 = Pa * V0 = nRT0/4
3. (Pb,T0/4, V1) to (P1,T1, V1) where Pb = (P1/T1) * (1/4)
a. Are these states of the system correct or did I mess them up?
b. If they are correct, I wonder (with appropriate reservations on my rusty math skills) why am I not able to prove this geometrically?
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- #46
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My motivaation was this: Case D was lower than C, because case C was at To and case D was at T1<To. So I could get the work as low as I wanted by expanding from Vo to V1 at lower temperature, while doing whatever temperature changes at constant volume.guhag said:
$$V_1=V_0\frac{P_0T_1}{P_1T_0}$$So,$$P_b=\frac{P_0V_0}{4}\frac{P_1T_0}{V_0P_0T_1}=\frac{T_0}{T_1}\frac{P_1}{4}$$guhag said:
$$P_b=\frac{T_0}{T_1}\frac{P_1}{4}$$guhag said:
What part are you trying to prove geometrically?guhag said:
- #47
guhag
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Thanks. Typo in my calculation:Chestermiller said:
1. (P0,T0,V0) to (Pa, T0/4, Vo) where Pa = P0/4
2. (Pa, T0/4, Vo) to (Pb,T0/4, V1) where Pb* V1 = Pa * V0 = nRT0/4
3. (Pb,T0/4, V1) to (P1,T1, V1) where Pb = (P1/T1) * (T0/4)
- #48
guhag
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I was trying to plot the process E on the P-V diagram and calculate the area to see how it compared with other reversible processes (A-D). I will try this on my own and ask you if I don't make any headway.Chestermiller said:
On a side note: I am trying to read a J.Chem.Edu article (attached) on 1st law. I hope you don't mind if I come back to you with questions on it.
- #49
guhag
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I am afraid I don't follow your insight here...Chestermiller said:
Secondly, this is not obvious to me. why you compare two reversible processes C and D instead of compare processes E, C (or D) and irreversible...
- #50
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OKguhag said:
- #51
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What I learned by comparing reversible processes C and D is that, if all the heat transfer is carried out at constant volume (no work in such steps) and all the work is carried out at constant temperature, then I can make the amount of work done in the overall process as low as I desire by carrying out the expansion at as low a temperature as I choose. In process E, all the heat transfer is carried out in steps 1 and 3 at constant volume, and I chose a low enough temperature for step 2 (To/4) so that the overall reversible work is very low (even lower than for the irreversible process between the same two end states).guhag said:
- #52
guhag
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This is a BEAUTIFUL explanation. Wow ! Thank you, Chet. I've learned a LOT already from you in this thread and there is still more to go :)Chestermiller said:
Is there a mathematical generalization behind this ? Or, it depends upon a particular situation, like this case?
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- #54
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I do't know what you mean.guhag said:
- #55
guhag
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Thank you. Mine looks like this....Chestermiller said:
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- #56
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Is nobody willing to attempt part (c), the entropy change for each of the 5 reversible paths that we have identified?guhag said:
- #57
guhag
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I haven't checked this thread in a while. Company layoffs looming ahead. I'll try part c) this week, for sure.Chestermiller said:
- #58
Lollo
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@ChetMiller I have not found any explicit treatment, in books, about irreversible adiabatic transformations. 1.1) Very often it is said that, in an adiabatic expansion, given the same initial conditions and final pressure (or final volume) the work done by an irreversible one is less than that done by a reversible one, but I have not yet found an explicit demonstration of this.
1.2) Also, it is said that in an adiabatic compression, the work done by an irreversible is greater than that done by a reversible, given the same initial conditions and final pressure (or final volume).
In short, it should hold $$|W_irr < W_rev|$$. How to prove it?
3) Moreover, it was not understood what inequalities should exist between pressure, volume and temperature between irreversibles and reversibles. Given the same initial conditions and final pressure, is the final temperature in an irreversible adiabatic transformation greater or less than the final temperature in a reversible adiabatic transformation? If yes, why?
1.2) Also, it is said that in an adiabatic compression, the work done by an irreversible is greater than that done by a reversible, given the same initial conditions and final pressure (or final volume).
In short, it should hold $$|W_irr < W_rev|$$. How to prove it?
3) Moreover, it was not understood what inequalities should exist between pressure, volume and temperature between irreversibles and reversibles. Given the same initial conditions and final pressure, is the final temperature in an irreversible adiabatic transformation greater or less than the final temperature in a reversible adiabatic transformation? If yes, why?
- #59
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Have you read over the posts in this thread. If you had, you would know that there are an infinite number of reversible paths between the initial and final states of the irreversible adiabatic process analyzed here and that the work along some of these reversible paths is greater than for the irreversible path while, along others, the work is less than for the irreversible path; however the entropy change is the same for all the reversible paths as for the irreversible path.Lollo said:
If you have any comments or questions about the analysis presented here, let's hear them.
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- #60
Lollo
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@Chestermiller I am very surprised that there are cases where, in reversible adiabatic processes, the work is less than in the corresponding irreversible path. I always thought that the work carried out on the system in irreversible adiabatic compression is greater than the work carried out on the system in reversible adiabatic compression, and that the work carried out on the system in irreversible adiabatic expansion is always less than the work carried out on the system in reversible adiabatic compression. Is this not so? I have seen several answers on StackExchange saying this, and they have never been refuted. See for example https://chemistry.stackexchange.com...rreversible-adiabatic-compression-greater-tha. "the entropy change is the same for all the reversible paths as for the irreversible path." How to prove that? It seems to me that point (c) has been left open. Can you post a solution for the benefit of the forum?