Thermodynamic Process: Is Work Conserved?

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Discussion Overview

The discussion revolves around the thermodynamic processes involved in a cycle consisting of isobaric and isothermal transitions. Participants analyze the work done during these processes and the implications for internal energy and heat transfer, exploring the relationships between pressure, volume, and temperature for ideal gases.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants present calculations for work done from B to C, expressing it in terms of pressures and volumes, and question the validity of their logic.
  • Others note that isothermal processes have no change in internal energy for an ideal gas and discuss the heat transfer required during isobaric expansion and compression.
  • Participants mention that the change in internal energy for the entire thermodynamic cycle should be zero, given that the initial and final states are identical.
  • There is a discussion on the specific changes in internal energy for monatomic and diatomic ideal gases during isobaric processes, with references to the equations governing these changes.
  • Some participants express confidence in their calculations and reasoning but seek confirmation from others regarding the correctness of their analyses.

Areas of Agreement / Disagreement

While there is some agreement on the calculations and the implications of the thermodynamic cycle, participants express uncertainty about specific aspects of the logic and calculations, particularly regarding the relationships between internal energy and work done. The discussion remains unresolved on certain points, with multiple viewpoints presented.

Contextual Notes

Participants acknowledge that the calculations depend on the assumptions made about the ideal gas behavior and the specific conditions of the processes. There are also references to potential dependencies on the type of gas involved.

Who May Find This Useful

This discussion may be useful for students and practitioners in thermodynamics, particularly those interested in the analysis of thermodynamic cycles and the behavior of ideal gases under various processes.

Ian Baughman
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What is wrong with this logic, if any? It does not seem like this should be true but maybe I'm mistaken.

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Assuming the process consists of two isobaric processes and two isothermal processes the work from B to C in terms of p1, p2, V1, and V2 is given by the following.

1) WBC=p2(VC-VB)
2) Isothermal from A → B so p1V1 = p2VB so VB=(p1V1)/p2
3) Similarly VC=(p1V2)/p2
4) WBC= p2((p1V2)/p2-(p1V1)/p2)
5) Factor out p1/p2 and simplify the you get;
WBC=p1(V2-V1)​
 
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Ian Baughman said:
What is wrong with this logic, if any? It does not seem like this should be true but maybe I'm mistaken.

Physics.png


Assuming the process consists of two isobaric processes and two isothermal processes the work from B to C in terms of p1, p2, V1, and V2 is given by the following.

1) WBC=p2(VC-VB)
2) Isothermal from A → B so p1V1 = p2VB so VB=(p1V1)/p2
3) Similarly VC=(p1V2)/p2
4) WBC= p2((p1V2)/p2-(p1V1)/p2)
5) Factor out p1/p2 and simplify the you get;
WBC=p1(V2-V1)​
The isothermal processes have no change in internal energy. for an ideal gas. Much heat must be added to the system in the isobaric expansion for two reasons: 1)work is being done as it expands 2) It is going to a higher temperature state. In the isobaric compression, much heat must be removed from the system for two reasons 1) The system is going to a lower temperature 2) Work is being done on the system. The energy from this work must also be removed by removing heat. I do think your calculations are correct. Thermodynamic problems can sometimes be tricky, but I believe you analyzed it correctly.
 
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It should also be noted that you can readily compute ## \Delta Q=\Delta W ## for each of the two isotherms, since each has ## \Delta U=0 ##. ## \\ ## Meanwhile ## \Delta U ## for each of the isobars will be ## \Delta U=\frac{3}{2}nR \Delta T ## for an ideal monatomic gas, and ## \Delta U=\frac{5}{2}nR \Delta T ## for an ideal diatomic gas (where ## \Delta T=\Delta (PV)/(nR) ##). You can thereby readily compute ## \Delta Q ## for each of the isobars, if you know the type of gas. ## \\ ## I think I have this completely correct, but it's always good to have someone double-check the solution. @Chestermiller Did we get everything correct?
 
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Charles Link said:
It should also be noted that you can readily compute ## \Delta Q=\Delta W ## for each of the two isotherms, since each has ## \Delta U=0 ##. ## \\ ## Meanwhile ## \Delta U ## for each of the isobars will be ## \Delta U=\frac{3}{2}nR \Delta T ## for an ideal monatomic gas, and ## \Delta U=\frac{5}{2}nR \Delta T ## for an ideal diatomic gas (where ## \Delta T=\Delta (PV)/(nR) ##). You can thereby readily compute ## \Delta Q ## for each of the isobars, if you know the type of gas. ## \\ ## I think I have this completely correct, but it's always good to have someone double-check the solution. @Chestermiller Did we get everything correct?

Since this is a thermodynamic cycle with the final state being identical to the initial state would the total change in internal energy be equal to zero?
 
Ian Baughman said:
Since this is a thermodynamic cycle with the final state being identical to the initial state would the total change in internal energy be equal to zero?
Yes. And if you will notice, for the two isotherms ## \Delta U=0 ## and for the transitions on the two isobars, ## \Delta U_1=-\Delta U_2 ##. ## \\ ## Editing... perhaps we haven't proved yet that ## \Delta (PV) ## is equal and opposite for these two cases...but that is obvious because on one isotherm, ## PV=nRT_1 ## and on the other isotherm ## PV=nRT_2 ##.Thereby ## \Delta (PV)=nR \Delta T ## and ## \Delta U=\frac{3}{2} nR \Delta T ## or ## \Delta U=\frac{5}{2} nR \Delta T ##.
 
Ian Baughman said:
What is wrong with this logic, if any? It does not seem like this should be true but maybe I'm mistaken.

Physics.png


Assuming the process consists of two isobaric processes and two isothermal processes the work from B to C in terms of p1, p2, V1, and V2 is given by the following.

1) WBC=p2(VC-VB)
2) Isothermal from A → B so p1V1 = p2VB so VB=(p1V1)/p2
3) Similarly VC=(p1V2)/p2
4) WBC= p2((p1V2)/p2-(p1V1)/p2)
5) Factor out p1/p2 and simplify the you get;
WBC=p1(V2-V1)​
Yes. So...

And yes, since this a thermodynamic cycle, the change in internal energy for the working fluid has to be zero.
 
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