Focus Problem for Entropy Change in Irreversible Adiabatic Process

[Chestermiller]

Not at all. I just prefer not to abuse your time. I'll try to read this thread a few times, check my books about thermodynamics and if I still can't catch it on my own then I'll definitely open that thread.
I just feel kind of guilty about using your and others' expertise too often lately by opening too many posts or opening questions in other's people threads.
To be honest, that post is most likely coming. This kind of misunderstanding has been rubbing on me for too long and I can't seem to get it right. It would be surprising if I find out how it works in the following days.

Helping other members is my pleasure, especially those as motivated as you are.

 

[guhag]

See my graph for B in the previous post. Your points should not be evenly spaced horizontally. I didn't ask you to change the numbers on the scale. I asked you to change the scale itself. Double click on the x axis. you should get a dialog box with a selection called scale, and showing linear. Just change the selection to Log.

B, C, D, and irreversible look correct, but not A. Pl\ease recheck your calculation for A. A should be lower than B.

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.
I have attached my working and the stacked graphs. Thanks and have a good weekend.

 

[Chestermiller]

There was a typo in the sign (sigh!) in calculations for process A. Also, I am using Freeoffice and not excel, so things took time to figure out. Apologies for the delay.
I have attached my working and the stacked graphs. Thanks and have a good weekend.

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]

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.

PROCESS C
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.

 

[guhag]

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.

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 excel

 

[Chestermiller]

 

[guhag]

View attachment 329568

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]

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.
View attachment 329576

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?

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?

 

[guhag]

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?

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?

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]

PROCESS C
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.

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]

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.

I got it, thanks ! Here is the corrected graph for all the processes, finally it seems to match with yours !

 

[Chestermiller]

I’m

I got it, thanks ! Here is the corrected graph for all the processes, finally it seems to match with yours !
View attachment 329614

Excellent. Next, please consider the questions I posed in post 39.

 

[guhag]

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?

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]

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 !

Consider this 3-step path E:

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

 

[guhag]

Consider this 3-step path E:

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

Dimensionless work, W/nRT0 = 0.25* ln(V1/V0) = W( for path C) /4. This is less than W(irrev).

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?

 

[Chestermiller]

Dimensionless work, W/nRT0 = 0.25* ln(V1/V0) = W( for path C) /4. This is less than W(irrev).

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):

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.

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

$$V_1=V_0\frac{P_0T_1}{P_1T_0}$$So,$$P_b=\frac{P_0V_0}{4}\frac{P_1T_0}{V_0P_0T_1}=\frac{T_0}{T_1}\frac{P_1}{4}$$

3. (Pb,T0/4, V1) to (P1,T1, V1) where Pb = (P1/T1) * (1/4)

$$P_b=\frac{T_0}{T_1}\frac{P_1}{4}$$

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?

What part are you trying to prove geometrically?

 

[guhag]

$$V_1=V_0\frac{P_0T_1}{P_1T_0}$$So,$$P_b=\frac{P_0V_0}{4}\frac{P_1T_0}{V_0P_0T_1}=\frac{T_0}{T_1}\frac{P_1}{4}$$$$P_b=\frac{T_0}{T_1}\frac{P_1}{4}$$

Thanks. Typo in my calculation:

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)

 

[guhag]

What part are you trying to prove geometrically?

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.

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.

 

[guhag]

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.

I am afraid I don't follow your insight here...
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...

 

[Chestermiller]

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.

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.

OK

 

[Chestermiller]

I am afraid I don't follow your insight here...
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...

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]

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).

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 :)

Is there a mathematical generalization behind this ? Or, it depends upon a particular situation, like this case?

 

[Chestermiller]

 

[Chestermiller]

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 :)

Is there a mathematical generalization behind this ? Or, it depends upon a particular situation, like this case?

I do't know what you mean.

 

[guhag]

View attachment 329674

Thank you. Mine looks like this....

 

[Chestermiller]

Thank you. Mine looks like this....

View attachment 329676

Is nobody willing to attempt part (c), the entropy change for each of the 5 reversible paths that we have identified?

 

[guhag]

Is nobody willing to attempt part (c), the entropy change for each of the 5 reversible paths that we have identified?

I haven't checked this thread in a while. Company layoffs looming ahead. I'll try part c) this week, for sure.

 

[Lollo]

@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?

 

[Chestermiller]

@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?

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.

If you have any comments or questions about the analysis presented here, let's hear them.

 

[Lollo]

@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?