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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:
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: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.
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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: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?
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: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.
I got it, thanks ! Here is the corrected graph for all the processes, finally it seems to match with yours !Chestermiller said: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.
Excellent. Next, please consider the questions I posed in post 39.guhag said:I got it, thanks ! Here is the corrected graph for all the processes, finally it seems to match with yours !
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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: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?
Consider this 3-step path E:guhag said: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 !
Dimensionless work, W/nRT0 = 0.25* ln(V1/V0) = W( for path C) /4. This is less than W(irrev).Chestermiller said: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
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: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):
$$V_1=V_0\frac{P_0T_1}{P_1T_0}$$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: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
$$P_b=\frac{T_0}{T_1}\frac{P_1}{4}$$guhag said:3. (Pb,T0/4, V1) to (P1,T1, V1) where Pb = (P1/T1) * (1/4)
What part are you trying to prove geometrically?guhag said: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?
Thanks. Typo in my calculation:Chestermiller said:$$V_1=V_0\frac{P_0T_1}{P_1T_0}$$So,$$P_b=\frac{P_0V_0}{4}\frac{P_1T_0}{V_0P_0T_1}=\frac{T_0}{T_1}\frac{P_1}{4}$$$$P_b=\frac{T_0}{T_1}\frac{P_1}{4}$$
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:What part are you trying to prove geometrically?
I am afraid I don't follow your insight here...Chestermiller said: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.
OKguhag said: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.
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: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...
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: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).
I do't know what you mean.guhag said: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?
Is nobody willing to attempt part (c), the entropy change for each of the 5 reversible paths that we have identified?guhag said:
I haven't checked this thread in a while. Company layoffs looming ahead. I'll try part c) this week, for sure.Chestermiller said:Is nobody willing to attempt part (c), the entropy change for each of the 5 reversible paths that we have identified?