# Focus Problem for Entropy Change in Irreversible Adiabatic Process

• I
• Chestermiller
In summary, the gas expands from To to Vo and the temperature and pressure change. The gas does work against the pressure to do this, and the change in volume and temperature is equal to the work done.
Chestermiller
Staff Emeritus
Homework Helper
2023 Award
TL;DR Summary
This is a thread I am posting to help members understand how to calculate entropy change in irreversible adiabatic processes
I have an ideal gas in a cylinder with a massless, frictionless piston, and the gas starts out at To, Po, Vo. The system is adiabatic. Initially, the gas is in equilibrium with an external pressure, also at Po. I initiate an irreversible process by instantly dropping the external pressure to P1, and hold this pressure constant while the gas expands and equilibrates with the new external pressure P1 at a new volume and temperate V1 and T1.

(a) Based on the 1st law of thermodynamics and the ideal gas law, what at the values of the new volume and temperature (in terms of Po, To, Vo, and P1)? How much irreversible work was done? What was the change in U?

(b) Describe at least 2 reversible process you have devised to transition from the same initial state to the same final state. How much work was done, how much heat was added in each, and what was the change in U for these reversible processes?

(c). What was the change in entropy for the reversible processes? What was the change in entropy of the irreversible process?

Last edited:
vanhees71 and hutchphd
What do you mean by "adiabatic" here? It cannot be ##\Delta S =0## since on the other hand it's supposed to be an irreversible process, for which by definition ##\Delta S>0##.

vanhees71 said:
What do you mean by "adiabatic" here? It cannot be ##\Delta S =0## since on the other hand it's supposed to be an irreversible process, for which by definition ##\Delta S>0##.
I mean that no heat is added or removed during the irreversible process .

vanhees71
So here it means ##\Delta Q=0##.

Let's see, if I can at least do (a):

According to the 1st Law
$$\Delta U=\frac{f}{2} N k \Delta T=-P_1 \Delta V,$$
because the pressure is held constant during the expansion, and that's the work done by the gas expanding against the constant pressure ##P_1##. Further we have
$$N k T_0=P_0 V_0, \quad N k T_1=P_1 V_1.$$
So
$$\frac{f}{2} \frac{P_0 V_0}{T_0} (T_1-T_0)=-P_1(V_1-V_0)=P_1 V_0 - \frac{P_0 V_0 T_1}{T_0}.$$
So we get [EDIT: Typo corrected in view of #6]
$$T_1=\frac{f P_0-2 P_1}{(2+f)P_0} T_0$$
and
$$V_1=\frac{P_0 V_0 T_1}{T_0 P_1}=\frac{f P_0-2P_1}{(2+f)P_1} V_0.$$

Last edited:
@chet

Continuing from stackexchange....

Since it is an ideal gas we have P1.V1 = P0.V0 or P1/P0 = V0/V1

or V1 = V0* (P0/P1) -(A)

Cv * (T1 - T0) = -P1 (V1-V0) -(B)

Plug in P1 = RT1/V1 as it is an ideal gas and rearrange to get

(Cv/R) * (1-T0/T1) = - (1-V0/V1) - (C)

Plug (A) into (C) and use γ= Cp/Cv ; Cp= Cv +R to get

(Cv/R) * [1-(T0/T1)] = 1-(P1/P0)

1-(T0/T1) = (γ-1) * [ 1- (P0/P1)]

vanhees71 said:
So here it means ##\Delta Q=0##.

Let's see, if I can at least do (a):

According to the 1st Law
$$\Delta U=\frac{f}{2} N k \Delta T=-P_1 \Delta V,$$
because the pressure is held constant during the expansion, and that's the work done by the gas expanding against the constant pressure ##P_1##. Further we have
$$N k T_0=P_0 V_0, \quad N k T_1=P_1 V_1.$$
So
$$\frac{f}{2} \frac{P_0 V_0}{T_0} (T_1-T_0)=-P_1(V_1-V_0)=P_1 V_0 - \frac{P_0 V_0 T_1}{T_0}.$$
So we get
$$T_1=\frac{f P_0-2 P_1}{(2+f)P_0}$$
Something looks wrong here. T1 should depend on To. Please check your algebra. But you starting equation is correct.

Working in terms of Cv and R, I get $$nC_v(T_1-T_0)=-P_1(V_1-V_0)=-nRT_1+nRT_0\frac{P_1}{P_0}$$So, $$(C_v+R)T_1=\left[C_v+R\frac{P_1}{P_0}\right]T_0$$
or $$\frac{T_1}{T_0}=\frac{1+(\gamma-1)P_1/P_0}{\gamma}$$

guhag said:
@chet

Continuing from stackexchange....

Since it is an ideal gas we have P1.V1 = P0.V0 or P1/P0 = V0/V1

or V1 = V0* (P0/P1) -(A)
This equation is incorrect. It applies only to initial and final temperatures being equal. In our adiabatic case, the final temperature is less than the initial temperature. See @vanhees71 analysis.

Chestermiller said:
Something looks wrong here. T1 should depend on To. Please check your algebra. But you starting equation is correct.
A factor ##T_0## was missing, and also the next formula was mistyped. I've corrected this.
Chestermiller said:
Working in terms of Cv and R, I get $$nC_v(T_1-T_0)=-P_1(V_1-V_0)=-nRT_1+nRT_0\frac{P_1}{P_0}$$So, $$(C_v+R)T_1=\left[C_v+R\frac{P_1}{P_0}\right]T_0$$
or $$\frac{T_1}{T_0}=\frac{1+(\gamma-1)P_1/P_0}{\gamma}$$
Let's see. We have ##C_V=f/2 k N## and ##C_P=C_V+N k = (f+2) k N/2##. Now ##\gamma=C_P/C_V=(f+2)/f##. So your formula agrees with my corrected formula :-).

guhag
@guhag Have you redone your analysis with the corrected volume relation. Are you now in agreement with @vanhees71 and me?

Are we ready to face the challenge of part(b)? So far I have devised 3 different reversible paths between the initial and final states. But there are infinitely more. I just want to give others a chance to chime in. This is the most important part of what I’m trying to convey.

vanhees71
I have identified the several 2-step reversible processes between the initial and final states of our irreversible process. Please check them over and let me know if you agree they are reversible or, if you disagree, explain why you feel that they are not reversible. If you agree, please calculate the reversible work done on the surroundings, the heat added to the system, and the change in internal energy for each. In these calculations, it is valid to express the results in terms of n, To, Po, Vo, T1, P1, and V1.

PROCESS A
1. Isobaric cooling at pressure Po from To to T1
2. Isothermal expansion at temperature T1 from Po to P1

PROCESS B
1. Isothermal expansion at temperature To from Po to P1
2. Isobaric cooling at pressure P1 from To to T1

PROCESS C
1. Isothermal expansion at temperature To from Vo to V1
2. Isochoric cooling at volume V1 from To to T1

PROCESS D
1. Isochoric cooling at volume Vo from To to T1
2. Isothermal expansion at temperature T1 from Vo to V1

guhag and vanhees71
Chestermiller said:
@guhag Have you redone your analysis with the corrected volume relation. Are you now in agreement with @vanhees71 and me?

Are we ready to face the challenge of part(b)? So far I have devised 3 different reversible paths between the initial and final states. But there are infinitely more. I just want to give others a chance to chime in. This is the most important part of what I’m trying to convey.

Chet, Thanks for helping me out with getting my concepts clear. I appreciate all your help and time.

Yes, I reworked my analysis and got T1/T0 = (P1/P0)* [1-(1/γ)] + (1/γ)

There will be a delay from my side in seeing/replying to your posts. Unfortunately, I can't check non-work related websites in the lab.

Last edited:
Chestermiller said:
I have identified the several 2-step reversible processes between the initial and final states of our irreversible process. Please check them over and let me know if you agree they are reversible or, if you disagree, explain why you feel that they are not reversible. If you agree, please calculate the reversible work done on the surroundings, the heat added to the system, and the change in internal energy for each. In these calculations, it is valid to express the results in terms of n, To, Po, Vo, T1, P1, and V1.

PROCESS A
1. Isobaric cooling at pressure Po from To to T1
2. Isothermal expansion at temperature T1 from Po to P1

PROCESS B
1. Isothermal expansion at temperature To from Po to P1
2. Isobaric cooling at pressure P1 from To to T1

PROCESS C
1. Isothermal expansion at temperature To from Vo to V1
2. Isochoric cooling at volume V1 from To to T1

PROCESS D
1. Isochoric cooling at volume Vo from To to T1
2. Isothermal expansion at temperature T1 from Vo to V1
Regarding reversibility or not: I don't know entropy based reasoning just yet... so, I am going to say that if a path exists between two states and the transformation is brought about slowly so that the system is in equilibrium and that it can be reversed, it should be reversible. I apologize if this is hand-wavy and can't tell what happens to the state of the surroundings (if they are going to get back to their original state or not). I would like to know how one tests reversibility of a process without second law (if possible).

When I thought about ways to achieve the transformation, I missed your Process C. And I have a Process E instead (isochoric cooling at V0 : change P0 to P1 followed by Isobaric expansion at P1, change V0 to V1)

I have attached a n image of my scratch sheet with my working for process B. Please let me know if I have messed up somewhere... I'll send my working for the other processes by tomorrow. Thanks.

#### Attachments

• IMG_20230719_213857176.jpg
33.6 KB · Views: 54
Last edited:
guhag said:
Regarding reversibility or not: I don't know entropy based reasoning just yet... so, I am going to say that if a path exists between two states and the transformation is brought about slowly so that the system is in equilibrium and that it can be reversed, it should be reversible. I apologize if this is hand-wavy and can't tell what happens to the state of the surroundings (if they are going to get back to their original state or not). I would like to know how one tests reversibility of a process without second law (if possible).
Your assessment regarding the system having to pass through a continuous sequence of thermodynamic equilibrium states for reversibility is correct. And it is always possible to find a corresponding path for the surroundings, such that the overall combination experiences a reversible process. And if this is the case, both the system and surroundings can be returned to their original states without affecting any other entity.

In classical thermodynamics, the only way we know of to determine the entropy change of a system is to devise a reversible path between the initial and final thermodynamic equilibrium state and to calculate the integral of dQ/T for that path. This can be used to determine the entropy change for an irreversible path between two end states also (i.e., using a reversible path). In part (c) we will show that the entropy change for all 4 reversible processes between the two end states is exactly the same.
guhag said:
When I thought about ways to achieve the transformation, I missed your Process C. And I have a Process E instead (isochoric cooling at V0 : change P0 to P1 followed by Isobaric expansion at P1, change V0 to V1)

I have attached a n image of my scratch sheet with my working for process B. Please let me know if I have messed up somewhere... I'll send my working for the other processes by tomorrow. Thanks.
I am not able to read your sheet (please click on LaTex Guide at the bottom of each window), but here is my analysis of reversible process B:

1. Isothermal expansion at temperature To from Po to P1

Since this step is isothermal, PV=const. Therefore, it follows that $$dW=PdV=\frac{nRT_0}{V}dV=-\frac{nRT_0}{P}dP$$So, $$W_1=-nRT_0\ln{\frac{P_1}{P_0}}=+nRT_0\ln{\frac{P_0}{P_1}}$$Since the expansion is isothermal, $$Q_1=W_1=+nRT_0\ln{\frac{P_0}{P_1}}$$and $$\Delta U_1=0$$

2. Isobaric cooling at pressure P1 from To to T1

$$W_2=\int{P_1dV}=P_1\Delta V=nR(T_1-T_0)$$
$$Q_2=nC_p(T_1-T_0)$$
$$\Delta U_2=nC_v(T_1-T_0)$$

OVERALL REVERSIBLE PROCESS B

$$W=W_1+W_2=nRT_0\ln{\frac{P_0}{P_1}}+nR(T_1-T_0)$$
$$Q=Q_1+Q_2=nRT_0\ln{\frac{P_0}{P_1}}+nC_p(T_1-T_0$$
$$\Delta U=nC_v(T_1-T_0)$$

Last edited:
guhag and vanhees71
Chestermiller said:
Your assessment regarding the system having to pass through a continuous sequence of thermodynamic equilibrium states for reversibility is correct. And it is always possible to find a corresponding path for the surroundings, such that the overall combination experiences a reversible process. And if this is the case, both the system and surroundings can be returned to their original states without affecting any other entity.

In classical thermodynamics, the only way we know of to determine the entropy change of a system is to devise a reversible path between the initial and final thermodynamic equilibrium state and to calculate the integral of dQ/T for that path. This can be used to determine the entropy change for an irreversible path between two end states also (i.e., using a reversible path). In part (c) we will show that the entropy change for all 4 reversible processes between the two end states is exactly the same.

I am not able to read your sheet (please click on LaTex Guide at the bottom of each window), but here is my analysis of reversible process B:

1. Isothermal expansion at temperature To from Po to P1

Since this step is isothermal, PV=const. Therefore, it follows that $$dW=PdV=\frac{nRT_0}{V}dV=-\frac{nRT_0}{P}dP$$So, $$W_1=-nRT_0\ln{\frac{P_1}{P_0}}=+nRT_0\ln{\frac{P_0}{P_1}}$$Since the expansion is isothermal, $$Q_1=W_1=+nRT_0\ln{\frac{P_0}{P_1}}$$and $$\Delta U_1=0$$

2. Isobaric cooling at pressure P1 from To to T1

$$W_2=\int{P_1dV}=P_1\Delta V=nR(T_1-T_0)$$
$$Q_2=nC_p(T_1-T_0)$$
$$\Delta U_2=nC_v(T_1-T_0)$$

OVERALL REVERSIBLE PROCESS B

$$W=W_1+W_2=nRT_0\ln{\frac{P_0}{P_1}}+nR(T_1-T_0)$$
$$Q=Q_1+Q_2=nRT_0\ln{\frac{P_0}{P_1}}+nC_p(T_1-T_0$$
$$\Delta U=nC_v(T_1-T_0)$$
Hi Chet, thank you for showing your analysis and explanation of reversibility of a process being tied to entropy. I haven't used LaTeX at all, so it is another thing to learn. You mentioned that you were not able to read the image of my scratchsheet. I have converted the image to a pdf and attached it here. Is this any better? Or, by unreadable, did you mean my handwriting?

#### Attachments

• IMG_20230719_213857176.pdf
1.9 MB · Views: 54
guhag said:
Hi Chet, thank you for showing your analysis and explanation of reversibility of a process being tied to entropy. I haven't used LaTeX at all, so it is another thing to learn. You mentioned that you were not able to read the image of my scratchsheet. I have converted the image to a pdf and attached it here. Is this any better? Or, by unreadable, did you mean my handwriting?
In Process B, the volume at the end of step 1 is not V1. It is ##V_0\frac{P_0}{P_1}##.

vanhees71
That's an embarrassing error ! I have corrected it. I have worked the U, Q, W for the other processes A,C,D and E. Please find my working attached. I will start learning to type in LaTeX for the future.

#### Attachments

• Process A, C, D and E.pdf
261.8 KB · Views: 59
guhag said:
That's an embarrassing error ! I have corrected it. I have worked the U, Q, W for the other processes A,C,D and E. Please find my working attached. I will start learning to type in LaTeX for the future.
What happened to the n in ##\Delta U=nC_v\Delta T## and ##\Delta H=nC_p\Delta T##? Please correct that in your writeup.

Please make a graph of the overall process work W as a function of Po/P1 (1.0 to 10.0) at constant ##\gamma = 1.4## for the 4 reversible paths and the irreversible path. Instead of W, please plot the dimensionless work ##\frac{W}{nRT_0}##. We want to see how the 4 amounts of reversible work compare with the irreversible work and show that the work is different between the reversible paths

Last edited:
Chestermiller said:
What happened to the n in ##\Delta U=nC_v\Delta T## and ##\Delta H=nC_p\Delta T##? Please correct that in your writeup.

Please make a graph of the overall process work W as a function of Po/P1 (1.0 to 10.0) at constant ##\gamma = 1.4## for the 4 reversible paths and the irreversible path. Instead of W, please plot the dimensionless work ##\frac{W}{nRT_0}##. We want to see how the 4 amounts of reversible work compare with the irreversible work and show that the work is different between the reversible paths
I switched between using 1 mole for some and n moles for others ! Actually, I haven't calculated delta H, but will do so. and correct the errors related to #moles of the gas. I'll plot these graphs and upload tonight.

guhag said:
I switched between using 1 mole for some and n moles for others ! Actually, I haven't calculated delta H, but will do so. and correct the errors related to #moles of the gas. I'll plot these graphs and upload tonight.
Delta H is equal to Q at constant pressure.
If you’re given Po, Vo, and To, n is determined .

Please plot all the curves on the same graph.

Last edited:
Chestermiller said:
Delta H is equal to Q at constant pressure.
If you’re given Po, Vo, and To, n is determined .

Please plot all the curves on the same graph.
Sorry I couldn't get to it today, a long day at work. I'll post it today (Saturday).=

guhag said:
Sorry I couldn't get to it today, a long day at work. I'll post it today (Saturday).=
Please select logarithmic for the Po/P1 axis scale.

I'm pretty late to the party but I'd like to understand the problem posted because I'm having quite some trouble to understand it. In this case, I don't think it is I don't understand the math or know the formulas, I think I am missing some key fundamentals but I don't know which.

The way I see it, the work done by the adiabatic expansion of the gas inside the piston must be equal to the work done in the isobaric compression of the gas outside the piston. Is that right? Energy has no other place to go since heat transfer is not allowed (adiabatic system) and any other path for energy has been restricted (frictionless, massless, etc.).
The gas inside the piston will expand until its pressure levels with the outside pressure (P0→P1).
The energy required for such expansion would come from the internal energy of the gas inside the piston which can be expressed as a function of temperature (ideal gas).

However, I don't see how those two Works (adiabatic expansion in the piston and isobaric expansion outside) can be the same.

I'd need to include something else. For example, if the piston had mass, its kinetic energy would be ##K_p = W_1-W_2##.

I feel like that's a realistic scenario. @Chestermiller you mention how you drop the external pressure to P1. I'll try to propose an equivalent description for the problem. Let's say the piston is locked. Outside the piston, there is the usual ##P_{atm}##. If the piston's pressure is increased and then the locking mechanism is released, then the difference between inner and outer pressure would result in the expansion of the chamber. The kinetic energy from the piston would come from the difference in the works I wrote.

What am I misinterpreting to get such a different answer when compared with the rest of the thread? I feel it's related to the assumption "the work done by the adiabatic expansion of the gas inside the piston will be equal to the work done in the isobaric compression of the gas outside the piston" but I don't see how that could be wrong.

#### Attachments

• 1690041884504.png
10.4 KB · Views: 85
Juanda said:
I'm pretty late to the party but I'd like to understand the problem posted because I'm having quite some trouble to understand it. In this case, I don't think it is I don't understand the math or know the formulas, I think I am missing some key fundamentals but I don't know which.

The way I see it, the work done by the adiabatic expansion of the gas inside the piston must be equal to the work done in the isobaric compression of the gas outside the piston. Is that right? Energy has no other place to go since heat transfer is not allowed (adiabatic system) and any other path for energy has been restricted (frictionless, massless, etc.).
The gas inside the piston will expand until its pressure levels with the outside pressure (P0→P1).
The energy required for such expansion would come from the internal energy of the gas inside the piston which can be expressed as a function of temperature (ideal gas).

However, I don't see how those two Works (adiabatic expansion in the piston and isobaric expansion outside) can be the same.
View attachment 329520

I'd need to include something else. For example, if the piston had mass, its kinetic energy would be ##K_p = W_1-W_2##.

I feel like that's a realistic scenario. @Chestermiller you mention how you drop the external pressure to P1. I'll try to propose an equivalent description for the problem. Let's say the piston is locked. Outside the piston, there is the usual ##P_{atm}##. If the piston's pressure is increased and then the locking mechanism is released, then the difference between inner and outer pressure would result in the expansion of the chamber. The kinetic energy from the piston would come from the difference in the works I wrote.

What am I misinterpreting to get such a different answer when compared with the rest of the thread? I feel it's related to the assumption "the work done by the adiabatic expansion of the gas inside the piston will be equal to the work done in the isobaric compression of the gas outside the piston" but I don't see how that could be wrong.
Is the piston still moving when the system equilibrates? If the piston has very little mass, does the gas still execute an adiabatic reversible expansion as suggested by your equation?

This post is a distraction from the issues and direction I wanted this thread to follow. Please resubmit these questions in a separate thread, and I'll be glad to answer. Thanks.

Chestermiller said:
Please select logarithmic for the Po/P1 axis scale.
Please find attached the graphs for reversible processes A through D alongwith the irreversible one. Apologies for the delay, Chet. Thanks so much for your time and help.

#### Attachments

• Process A, C, D and E - work.pdf
113 KB · Views: 71
guhag said:
Please find attached the graphs for reversible processes A through D alongwith the irreversible one. Apologies for the delay, Chet. Thanks so much for your time and help.
I haven't checked the results for case A yet, but B looks correct. But cases C and D are not. Dimensionless work for C should be ln(V1/V0). The work for the irreversible case is definitely wrong; at Po/P1 =1, dimensionless work should be zero.

When I asked for a log scale on the abscissa, I didn't mean to show equal increments along the horizontal axis; I meant for you to select logarithmic scale on your graphics software. For case B, doing this I got

I didn't follow your remark on the logarithms. I used log(po/p1) for the X-coordinate.

I reworked the calculations. Seems I was using the sign convention for W and Q followed in chemistry which is opposite to that followed in engineering !

#### Attachments

• Process A, C, D and E - work.pdf
107.8 KB · Views: 65
guhag said:
I didn't follow your remark on the logarithms. I used log(po/p1) for the X-coordinate.
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.
guhag said:
I reworked the calculations. Seems I was using the sign convention for W and Q followed in chemistry which is opposite to that followed in engineering !
B, C, D, and irreversible look correct, but not A. Pl\ease recheck your calculation for A. A should be lower than B.

This post deviates from the original intention of the thread. Disregard it.

Chestermiller said:
Is the piston still moving when the system equilibrates? If the piston has very little mass, does the gas still execute an adiabatic reversible expansion as suggested by your equation?
I don't see how the system equilibrates. I see it would keep bouncing like a spring system. I guess that's where the irreversibility comes into place since it'd be the gas consuming the energy. But I don't see how the gas consumes that energy. Probably it's eventually transformed into heat but that heat cannot go anywhere (adiabatic) so then it'll be thermal energy in the gas.

Chestermiller said:
This post is a distraction from the issues and direction I wanted this thread to follow. Please resubmit these questions in a separate thread, and I'll be glad to answer. Thanks.
Sorry about that. It was not my intention. I'll leave it here and just read the comments in the thread. Maybe I'll find the issue later. If not, I'll try to open that separate thread as you suggested.

Juanda said:
This post deviates from the original intention of the thread. Disregard it.I don't see how the system equilibrates. I see it would keep bouncing like a spring system. I guess that's where the irreversibility comes into place since it'd be the gas consuming the energy. But I don't see how the gas consumes that energy. Probably it's eventually transformed into heat but that heat cannot go anywhere (adiabatic) so then it'll be thermal energy in the gas.Sorry about that. It was not my intention. I'll leave it here and just read the comments in the thread. Maybe I'll find the issue later. If not, I'll try to open that separate thread as you suggested.

Chestermiller said:
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.

Juanda said:
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.

SammyS and Juanda
Chestermiller said:
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.

#### Attachments

• Process A, C, D and E - work.pdf
105.5 KB · Views: 65
Last edited:
guhag said:
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.

Chestermiller said:
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.

Chestermiller said:
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

Last edited:

• Thermodynamics
Replies
22
Views
2K
• Thermodynamics
Replies
12
Views
2K
• Thermodynamics
Replies
2
Views
2K
• Thermodynamics
Replies
69
Views
4K
• Thermodynamics
Replies
4
Views
1K
• Thermodynamics
Replies
4
Views
1K
• Thermodynamics
Replies
30
Views
2K
• Thermodynamics
Replies
5
Views
3K
• Biology and Chemistry Homework Help
Replies
1
Views
245
• Thermodynamics
Replies
1
Views
1K