Conceptualizing entropy change of surroundings

In summary: However, this path is known to be isothermal. The entropy change (of the system) in going from A-->D is simply the sum of the individual entropy changes (of the system) in each of the above steps (since entropy is a state function and since each step is reversible). Now comes the confusing part:The entropy change of the surroundings in going from A-->D is calculated by summing up the heat transferred to the surroundings in each of the individual steps, and then divided by the temperature of the isothermal process A-->D. This confuses me greatly. I don't understand why this approach is used
  • #1
Bipolarity
776
2
I asked a question on this forum a few days ago about the entropy change of the surroundings, and am grateful for the insight provided. However, something faulty in my conceptualization is preventing me from solving this problem.

Let's say you have a set processes shown in the following diagram, where each of the arrows indicates a reversible process for which q and w can all be calculated:

A --> B --> C --> D

Suppose that there is a direct path A-->D, but this path is irreversible and the heat of this process is unknown. However, this path is known to be isothermal.

The entropy change (of the system) in going from A-->D is simply the sum of the individual entropy changes (of the system) in each of the above steps (since entropy is a state function and since each step is reversible).

Now comes the confusing part:
The entropy change of the surroundings in going from A-->D is calculated by summing up the heat transferred to the surroundings in each of the individual steps, and then divided by the temperature of the isothermal process A-->D.

This confuses me greatly. I don't understand why this approach is used in calculating the surroundings' entropy change.

Perphaps someone more experienced in thermo can explain?

BiP
 
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  • #2
Bipolarity said:
The entropy change (of the system) in going from A-->D is simply the sum of the individual entropy changes (of the system) in each of the above steps (since entropy is a state function and since each step is reversible).
You have to be able to define states A and D first. Once you do that, you find a reversible path between states A and D, determine the heatflow for that path (ie the reversible heatflow) and then divide by T to determine the change in entropy.

AM
 
  • #3
Maybe I don't quite understand your setup, nevertheless, I'll try to give quite trivial a counter example: Let's say my system consists of a resistor which is well cooled.
The states A, B, C, D of the system are taken to be all the same and the entropy, heat, work etc to go from one to the other are all trivially =0. Nevertheless I can go from A to D irreversibly creating an arbitrary amount of heat by letting flow a current through the resistor for an arbitrary amount of time.
 
  • #4
Andrew Mason said:
You have to be able to define states A and D first. Once you do that, you find a reversible path between states A and D, determine the heatflow for that path (ie the reversible heatflow) and then divide by T to determine the change in entropy.

AM

OK here is an example of my states:

- All states have 1 atm pressure.
- All states have the same volume.

A: water at 283K
B: water at 273K
C: ice at 273L
D: ice at 283K

As you can see, the freezing of water into ice is only reversible if the process is done at equilibrium conditions, i.e. 273K because that is the freezing point of water, at which the Gibbs energy change is 0. That is, B-->C represents the reversible freezing of water, whereas A--D represents the irreversible freezinong of water.

So to calculate the system's entropy change from A-->D, we would construct the reversible path A-->B-->C-->D, and then sum up the entropy change in each (reversible) step. Right?

Now what about the surroundings? The surrounding's entropy change should just be the heat (transferred to surroundings) of the process divided by 283K. Does "heat of the process" refer to heat in process A + heat in process B + heat in process C?

I seem to have reached some paradox. The entropy change of the universe in A-->D must be >0 because the overall process is irreversible (by second law). Thus, the entropy change of the surroundings in A-->D must be greater in magnitude than that of the system (if the system's entropy change is negative).

But in each process A-->B, B--C, C--D, the entropy change of the surroundings has equal magnitude to that of the system. What's the deal here? Is the entropy change of the surroundings not a state function?

BiP
 
  • #5
There won't be ice at 283 K at standard pressure.
 
  • #6
DrDu said:
There won't be ice at 283 K at standard pressure.

Yes, it's not a spontaneous process obviously, but what if we were to calculate the entropy change of this process? It's still a process, whether or not it actually occurs...

BiP
 
  • #7
Bipolarity said:
I asked a question on this forum a few days ago about the entropy change of the surroundings, and am grateful for the insight provided. However, something faulty in my conceptualization is preventing me from solving this problem.

Let's say you have a set processes shown in the following diagram, where each of the arrows indicates a reversible process for which q and w can all be calculated:

A --> B --> C --> D

Suppose that there is a direct path A-->D, but this path is irreversible and the heat of this process is unknown. However, this path is known to be isothermal.

The entropy change (of the system) in going from A-->D is simply the sum of the individual entropy changes (of the system) in each of the above steps (since entropy is a state function and since each step is reversible).

Now comes the confusing part:
The entropy change of the surroundings in going from A-->D is calculated by summing up the heat transferred to the surroundings in each of the individual steps, and then divided by the temperature of the isothermal process A-->D.

BiP

No. The sum of the heats transferred in the individual reversible steps does not necessarily sum to the heat transferred in the irreversible isothermal process. Just take the simple one-step example of a spontaneous isobaric expansion (with heat supplied by a constant temperature reservoir held at the initial temperature) compared to a reversible isothermal expansion at the initial temperature (exchanging heat with a constant temperature reservoir at the same initial temperature). The heats transferred and the works done in these two processes will be different, although, for an ideal gas, the change in internal energy and change in entropy for the system will be the same. But, the entropy change in the surroundings for the irreversible process will be higher. Calculate it in an example and see.

Chet
 
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  • #8
Bipolarity said:
OK here is an example of my states:

- All states have 1 atm pressure.
- All states have the same volume.

A: water at 283K
B: water at 273K
C: ice at 273L
D: ice at 283K
Did you mean 263K for D?
As you can see, the freezing of water into ice is only reversible if the process is done at equilibrium conditions, i.e. 273K because that is the freezing point of water, at which the Gibbs energy change is 0. That is, B-->C represents the reversible freezing of water, whereas A--D represents the irreversible freezinong of water.

So to calculate the system's entropy change from A-->D, we would construct the reversible path A-->B-->C-->D, and then sum up the entropy change in each (reversible) step. Right?
Correct. What would a reversible path from A to B be? hint: try a combination of adiabatic and isothermal changes (compression or expansions).

AM
 
  • #9
Oh damn that was a mistake. Andrew it is indeed 263K you are correct.

I will restate the problem:
A: water at 263K
B: water at 273K
C: ice at 273K
D: ice at 263k

Water freezing to ice at 263K (i.e. process A-->D) is not reversible. Its entropy change of system can be calculated by summing up individual entropy changes of the system in each reversible step.

How would you calculate the entropy change of the surroundings in freezing a certain amount of water at 263K?

BiP
 
  • #10
Tell us first how you would obtain liquid water at 263°K
 
  • #11
Studiot said:
Tell us first how you would obtain liquid water at 263°K

There are supercooled water droplets below freezing point in the stratosphere...
But I don't see how that has any bearing on the problem.

The question asks to find the entropy change of the surroundings of water freezing to ice at 263K. Easily the Gibbs energy change of this reaction can be calculated, and it must be a spontaneous process.

But I am asking about the entropy change of the surroundings as the water isothermally freezes at this temperature. We can't use the latent heat of fusion because that assumes the water is in equilibrium while freezing (273K and 1atm).

Perhaps it's impossible to calculate the entropy change of the surroundings in this process? Maybe it needs to be empirically determined using a calorimeter and that's the only way?

Ideas? Thanks!

BiP
 
  • #12
Supercooled maybe, but have you checked the phase diagram for the conditions under which liquid water is possible at 263°K ?
 
  • #13
Studiot said:
Supercooled maybe, but have you checked the phase diagram for the conditions under which liquid water is possible at 263°K ?

Yes, water doesn't naturally exist at that point. It will turn into ice. And that is the point of my question... when it does turn into ice, what will the entropy change of the surroundings be for this process?

BiP
 
  • #14
If it can't exist how can it be there to turn into ice?
 
  • #15
Bipolarity said:
There are supercooled water droplets below freezing point in the stratosphere...
But I don't see how that has any bearing on the problem.

The question asks to find the entropy change of the surroundings of water freezing to ice at 263K. Easily the Gibbs energy change of this reaction can be calculated, and it must be a spontaneous process.

But I am asking about the entropy change of the surroundings as the water isothermally freezes at this temperature. We can't use the latent heat of fusion because that assumes the water is in equilibrium while freezing (273K and 1atm).

Perhaps it's impossible to calculate the entropy change of the surroundings in this process? Maybe it needs to be empirically determined using a calorimeter and that's the only way?

Ideas? Thanks!

BiP

Yes. The entropy change of the surroundings is not unique. Think of the surroundings as a separate system. In one case you can employ a high temperature gradient in the surroundings (holding the temperature at the boundary with your system fixed), and in another case you can employ a low temperature gradient in the surroundings (holding the temperature at the boundary with your system fixed at the same value). The entropy changes in the surroundings in these two cases differ, even though the entropy changes for the system are the same.
 
  • #16
If you really want the answer to your question get hold of a Mollier Diagram that goes down to your temperature range.

Mollier diagrams plot (specific) entropy against (specific) enthalpy for different pressure and temperature isolines.
 
  • #17
Bipolarity said:
Oh damn that was a mistake. Andrew it is indeed 263K you are correct.

I will restate the problem:
A: water at 263K
B: water at 273K
C: ice at 273K
D: ice at 263k

Water freezing to ice at 263K (i.e. process A-->D) is not reversible. Its entropy change of system can be calculated by summing up individual entropy changes of the system in each reversible step.
I am still confused. Is the liquid water initially at 263K or 283K? How are you getting water at 263K? Is this under high pressure or super-cooled conditions?

There is a reversible path between two normal states. So, for example, you could get the temperature changes needed by running a Carnot heat pump or a Carnot engine between the surroundings and the water/ice.

AM
 
  • #18
Ok, so all this confusing stuff about states A, B, C, D is really about freezing of water at -10 deg C.
You could have stated this at the very beginning!
I think there is no problem in supercooling water in that range.
The change in enthalpy is [itex]\Delta H(263 K)=\Delta H(273 K)+\int_{273 K}^{263 K} dT (C_{P, solid}(T)-C_{P, liquid}(T))[/itex].
The entropy change of the surrounding is ΔH/263K.
The change in free enthalpy of the system can also be calculated using the Gibbs-Helmholtz equation. From the free enthalpy change and the enthalpy change, also the entropy change of the system can be calculated.
 
  • #19
The path directly from A-D is irreversible but there is a reversible path from A-B-C-D.

[itex]\Delta S_{AD} = \Delta S_{AB} + \Delta S_{BC} + \Delta S_{CD} = \int_{263}^{273} mC_{water}dT/T - mL/273 + \int_{273}^{263}mC_{ice} dT/T [/itex]

[itex]\Delta S_{AD} = mC_{water}\ln{\frac{273}{263}} - mL/273 - mC_{ice}\ln{\frac{273}{263}}[/itex]

The change in entropy of the surroundings is just:

[itex]\Delta S_{surr} = (-10mC_{water} + ml + 10mC_{ice})/273[/itex]

AM
 
  • #20
DrDu said:
Ok, so all this confusing stuff about states A, B, C, D is really about freezing of water at -10 deg C.
You could have stated this at the very beginning!
I think there is no problem in supercooling water in that range.
The change in enthalpy is [itex]\Delta H(263 K)=\Delta H(273 K)+\int_{273 K}^{263 K} dT (C_{P, solid}(T)-C_{P, liquid}(T))[/itex].
The entropy change of the surrounding is ΔH/263K.
The change in free enthalpy of the system can also be calculated using the Gibbs-Helmholtz equation. From the free enthalpy change and the enthalpy change, also the entropy change of the system can be calculated.

Andrew Mason said:
The path directly from A-D is irreversible but there is a reversible path from A-B-C-D.

[itex]\Delta S_{AD} = \Delta S_{AB} + \Delta S_{BC} + \Delta S_{CD} = \int_{263}^{273} mC_{water}dT/T - mL/273 + \int_{273}^{263}mC_{ice} dT/T [/itex]

[itex]\Delta S_{AD} = mC_{water}\ln{\frac{273}{263}} - mL/273 - mC_{ice}\ln{\frac{273}{263}}[/itex]

The change in entropy of the surroundings is just:

[itex]\Delta S_{surr} = (-10mC_{water} + ml + 10mC_{ice})/273[/itex]

AM

Thank you all so much! I just want to make sure I understood this correctly:
- The entropy change of the system in going from A to D is simply the sum of the entropy changes of the system in each reversible step A to B, B to C, and C to D.
- The entropy change of the surroundings in going from A to D is NOT equal to the sum of the entropy changes of the surroundings in each reversible step A to B, B to C, and C to D.
- Rather, the entropy change of the surroundings must be calculated by finding the heat of the reaction A to D.
- Since heat is a path function, we need some way to calculate it in this strange path of water freezing at 263K.
- We know that at constant pressure, heat = enthalpy which does happen to be a state function.
- We can easily calculate enthalpy for this reaction A to D. We do so, and equate with heat.
- Finally, we divide the enthalpy of A to D by the temperature of this reaction to get the entropy change of the surroundings.
- We note that whatever value we get must be such that the entropy of the universe is positive.

If anywhere I am mistaken pleease correct me! Thanks all for the help!

BiP
 
  • #21
Bipolarity said:
Thank you all so much! I just want to make sure I understood this correctly:
- The entropy change of the system in going from A to D is simply the sum of the entropy changes of the system in each reversible step A to B, B to C, and C to D.
Correct.
- The entropy change of the surroundings in going from A to D is NOT equal to the sum of the entropy changes of the surroundings in each reversible step A to B, B to C, and C to D.
How do you conclude that? Assume that you effect each of the changes by Carnot cycles between the water/ice and the constant temperature surroundings (surroundings have infinite heat capacity).

- Rather, the entropy change of the surroundings must be calculated by finding the heat of the reaction A to D.
No. The entropy of the surroundings is the sum of all the heat flows to and from the water/ice in the reversible path A-B-C-D, divided by the temperature of the surroundings (273K).

- Since heat is a path function, we need some way to calculate it in this strange path of water freezing at 263K.
To find the entropy change you have to use the reversible path A-B-C-D.

- We note that whatever value we get must be such that the entropy of the universe is positive.
And it does - see my calculation: Cwater>Cice and ln(273/263)>10/273

AM
 
  • #22
Andrew Mason said:
The path directly from A-D is irreversible but there is a reversible path from A-B-C-D.

[itex]\Delta S_{AD} = \Delta S_{AB} + \Delta S_{BC} + \Delta S_{CD} = \int_{263}^{273} mC_{water}dT/T - mL/273 + \int_{273}^{263}mC_{ice} dT/T [/itex]

[itex]\Delta S_{AD} = mC_{water}\ln{\frac{273}{263}} - mL/273 - mC_{ice}\ln{\frac{273}{263}}[/itex]

The change in entropy of the surroundings is just:

[itex]\Delta S_{surr} = (-10mC_{water} + ml + 10mC_{ice})/273[/itex]

AM
You mean divided by 263, not 273, in the last step.
 
  • #23
DrDu said:
You mean divided by 263, not 273, in the last step.
I thought the surroundings were at 273K.

If the surroundings are at 263K, one has to postulate an arbitrarily large reservoir at 273K that is in thermal contact with the surroundings and the system (via Carnot devices) in order to have the reversible path A-B-C-D.

But you are right if the surroundings are at 263K. The change in entropy of the surroundings would be:

[itex]\Delta S_{surr} = (-10mC_{water} + mL + 10mC_{ice})/263[/itex]

AM
 
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  • #24
Bipolarity said:
Thank you all so much! I just want to make sure I understood this correctly:
- The entropy change of the system in going from A to D is simply the sum of the entropy changes of the system in each reversible step A to B, B to C, and C to D.
- The entropy change of the surroundings in going from A to D is NOT equal to the sum of the entropy changes of the surroundings in each reversible step A to B, B to C, and C to D.
- Rather, the entropy change of the surroundings must be calculated by finding the heat of the reaction A to D.
- Since heat is a path function, we need some way to calculate it in this strange path of water freezing at 263K.
- We know that at constant pressure, heat = enthalpy which does happen to be a state function.
- We can easily calculate enthalpy for this reaction A to D. We do so, and equate with heat.
- Finally, we divide the enthalpy of A to D by the temperature of this reaction to get the entropy change of the surroundings.
- We note that whatever value we get must be such that the entropy of the universe is positive.

If anywhere I am mistaken pleease correct me! Thanks all for the help!

BiP
Sounds all correct to me.
 
  • #25
Bipolarity said:
- The entropy change of the surroundings in going from A to D is NOT equal to the sum of the entropy changes of the surroundings in each reversible step A to B, B to C, and C to D.
- Rather, the entropy change of the surroundings must be calculated by finding the heat of the reaction A to D.
DrDu and I seem to disagree on this part. In my view, this cannot be correct. The entropy change of the surroundings in going from A to D cannot depend on the path.

AM
 
  • #26
I would like to quote the words of my textbook, in case it helps anyone:
"So far, our discussion of entropy changes has focused on the system. For a proper understanding of entropy, we must also examine what happens to the surroundings. Because of its size and amount of material it contains, the surroundings can be thought of as an infinitely large heat reservoir. Therefore, the exchange of hat and work between a system and its surroundings alters the properties of the surroundings by only an infinitesimal amount. Because infinitesimal changes are characteristic of reversible processes, it follows that any process has the same effect on the surroundings as a reversible process. Thus, regardless of whether a process is reversible or irreversible with respect to the system, we can write the heat change in the surroundings as

[tex] (dq_{sur})_{rev} = (dq_{sur})_{irev} = dq_{sur} [/tex]



BiP
 
  • #27
Bipolarity said:
I would like to quote the words of my textbook, in case it helps anyone:
"So far, our discussion of entropy changes has focused on the system. For a proper understanding of entropy, we must also examine what happens to the surroundings. Because of its size and amount of material it contains, the surroundings can be thought of as an infinitely large heat reservoir. Therefore, the exchange of hat and work between a system and its surroundings alters the properties of the surroundings by only an infinitesimal amount. Because infinitesimal changes are characteristic of reversible processes, it follows that any process has the same effect on the surroundings as a reversible process. Thus, regardless of whether a process is reversible or irreversible with respect to the system, we can write the heat change in the surroundings as

[tex] (dq_{sur})_{rev} = (dq_{sur})_{irev} = dq_{sur} [/tex]
Correct. The change in entropy of the surroundings is the negative of the actual heat flow into the system that occurs during the actual process divided by the temperature of the surroundings.

If that is what you mean by " Rather, the entropy change of the surroundings must be calculated by finding the heat of the reaction A to D" then you are correct. But if the actual path is A-B-C-D for the system, which is what I understood from your example, then the change in entropy of the surroundings will be the sum of the heat flows out of the system/Tsurr: [itex]\Delta S_{surr} = -(Q_{AB} + Q_{BC} + Q_{CD})/T_{surr}[/itex]

AM
 
  • #28
Andrew Mason said:
Correct. The change in entropy of the surroundings is the negative of the actual heat flow into the system that occurs during the actual process divided by the temperature of the surroundings.

If that is what you mean by " Rather, the entropy change of the surroundings must be calculated by finding the heat of the reaction A to D" then you are correct. But if the actual path is A-B-C-D for the system, which is what I understood from your example, then the change in entropy of the surroundings will be the sum of the heat flows out of the system/Tsurr: [itex]\Delta S_{surr} = -(Q_{AB} + Q_{BC} + Q_{CD})/T_{surr}[/itex]

AM

I am sorry for the confusion Andrew, the "actual" reaction that the problem asked about was the irreversible direct path A-->D. I hope that resolves everything?

BiP
 
  • #29
Bipolarity said:
I would like to quote the words of my textbook, in case it helps anyone:
"So far, our discussion of entropy changes has focused on the system. For a proper understanding of entropy, we must also examine what happens to the surroundings. Because of its size and amount of material it contains, the surroundings can be thought of as an infinitely large heat reservoir. Therefore, the exchange of hat and work between a system and its surroundings alters the properties of the surroundings by only an infinitesimal amount. Because infinitesimal changes are characteristic of reversible processes, it follows that any process has the same effect on the surroundings as a reversible process. Thus, regardless of whether a process is reversible or irreversible with respect to the system, we can write the heat change in the surroundings as

[tex] (dq_{sur})_{rev} = (dq_{sur})_{irev} = dq_{sur} [/tex]



BiP

In my judgement, this is basically incorrect.

Of course, we must always have that

(dQsys) + (dQsurr) =0

Thus, (dQsys)rev + (dQsurr)rev =0

and (dQsys)irrev + (dQsurr)irrev =0

But since, in general,
(dQsys)rev≠ (dQsys)irrev

we must also have that

(dQsurr)rev≠ (dQsurr)irrev

Also, none of these relationship say anything about the change in entropy of the surroundings, since we don't know the details of how the heat from the surroundings was supplied, which, of course, determines the entropy change of the surroundings. We can always think of processes in which

1. the path for the system and the path for the surroundings is reversible
2. the path for the system is irreversible, but the path for the surroundings is reversible
3. the path for the system is reversible, but the path for the surroundings is irreversible
4. the paths for the system and the the path for surroundings is irreversible
 
  • #30
Chestermiller said:
In my judgement, this is basically incorrect.

Of course, we must always have that

(dQsys) + (dQsurr) =0

Thus, (dQsys)rev + (dQsurr)rev =0

and (dQsys)irrev + (dQsurr)irrev =0

But since, in general,
(dQsys)rev≠ (dQsys)irrev

we must also have that

(dQsurr)rev≠ (dQsurr)irrev

Also, none of these relationship say anything about the change in entropy of the surroundings, since we don't know the details of how the heat from the surroundings was supplied, which, of course, determines the entropy change of the surroundings. We can always think of processes in which

1. the path for the system and the path for the surroundings is reversible
2. the path for the system is irreversible, but the path for the surroundings is reversible
3. the path for the system is reversible, but the path for the surroundings is irreversible
4. the paths for the system and the the path for surroundings is irreversible

Everything you say makes perfect sense Chet, which continues to boggle me because then it implies that the the entropy change of the universe must be 0 for all processes reversible and irreversible, which is a direct contradiction to the second law.

I thank everyone for their insights provided so far, but it isn't over yet. I encourage all to continue posting their thoughts so we can deliberate on this matter further and reach a ground where everyone can agree.

The question then, is it reconcile Chet's argument with the second law of thermodynamics, to come up with a precise definition of the surrounding's entropy change. Only then can I (or others) solve textbook problems involving the surrounding's entropy change without having a fuzzy notion of what we are talking about.

BiP
 
  • #31
Bipolarity said:
Everything you say makes perfect sense Chet, which continues to boggle me because then it implies that the the entropy change of the universe must be 0 for all processes reversible and irreversible, which is a direct contradiction to the second law.

That is not how you calculate entropy change. When the system process is irreversible, a reversible path between the beginning and end points results in a different heat flow than the actual path. You use a reversible path BETWEEN THE BEGINNING AND END STATES for the system to calculate entropy change for the system.

Similarly, you use a reversible path BETWEEN THE BEGINNING AND END STATES for the surroundings to calculate entropy change for the surroundings.

The reversible paths for the surroundings and the system are the SAME for both ONLY if the actual path is reversible. If the actual path is irreversible, the reversible path for the system between its beginning and end states IS DIFFERENT than the reversible path between the surrounding's beginning and end states.

Example: an adiabatic free expansion of an ideal gas from (P,V,T) to (P/2,2V,T). There is no Q and no work is done on or by the surroundings by or on the gas. Since ΔU = Q-W = 0, there is no change in T for the gas.

The change in entropy of the gas, however, is determined by the reversible path between the beginning and end states. The reversible path from (P,V,T) to (P/2,2V,T) for the ideal gas is a quasi-static isothermal expansion in which work is done against an external pressure. Since ΔU=Q-W = 0 and W>0 then Qrev>0 so ΔS = ∫dQrev/T > 0.

HOWEVER, for the surroundings the beginning and end states of the surroundings, UNLIKE the gas, ARE IDENTICAL. There has been no heatflow to/from the surroundings. So there is no change in entropy of the surroundings.

AM
 
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  • #32
Andrew Mason said:
HOWEVER, for the surroundings the beginning and end states of the surroundings, UNLIKE the gas, ARE IDENTICAL. There has been no heatflow to/from the surroundings. So there is no change in entropy of the surroundings.

AM

Can't this argument be extended to any finite process which occurs in the laboratory, whether reversible or irreversible? In other words, wouldn't the entropy change of the surroundings always be 0 according to your logic, even for a reversible process?

BiP
 
  • #33
an adiabatic free expansion of an ideal gas from (P,V,T) to (P/2,2V,T).

Adiabatic or isothermal?

What do you mean by free expansion?

Please explain.
 
  • #34
Bipolarity said:
Can't this argument be extended to any finite process which occurs in the laboratory, whether reversible or irreversible? In other words, wouldn't the entropy change of the surroundings always be 0 according to your logic, even for a reversible process?
??No! If there is heatflow to/from the surroundings the entropy of the surroundings changes.

In the quasi static isothermal expansion of a gas, for example, there is heatflow from the surroundings to the system so the entropy of the surroundings decreases by the same amount that the entropy of the gas increases so the total entropy change is 0. In the free expansion of the gas, the entropy of the gas increases while the entropy of the surroundings is 0 so the total entropy change is > 0.

AM
 
  • #35
Bipolarity said:
Everything you say makes perfect sense Chet, which continues to boggle me because then it implies that the the entropy change of the universe must be 0 for all processes reversible and irreversible, which is a direct contradiction to the second law.

BiP

It doesn't imply this at all. At the boundary between the system and surroundings, the rate of heat flow from the surroundings to the system is always minus the rate of heat flow from the system to the surroundings, and the temperature at the boundary is the same for both the system and the surroundings, say TB. (This assumes that there is only one location where heat is being exchanged between the system and the surroundings). This all implies that dQ/TB for the system is always equal to -dQ/TB for the surroundings. But this definitely does not mean that the entropy change for the system has to be minus the temperature change for the surroundings. The entropy change for the system is only equal to dQ/TB if the system is undergoing a reversible path, and the entropy change for the surroundings is only equal to -dQ/TB if the surroundings are undergoing a reversible path. We can regard the system and the surroundings as really two separate systems that are mutually constrained at the boundary between them.

Now, for the system, we have that

(dS)syst = dQ/TB for a reversible path

(dS)syst > dQ/TB for an irreversible path

The second relationship represents the simplified version of the so-called Cauchy inequality.

And, for the surroundings,

(dS)surr = -dQ/TB for a reversible path

(dS)surr > -dQ/TB for an irreversible path

So if both system and surroundings undergo reversible paths, then

(dS)syst + (dS)surr = 0

If either the system or the surroundings or both undergo irreversible paths in the actual process, then

(dS)syst + (dS)surr > 0

BiP, yesterday I sent you an analysis I did of the irreversible heating of a metal bar by means of heat supplied by an isothermal reservoir (comprising the surroundings). This is a case where the system is undergoing a irreversible path, while the surroundings are experiencing a reversible path. I showed how to precisely calculate the change in entropy for both the system and the surroundings from the solution to the transient heat conduction equation, and how the positive definiteness of the sum total of the entropy changes for the system and the surroundings comes about (for this problem). I have attached that analysis to this reply for others involved in this thread to examine. This provides a specific example of how the Cauchy inequality evolved, and why the sum of the entropy changes for the system and surroundings is always greater than zero for an irreversible process.

Chet
 

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  • Irreversible heating of a Bar.doc
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<h2>1. What is entropy change of surroundings?</h2><p>Entropy change of surroundings is a measure of the increase or decrease in the disorder or randomness of the environment surrounding a system. It is often used to describe the change in heat transfer or energy distribution between a system and its surroundings.</p><h2>2. How is entropy change of surroundings calculated?</h2><p>The entropy change of surroundings can be calculated using the equation ΔS = q/T, where ΔS is the change in entropy, q is the heat transferred between the system and surroundings, and T is the temperature of the surroundings.</p><h2>3. What factors affect the entropy change of surroundings?</h2><p>The entropy change of surroundings is affected by the temperature of the surroundings, the amount of heat transferred, and the nature of the system and its surroundings. It is also influenced by the type of process (reversible or irreversible) and the efficiency of the system.</p><h2>4. How does the entropy change of surroundings relate to the second law of thermodynamics?</h2><p>The second law of thermodynamics states that the total entropy of a closed system and its surroundings always increases in a spontaneous process. This means that the entropy change of the surroundings will always be positive, as the system and surroundings tend towards a state of maximum disorder.</p><h2>5. Can the entropy change of surroundings be negative?</h2><p>Yes, the entropy change of surroundings can be negative if the system is undergoing a reversible process and the heat transfer is from the surroundings to the system. In this case, the surroundings become more ordered and the entropy decreases, while the system becomes more disordered and the entropy increases.</p>

1. What is entropy change of surroundings?

Entropy change of surroundings is a measure of the increase or decrease in the disorder or randomness of the environment surrounding a system. It is often used to describe the change in heat transfer or energy distribution between a system and its surroundings.

2. How is entropy change of surroundings calculated?

The entropy change of surroundings can be calculated using the equation ΔS = q/T, where ΔS is the change in entropy, q is the heat transferred between the system and surroundings, and T is the temperature of the surroundings.

3. What factors affect the entropy change of surroundings?

The entropy change of surroundings is affected by the temperature of the surroundings, the amount of heat transferred, and the nature of the system and its surroundings. It is also influenced by the type of process (reversible or irreversible) and the efficiency of the system.

4. How does the entropy change of surroundings relate to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system and its surroundings always increases in a spontaneous process. This means that the entropy change of the surroundings will always be positive, as the system and surroundings tend towards a state of maximum disorder.

5. Can the entropy change of surroundings be negative?

Yes, the entropy change of surroundings can be negative if the system is undergoing a reversible process and the heat transfer is from the surroundings to the system. In this case, the surroundings become more ordered and the entropy decreases, while the system becomes more disordered and the entropy increases.

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