Thermo - PV work and mechanical work

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  • #51
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Hello,
If I am not wrong the issue is why displacement work is an inexact differential whereas mechanical work is not always denoted as an exact differential?:biggrin:

Short answer: W=∫Fdx is the generalized equation for mechanical work ,where,

It's most appropriate to write W=∫δw and NOT ∫dw as work is a path dependent process.
So, writing W=∫dw or dw=Fdx is a serious malpractice.

Explanation: You need to specify the properties pressure and volume for a series of quasi static or equilibrium states to specify the path upon which work can be calculated. As work is defined as the area under the curve between the properties P and V in a displacement process one needs to add up all the elemental work for each equilibrium states to get total work done.As there can be more than one quasi static paths possible between two end states of a thermodynamic process different work transfers are possible.Thus work is essentially a path dependent process.

Even when considering mechanical extension or compression of a spring you need intermediate equilibrium positions to be achieved after each oscillation restores to a new equilibrium position.Thus to obtain work using integration we need a series of equilibrium states to specify the path upon which work is to be calculated hence work is practically always an inexact differential.

So it's always appropriate to write δw=Fdx or δw=PdV as it demands the application of mathematical integration to obtain total work.

Hope this helps
Best regards
 
  • #52
21,489
4,867
Hello,
If I am not wrong the issue is why displacement work is an inexact differential whereas mechanical work is not always denoted as an exact differential?:biggrin:

Short answer: W=∫Fdx is the generalized equation for mechanical work ,where,

It's most appropriate to write W=∫δw and NOT ∫dw as work is a path dependent process.
So, writing W=∫dw or dw=Fdx is a serious malpractice.

Explanation: You need to specify the properties pressure and volume for a series of quasi static or equilibrium states to specify the path upon which work can be calculated. As work is defined as the area under the curve between the properties P and V in a displacement process one needs to add up all the elemental work for each equilibrium states to get total work done.As there can be more than one quasi static paths possible between two end states of a thermodynamic process different work transfers are possible.Thus work is essentially a path dependent process.

Even when considering mechanical extension or compression of a spring you need intermediate equilibrium positions to be achieved after each oscillation restores to a new equilibrium position.Thus to obtain work using integration we need a series of equilibrium states to specify the path upon which work is to be calculated hence work is practically always an inexact differential.

So it's always appropriate to write δw=Fdx or δw=PdV as it demands the application of mathematical integration to obtain total work.

Hope this helps
Best regards
For an irreversible process, you can't assume that the work can be obtained by integrating over a quasistatic sequence of equilibrium states. This is only correct to do for a reversible process.

Chet
 
  • #53
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0
Hi!

I've read part of the discussion, and I'd have a related question.
I believe you can always (?) choose the environment so that its changes occur reversibly. Why is this i.e. why can you choose it that way?
 
  • #54
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Hi!

I've read part of the discussion, and I'd have a related question.
I believe you can always (?) choose the environment so that its changes occur reversibly. Why is this i.e. why can you choose it that way?
I don't quite understand this question. Are you saying that you can always find a reversible path between the initial and final equilibrium states of a (closed) system? Or, are you saying that, even for an irreversible path between the initial and final equilibrium states of a closed system, you can always somehow view the irreversible path from some perspective as being a reversible path?

Have you read my Physics Forums blog?

Chet
 
  • #55
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I don't quite understand this question. Are you saying that you can always find a reversible path between the initial and final equilibrium states of a (closed) system? Or, are you saying that, even for an irreversible path between the initial and final equilibrium states of a closed system, you can always somehow view the irreversible path from some perspective as being a reversible path?

Have you read my Physics Forums blog?

Chet
Thanks for the fast reply!
I'm actually looking at your blog right now.
And sorry. But I feel like I can't be more specific with the question before I understand the answer.
But I'll try, this is something I read from a book by Reiss. Reiss uses the wording "the environment (surroundings, outside the system in which there might occur an irreversible change) has been chosen to always behave reversibly". He refers to one of his chapters for a proof of this, but in that chapter I only find an example of a single process, hardly a proof.

Now that I think of it, I am not sure whether the statement requires for the system to be closed or not (the system is of course not isolated). But in the situations in which this is supposed to apply, this is supposed to help the examination of the system, if we know anything about the surroundings. For if the changes in the surroundings are reversible, then it is supposed to have well-defined quantities, and I can always define the entropy change, work/heat change, related to the possibly irreversible changes of the system.

I would appreciate any clarification:
-is that statement true (surroundings behave reversibly), and when?
-if so, how can you motivate/prove that it's true

EDIT: The idea of a reversible behaviour of the surroundings sounds like a very general concept, as he uses this, for example, to show that the work done reversibly is always greater than work done irreversibly (for the same change in a system)
[Here, he assumes an irreversible change in a system. For this change, there is a corresponding change in heat, [itex]Dq[/itex] (he uses D for not necessarily reversible changes). Here he states that as the surroundings behave reversibly, [itex]Dq = dq_{surr} = TdS_{surr}[/itex]. From here he goes onto writing the entropy change for system + surroundings, and so on..]
 
Last edited:
  • #56
21,489
4,867
Thanks for the fast reply!
I'm actually looking at your blog right now.
And sorry. But I feel like I can't be more specific with the question before I understand the answer.
But I'll try, this is something I read from a book by Reiss. Reiss uses the wording "the environment (surroundings, outside the system in which there might occur an irreversible change) has been chosen to always behave reversibly". He refers to one of his chapters for a proof of this, but in that chapter I only find an example of a single process, hardly a proof.

Now that I think of it, I am not sure whether the statement requires for the system to be closed or not (the system is of course not isolated). But in the situations in which this is supposed to apply, this is supposed to help the examination of the system, if we know anything about the surroundings. For if the changes in the surroundings are reversible, then it is supposed to have well-defined quantities, and I can always define the entropy change, work/heat change, related to the possibly irreversible changes of the system.

I would appreciate any clarification:
-is that statement true (surroundings behave reversibly), and when?
-if so, how can you motivate/prove that it's true

EDIT: The idea of a reversible behaviour of the surroundings sounds like a very general concept, as he uses this, for example, to show that the work done reversibly is always greater than work done irreversibly (for the same change in a system)
[Here, he assumes an irreversible change in a system. For this change, there is a corresponding change in heat, [itex]Dq[/itex] (he uses D for not necessarily reversible changes). Here he states that as the surroundings behave reversibly, [itex]Dq = dq_{surr} = TdS_{surr}[/itex]. From here he goes onto writing the entropy change for system + surroundings, and so on..]
I'm not familiar with Reiss' book, so I can't comment intelligently. I'm going to try to articulate what you are saying to see if I understand correctly.

He is saying that, even if a (closed) system experiences an irreversible process, it is possible to identify one or more reversible paths for the surroundings that imposes the exact same irreversible path on the system. Correct?

Chet
 
  • #57
8
0
I'm not familiar with Reiss' book, so I can't comment intelligently. I'm going to try to articulate what you are saying to see if I understand correctly.

He is saying that, even if a (closed) system experiences an irreversible process, it is possible to identify one or more reversible paths for the surroundings that imposes the exact same irreversible path on the system. Correct?

Chet

Yes! Or, rather (the bolded part) irreversible change (I think). Actually quite like you suggested earlier, I think Reiss says that between any two points on the state space one can find the reversible path between them.
The most counterintuitive part is that we would be able to always "find" that path from the surroundings.
[Now that I think of it it's actually kind of funny that we always "end up" on a point in the state space even with an irreversible change. I guess that's because of the "static" nature of thermodynamics - in the end of a process we are at some kind of equilibrium].
 
  • #58
183
2
For an irreversible process, you can't assume that the work can be obtained by integrating over a quasistatic sequence of equilibrium states. This is only correct to do for a reversible process.

Chet

Hi friend,
You are correct!But I already mentioned about the reversible path when I say quasi static process nevertheless I was speaking solely about reversible work.All quasi static processes are reversible.

For an irreversible process we cannot use integration.
 
  • #59
21,489
4,867
Hi friend,
You are correct!But I already mentioned about the reversible path when I say quasi static process nevertheless I was speaking solely about reversible work.All quasi static processes are reversible.

For an irreversible process we cannot use integration.
All quasi static processes are not reversible, because you can still have irreversible heat transfer (involving temperature gradients) occurring within the system. Just consider the transient heating of a solid by conduction. The solid is not deforming, so the process is quasi static, but it definitely is not reversible.

Chet
 
  • #60
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0
Chestermiller: So, do you know whether the statement holds?
And moreover, why does/doesn't it?
["He is saying that, even if a (closed) system experiences an irreversible process, it is possible to identify one or more reversible paths for the surroundings that imposes the exact same irreversible path on the system."]
 
  • #61
21,489
4,867
Chestermiller: So, do you know whether the statement holds?
And moreover, why does/doesn't it?
["He is saying that, even if a (closed) system experiences an irreversible process, it is possible to identify one or more reversible paths for the surroundings that imposes the exact same irreversible path on the system."]
No, I really don't know. I've never thought much about the path that the surroundings experiences. In my judgement, the focus should always be on the system. And, to get the entropy change of the system, I think we can always dream up a reversible path to get the entropy change. I guess I've never had much motivation for looking at the process experienced by the surroundings.

Maybe you can pick a specific example of an irreversible path for the system, and we can see whether we can dream up a reversible path for the surroundings that puts the system through the exact same irreversible path.

Chet
 
  • #62
8
0
No, I really don't know. I've never thought much about the path that the surroundings experiences. In my judgement, the focus should always be on the system. And, to get the entropy change of the system, I think we can always dream up a reversible path to get the entropy change. I guess I've never had much motivation for looking at the process experienced by the surroundings.

Maybe you can pick a specific example of an irreversible path for the system, and we can see whether we can dream up a reversible path for the surroundings that puts the system through the exact same irreversible path.

Chet

I also have a difficult time believing that this should always be the case.

But perhaps this should be viewed as just a "trick"/approximation to "at least be able to calculate something". [For example, I'm pretty sure my lecturer thinks that this is an "ok approach", and I just feel cognitive pain trying to understand/motivate it to myself.]

I believe there tries to be a philosophy like this:
If the system undergoes an irreversible change, some of its thermodynamical variables are not well-defined, right?
However, in the process the energy of the system changes, and work and heat are transferred between the system and its surroundings. And what leaves the system, enters its surroundings.
Now if the environment is big, one can approximate that its thermodynamic variables don't change much (at least the convenient ones...).
On the other hand one has to think that the environment behaves reversibly, but at the same time the environment should have super fast "relaxation times". This would allow me to make a macroscopic change in the system at a finite rate while only making reversible changes in the surroundings.
And, with this logic I get the changes in the system indirectly from the changes that happened in the surroundings.

I'm not sure, this is just how I've tried to understand it, now having read thermodynamics from a few different sources. Although most sources don't explicitly talk much about this. Some just use this and some just don't introduce problems where this kind of approach could be used.

For example, this seems to give ok results (well, at least results) in the adiabatic, irreversible compression of an ideal gas in a cylinder. In the problem we only know the initial state of the gas, the pressure of the surrounding air (and the area of the piston and the weight we suddenly put on it), and want to calculate work and the change in height.
[I've understood that the relation [itex]PV^{γ}=constant[/itex] only works for reversible processes, which is why this is not straightforward.]

[Sorry about my bad english, btw.]
 

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