Understanding reversible and irreversible process

[Soumalya]

Hello,
It is said that any process accompanied by dissipating effects is an irreversible process.For instance, a piston cylinder arrangement undergoing an expansion of a gas with heat lost as friction between surfaces is an irreversible process.Here the system does work on the surroundings with heat being rejected to the surroundings as a result of friction.The process can be summed up as:

Work transfer for the system= -W
Work transfer to the surroundings=+W

Heat lost from the system=-Q
Heat gained by the surroundings=+Q

Then logically the process could be reversed at any stage if external work is done by the surroundings on the system equal to W and heat lost as friction be added to the system so that both system and surroundings are restored to their initial conditions.So why it's still called an irreversible process?

 

[UltrafastPED]

Because you are injecting additional energy; to keep it going indefinitely would require an unlimited energy budget.

OTOH, a reversible system can recycle the energy already present - converting it repeatedly from potential to kinetic and back again, ad infinitum.

 

[Soumalya]

Because you are injecting additional energy; to keep it going indefinitely would require an unlimited energy budget.

OTOH, a reversible system can recycle the energy already present - converting it repeatedly from potential to kinetic and back again, ad infinitum.

Could you please explain it throughly when you say we are injecting additional energy?

 

[UltrafastPED]

Could you please explain it throughly when you say we are injecting additional energy?

You are applying work to the system.

 

[Soumalya]

Could you illustrate me a reversible process with an example and show how the process can be reversed without additional work?

 

[UltrafastPED]

Could you illustrate me a reversible process with an example and show how the process can be reversed without additional work?

You can find these in your textbook.

 

[Chestermiller]

Hello,
It is said that any process accompanied by dissipating effects is an irreversible process.For instance, a piston cylinder arrangement undergoing an expansion of a gas with heat lost as friction between surfaces is an irreversible process.Here the system does work on the surroundings with heat being rejected to the surroundings as a result of friction.The process can be summed up as:

Work transfer for the system= -W
Work transfer to the surroundings=+W

Heat lost from the system=-Q
Heat gained by the surroundings=+Q

Then logically the process could be reversed at any stage if external work is done by the surroundings on the system equal to W and heat lost as friction be added to the system so that both system and surroundings are restored to their initial conditions.So why it's still called an irreversible process?

Let's be precise about this. What exactly is the system you are referring to? (a) gas in cylinder or (b) combination of gas in cylinder plus piston. What makes you think that heat is lost as friction? If heat is generated as a result of friction, where does it go to? (a) into the gas within the cylinder or (b) into the surroundings.

Chet

 

[Andrew Mason]

Let's be precise about this. What exactly is the system you are referring to? (a) gas in cylinder or (b) combination of gas in cylinder plus piston. What makes you think that heat is lost as friction? If heat is generated as a result of friction, where does it go to? (a) into the gas within the cylinder or (b) into the surroundings.

Chet

The question says that the heat is lost to the surroundings, so the heat is not retained in the system.

If the work done by a process is not fully recoverable as useful work that can be applied to the system when the process is run in reverse, the system cannot return to its initial state. An example would be a Carnot engine that is used to lift a weight or stretch a spring, thereby storing the output as energy that can be completely converted to useful work. By reversing the direction of the engine (ie. turn it into a heat pump) that stored energy can be used to reverse the heat flows and return the system and surroundings to its initial state.

In this case, even if the work output of the otherwise reversible process was used to produce heat (ie. friction) that was retained in the system, that heat cannot be completely converted into useful work to reverse the process to return to the initial state. So it would necessariiy be irreversible.

AM

 

[Soumalya]

Let's be precise about this. What exactly is the system you are referring to? (a) gas in cylinder or (b) combination of gas in cylinder plus piston. What makes you think that heat is lost as friction? If heat is generated as a result of friction, where does it go to? (a) into the gas within the cylinder or (b) into the surroundings.

Chet

I am assuming the gas in the cylinder with the piston all together as the system and the heat generated is due to friction between the piston and the cylinder walls which gets lost to the surroundings through convection.

So the gas expands doing boundary work 'W' with 'Q' amount of energy lost as heat generated due to friction to the surroundings.

Then work done by the system is work received by the surroundings.Again heat lost by the system is heat received by the surroundings.

Theoretically the process should be reversed if 'W' amount of work is delivered externally to the system with addition of 'Q' amount of heat to the system.

Regardless of arguments it is said that "heat lost as friction is unrecoverable".

While it is true that you need additional work to make heat flow back to the system i.e, from a low temperature reservoir to a high temperature reservoir it would render the system back to it's initial state but would result in a change in the surroundings after the reversal.

Since for a reversible process, we must have both the system and surroundings restored to it's initial state the reversal should be without additional expenditure of energy.

But taking the example of a reversible isothermal process, for isothermal compression of a gas energy needs to be removed from the system as the gas is compressed to maintain a constant internal energy and hence temperature.If one needs to reverse the process at any stage the system must be expanded with addition of energy as heat which again would need additional expenditure of work.So any process associated with a heat transfer through finite temperature difference seems to be spontaneously irreversible!

But we assume a frictionless isothermal process to be perfectly reversible.Why?

 

[Chestermiller]

But taking the example of a reversible isothermal process, for isothermal compression of a gas energy needs to be removed from the system as the gas is compressed to maintain a constant internal energy and hence temperature.If one needs to reverse the process at any stage the system must be expanded with addition of energy as heat which again would need additional expenditure of work.So any process associated with a heat transfer through finite temperature difference seems to be spontaneously irreversible!

But we assume a frictionless isothermal process to be perfectly reversible.Why?

You described a reversible isothermal process. During the compression heat is removed from the system, and during the expansion, heat is added back to the system. A finite temperature difference is not used here. During the compression, the surroundings are only slightly lower in temperature than the system, and during the expansion, the surroundings are only slightly higher in temperature than the system. But the slight differences are insignificant.

We need to be more precise about what we call a reversible process applied to a system. It is not good enough to say that you can added Q and W to a system in a process, and then remove Q and W to bring the system back to its original state, and call that a reversible process. That is a necessary, but not sufficient condition, for the process path to be considered reversible.

In order for a process path to be considered reversible, it must pass through a continuous sequence of thermodynamic equilibrium states, differentially separated from one another (along the path). For such paths, the temperature and pressure within the system will be uniform. Any heat that is transferred between the system and surroundings will take place with only a slight differential temperature driving force, and any work done by the system on the surroundings will take place with only a slight differential pressure driving force. The process can reversed at any point along the path by just retracing the original path, reversing the slight differential driving forces. For such paths, if we evaluate the integral of dq/T_I along the path (where T_I is the temperature at the boundary between the system and surroundings), we will find that it is higher than any other (irreversible) path. We call the integral for the reversible path the change in entropy for the process. It depends only on the initial and final equilibrium states.

Chet

 

[Soumalya]

In order for a process path to be considered reversible, it must pass through a continuous sequence of thermodynamic equilibrium states, differentially separated from one another (along the path). For such paths, the temperature and pressure within the system will be uniform. Any heat that is transferred between the system and surroundings will take place with only a slight differential temperature driving force, and any work done by the system on the surroundings will take place with only a slight differential pressure driving force. The process can reversed at any point along the path by just retracing the original path, reversing the slight differential driving forces. For such paths, if we evaluate the integral of dq/T_I along the path (where T_I is the temperature at the boundary between the system and surroundings), we will find that it is higher than any other (irreversible) path. We call the integral for the reversible path the change in entropy for the process. It depends only on the initial and final equilibrium states.

Chet

Why is it necessary for a reversible process to pass through a continuous sequence of equilibrium states?

What you describe is the precise condition for a quasi equilibrium process where the process should proceed infinitely slowly or in stages of thermodynamic equilibrium states to fix the properties at each stage for plotting the entire path of the process.

"A reversible process is one where both system and surroundings could be restored to their initial states without any traces of the changes that once occurred".From this definition how can we arrive at the conclusion that for a reversible process it must proceed quasi statically?

 

[Chestermiller]

Why is it necessary for a reversible process to pass through a continuous sequence of equilibrium states?

What you describe is the precise condition for a quasi equilibrium process where the process should proceed infinitely slowly or in stages of thermodynamic equilibrium states to fix the properties at each stage for plotting the entire path of the process.

"A reversible process is one where both system and surroundings could be restored to their initial states without any traces of the changes that once occurred".From this definition how can we arrive at the conclusion that for a reversible process it must proceed quasi statically?

In my judgement, this is not a very good definition of a reversible process, although it is true. All this definition has done is cause uncertainty and confusion for you. The definition I presented is much more precise, although, I should have added that, once you get to the final state, you can return to the original state using any quasistatic path (for both heat flow and work), and not just the reverse of the path you followed to get to the final state.

If you think that the two definitions are incompatible, please identify for me a system and process where the second definition is satisfied and the first definition is not.

Chet

 

[Soumalya]

In my judgement, this is not a very good definition of a reversible process, although it is true. All this definition has done is cause uncertainty and confusion for you. The definition I presented is much more precise, although, I should have added that, once you get to the final state, you can return to the original state using any quasistatic path (for both heat flow and work), and not just the reverse of the path you followed to get to the final state.

If you think that the two definitions are incompatible, please identify for me a system and process where the second definition is satisfied and the first definition is not.

Chet

I would like to know if it's possible to show that for a process which is reversible it must follow a quasi equilibrium path.Or conversely if it is possible to show that if a change is state is achieved following a non quasi equilibrium path the process is irreversible.

 

[Andrew Mason]

I would like to know if it's possible to show that for a process which is reversible it must follow a quasi equilibrium path.Or conversely if it is possible to show that if a change is state is achieved following a non quasi equilibrium path the process is irreversible.

A quasi-static process is one that proceeds very slowly - so slowly that there is an infinitesimal or arbitrarily small change in the system over a finite time period.

A reversible process, such as one involving heat transfer or change in volume will take an arbitrarily long period to complete. So a reversible process is necessarily quasi-static.

But it doesn't always work the other way around. A process that may take an arbitrarily long time to complete may not be reversible. A process during which the system and surroundings are not in thermodynamic equilibrium all times is not reversible regardless of how slowly it proceeds. An example would be a large volume of gas in a container under higher pressure than the surroundings where the gas is let out into the surroundings through a tiny hole - so tiny that molecules of gas exit the container at an arbitrarily slow rate.

AM

 

[Chestermiller]

I would like to know if it's possible to show that for a process which is reversible it must follow a quasi equilibrium path.Or conversely if it is possible to show that if a change is state is achieved following a non quasi equilibrium path the process is irreversible.

To address this, let's consider a specific example that you yourself raised today in another related thread, to wit:

What is the difference if we make a heat transfer between the system and surroundings with a considerably large temperature difference?Why won't the system and surroundings be restored to their initial states?

If I can show you why the system and surroundings can't "be restored to their initial states without any traces of the changes that once occurred," will this satisfy your doubts?

Chet

 

[Soumalya]

To address this, let's consider a specific example that you yourself raised today in another related thread, to wit:

If I can show you why the system and surroundings can't "be restored to their initial states without any traces of the changes that once occurred," will this satisfy your doubts?

Chet

Yes yes I would be very relieved

But I do have some related doubts coming after this LOL

 

[Chestermiller]

Yes yes I would be very relieved

But I do have some related doubts coming after this LOL

Oh well. Here goes.

Suppose you have a gas in a cylinder (cylinder has negligible heat capacity), and its initial temperature is T₀. You want to raise its temperature to T₁ using a constant volume irreversible process (i.e., no work done). So you put the cylinder in contact with a constant temperature heat reservoir (a part of the surroundings) at temperature T₁ and wait until the gas is at temperature T₁. During this heating operation with finite driving force, the temperature of the gas within the cylinder will be not be uniform, since conductive heat transfer is taking place within the gas (involving temperature gradients). The average temperature of the gas at any time during the heating operation will lie somewhere between T₀ and T₁. In the end, the amount of heat transferred from the reservoir to the gas will be Q = mC_v(T₁-T₀). Now you want to return both the gas and the surroundings to their original states, "without any traces that any change had occurred." What is your detailed game plan for accomplishing this?

Chet

 

[Soumalya]

Well I will place the tank with the gas at T₁ with another reservoir at temperature T<T₁:tongue:

By the way "It's not wise to share your gameplans"

Ok.It's not possible to reverse the heat transfer spontaneously according to Kelvin Planck statement.We would need a heat pump to reverse the heat flow in expense of additional work.

Now show me how is it reversible if it was done infinitesimally slowly so that the difference in temperature between the tank and the reservoir was dT.

By the way "It's not wise to share your game plans"

 

[Chestermiller]

Well I will place the tank with the gas at T₁ with another reservoir at temperature T<T₁:tongue:

By the way "It's not wise to share your gameplans"

Ok.It's not possible to reverse the heat transfer spontaneously according to Kelvin Planck statement.We would need a heat pump to reverse the heat flow in expense of additional work.

Now show me how is it reversible if it was done infinitesimally slowly so that the difference in temperature between the tank and the reservoir was dT.

By the way "It's not wise to share your game plans"

OK. I think we've made some progress.

Let me summarize where we are. In our irreversible process, we have taken the gas from temperature T₀ to T₁ by bringing it into contact with a reservoir at T₁. The amount of heat transferred was Cv(T₁-T₀). Now suppose we decide to take the system back down to temperature T₀ by putting it into contact with a second reservoir at T₀ and waiting long enough for everything to equilibrate. In this reverse process, the transfer of heat from the system to the second reservoir is Cv(T₁-T₀). The system is now back in its original state, but, in the surroundings, we have brought about a net transfer of heat in the amount Cv(T₁-T₀) from the hot reservoir to the cold reservoir. So there has been more than a "trace of a change" for the surroundings.

Next, let's consider a slightly "less irreversible" process. In this case, we introduce a third reservoir at temperature (T₀+T₁)/2. And we carry out the process in two stages, by first contacting the gas with the reservoir at (T₀+T₁)/2 and letting the two of them equilibrate, and then removing the second reservoir, and contacting the gas with the reservoir at T₁. How much heat is transferred in each of these steps, and what is the total amount of heat transferred to the system over the two steps?

Next, let's return the gas back to its original state, by using a similar two-step process (involving the 3 reservoirs). I would you to do the calculations for this return path. How much heat is transferred in each step of the return path? What is the total amount of heat transferred from the gas in this return path? For the combination of the forward process path plus the return path, what is the net amount of heat transferred to or from each of the three reservoirs (comprising the surroundings). How does this compare with the previous situation where we only used two reservoirs?

Chet

 

[Soumalya]

OK. I think we've made some progress.

Let me summarize where we are. In our irreversible process, we have taken the gas from temperature T₀ to T₁ by bringing it into contact with a reservoir at T₁. The amount of heat transferred was Cv(T₁-T₀). Now suppose we decide to take the system back down to temperature T₀ by putting it into contact with a second reservoir at T₀ and waiting long enough for everything to equilibrate. In this reverse process, the transfer of heat from the system to the second reservoir is Cv(T₁-T₀). The system is now back in its original state, but, in the surroundings, we have brought about a net transfer of heat in the amount Cv(T₁-T₀) from the hot reservoir to the cold reservoir. So there has been more than a "trace of a change" for the surroundings.

Chet

How do you create the reservoir at temperature T₀ without expenditure of additional energy from the reservoir at T₁ which was assumed to be the surroundings?

Note:T₁>T₀.

OK. I think we've made some progress.


Next, let's consider a slightly "less irreversible" process. In this case, we introduce a third reservoir at temperature (T₀+T₁)/2. And we carry out the process in two stages, by first contacting the gas with the reservoir at (T₀+T₁)/2 and letting the two of them equilibrate, and then removing the second reservoir, and contacting the gas with the reservoir at T₁. How much heat is transferred in each of these steps, and what is the total amount of heat transferred to the system over the two steps?

Chet

Could you please elaborate on this part once again?

You introduced a third reservoir with temperature (T₀+T₁)/2 and brought the gas within the container which is currently at a temperature of T₁ to this reservoir so that thermal equilibrium is achieved.So the temperature of the gas after this process would be (T₀+T₁)/2 and Q=C_v[T₁-(T₀+T₁)/2]=C_v.(T₁-T₀)/2

Note:T₁>(T₀+T₁)/2>T₀.

Now do you wish to place the gas in contact with reservoir at T₁ again?

In doing so the gas temperature would be again T₁ whereas we want it to reverse to T₀.

I think you meant to say placing the gas at (T₀+T₁)/2 to another reservoir at T₀.Correct?

In such case final temperature of the gas is T₀ as similar to it's initial temperature.The system is restored and Q=C_v(T₁-T₀)/2 again.

So the overall result is a net heat transfer of C_v(T₁-T₀) achieved over two stages as was done in a single step earlier.

But the surroundings is not unaffected as you would need expenditure of work to create those two reservoirs you have assumed from a single reservoir at T₁ initially.

Next, let's return the gas back to its original state, by using a similar two-step process (involving the 3 reservoirs). I would you to do the calculations for this return path. How much heat is transferred in each step of the return path? What is the total amount of heat transferred from the gas in this return path? For the combination of the forward process path plus the return path, what is the net amount of heat transferred to or from each of the three reservoirs (comprising the surroundings). How does this compare with the previous situation where we only used two reservoirs?

You may use infinite number of reservoirs to reverse the process but the transfer of heat in the reverse process would the same as C_v(T₁-T₀) as was in the forward process.

Total heat transfer for the cycle is zero.

If I am wrong please do the calculations once again...

Soumalya

 

[Chestermiller]

How do you create the reservoir at temperature T₀ without expenditure of additional energy from the reservoir at T₁ which was assumed to be the surroundings?

Note:T₁>T₀.

The reservoirs at T₀ and T₁ were already present within the surroundings before we started doing anything to the system.

Could you please elaborate on this part once again?

In this second situation we are studying, there are three reservoirs present within the surroundings before we even start doing anything to the system.

You introduced a third reservoir with temperature (T₀+T₁)/2 and brought the gas within the container which is currently at a temperature of T₁ to this reservoir so that thermal equilibrium is achieved.So the temperature of the gas after this process would be (T₀+T₁)/2 and Q=C_v[T₁-(T₀+T₁)/2]=C_v.(T₁-T₀)/2

Note:T₁>(T₀+T₁)/2>T₀.

Now do you wish to place the gas in contact with reservoir at T₁ again?

Yes. In this new problem, we want to bring the final temperature up to T₁ again, just as in the previous problem. We want to do this before we start the return path. So, in this new problem, we first take the gas from T₀ to (T₀+T₁)/2, and then we take it from (T₀+T₁)/2 to T₁.

In doing so the gas temperature would be again T₁ whereas we want it to reverse to T₀.

As I said, in this problem, we'll start the reverse path once we get the gas to T₁.

I think you meant to say placing the gas at (T₀+T₁)/2 to another reservoir at T₀.Correct?

No.

So the overall result is a net heat transfer of C_v(T₁-T₀) achieved over two stages as was done in a single step earlier.

Yes.

But the surroundings is not unaffected as you would need expenditure of work to create those two reservoirs you have assumed from a single reservoir at T₁ initially.

No. All three reservoirs are assumed to be initially part of the surroundings. Creating these reservoirs is not part of what we are doing in our problem. They already exist when we start.

So, here's where we stand in Problem #2:

In Step 1, we have transferred Cv(T₁-T₂)/2 from the reservoir at (T₁+T₂)/2 to the gas to raise its temperature from T₀ to (T₁+T₂)/2.

In Step 2, we have transferred Cv(T₁-T₂)/2 from the reservoir at T₁ to the gas to raise its temperature from (T₁+T₂)/2 to T₁.

We are now ready to start the return path.

You may use infinite number of reservoirs to reverse the process but the transfer of heat in the reverse process would the same as C_v(T₁-T₀) as was in the forward process.

Please slow down, and don't jump ahead. After we finish discussing the return path, and the overall net outcome of the forward and return paths, all your doubts will be cleared up.

Chet

 

[Soumalya]

So, here's where we stand in Problem #2:

In Step 1, we have transferred Cv(T₁-T₂)/2 from the reservoir at (T₁+T₂)/2 to the gas to raise its temperature from T₀ to (T₁+T₂)/2.

In Step 2, we have transferred Cv(T₁-T₂)/2 from the reservoir at T₁ to the gas to raise its temperature from (T₁+T₂)/2 to T₁.

We are now ready to start the return path.

Chet

Which temperature does T₂ represent here?

 

[Chestermiller]

Which temperature does T₂ represent here?

Uh Oh. I meant T₀ and T₁. Please note that correction.

Now for the return path.

Step 1: Contact the gas at T₁ with the reservoir at (T₀+T₁)/2 to bring its temperature back down to (T₀+T₁)/2.

Step 2: Contact the gas at (T₀+T₁)/2 with the reservoir at T₀ to bring its temperature back down to T₀.

What is the heat transferred from the gas to the reservoir in each of these steps? What is the total heat transferred from the gas over both these steps?

Is the gas returned to its original state?

What is the net heat flow from the reservoir at T₁ over the combined forward and return paths?

What is the net heat flow from the reservoir at (T₀+T₁)/2 over the combined forward and return paths?

What is the net heat flow from the reservoir at T₀ over the combined forward and return paths?

Chet

 

[Soumalya]

What is the heat transferred from the gas to the reservoir in each of these steps? What is the total heat transferred from the gas over both these steps?

Chet

Q_{each step}=C_V(T₁-T₀)/2

Q_total=C_v(T₁-T₀)

Is the gas returned to its original state?

Yes

What is the net heat flow from the reservoir at T₁ over the combined forward and return paths?

Q_total=C_v(T₁-T₀)/2

What is the net heat flow from the reservoir at (T₀+T₁)/2 over the combined forward and return paths?

Q_total=0

What is the net heat flow from the reservoir at T₀ over the combined forward and return paths?

Q_total=C_v(T₁-T₀)/2

Then?

 

[Chestermiller]

Good job, except for the reservoir at T₀. The net heat flow from this reservoir over the combined forward and return paths is
Q_total= - C_v(T₁-T₀)/2. (The reservoir at T₀ removes heat from the gas.)

So, in the cases of using either 2 or 3 reservoirs, we found that the combined forward and return paths resulted in no change in the system, a transfer of C_v(T₁-T₀) from the surroundings to the system during the forward path, and a change in temperature for the system from T₀ to T₁ during the forward path. However, in the case of using 2 reservoirs, we found that the net effect of the forward and return paths was a transfer of C_v(T₁-T₀) from the reservoir at T₁ to the reservoir at T₀, while in the case of using 3 reservoirs, we found that the net effect of the forward and return paths was a transfer of only C_v(T₁-T₀)/2 from the reservoir at T₁ to the reservoir at T₀, and no other change in either the system or the surroundings. (Recall that the change in the middle reservoir was zero). So the conclusion here is that the case of using 3 reservoirs was "less irreversible" than the case of using only 2 reservoirs.

What do you think the results would be if we used 4 reservoirs, 5 reservoirs, or n reservoirs?
In the limit of an infinite number of reservoirs, what would be the net heat effect of the forward and return paths?

Chet

 

[Soumalya]

However, in the case of using 2 reservoirs, we found that the net effect of the forward and return paths was a transfer of C_v(T₁-T₀) from the reservoir at T₁ to the reservoir at T₀, while in the case of using 3 reservoirs, we found that the net effect of the forward and return paths was a transfer of only C_v(T₁-T₀)/2 from the reservoir at T₁ to the reservoir at T₀, and no other change in either the system or the surroundings. (Recall that the change in the middle reservoir was zero). So the conclusion here is that the case of using 3 reservoirs was "less irreversible" than the case of using only 2 reservoirs.

Chet

How is it so?
'
In case of 3 reservoirs at T₁,(T₀+T₁)/2 and T₀ together taken as the surroundings the net heat transfer was zero.So the system and surroundings were both returned to their initial states.

In case of 2 reservoirs at T₁ and T₀ together taken as surroundings net heat transfer is again zero.

What we should consider is all the reservoirs are part of the surroundings hence the total heat transfer for all the reservoirs combined is zero.

If I am wrong please clarify...

 

[Chestermiller]

How is it so?
'
In case of 3 reservoirs at T₁,(T₀+T₁)/2 and T₀ together taken as the surroundings the net heat transfer was zero.So the system and surroundings were both returned to their initial states.

In case of 2 reservoirs at T₁ and T₀ together taken as surroundings net heat transfer is again zero.

What we should consider is all the reservoirs are part of the surroundings hence the total heat transfer for all the reservoirs combined is zero.

If I am wrong please clarify...

No. The surroundings have not returned to their initial state "without a trace of a change." There is a definite detectable change in the surroundings. In both cases, the reservoir at T₁ has lost heat and the reservoir at T₀ has gained heat. If each of these reservoirs were set up initially so that they consisted of a combination of a solid and liquid in equilibrium at their respective melting points T₁ and T₀, then the reservoir at T₁ would contain more solid and less liquid in the end, and the reservoir at T₀ would contain more liquid and less solid in the end. This is certainly more than a trace of a change in the surroundings.

Chet

 

[Soumalya]

No. The surroundings have not returned to their initial state "without a trace of a change." There is a definite detectable change in the surroundings. In both cases, the reservoir at T₁ has lost heat and the reservoir at T₀ has gained heat. If each of these reservoirs were set up initially so that they consisted of a combination of a solid and liquid in equilibrium at their respective melting points T₁ and T₀, then the reservoir at T₁ would contain more solid and less liquid in the end, and the reservoir at T₀ would contain more liquid and less solid in the end. This is certainly more than a trace of a change in the surroundings.

Chet

So you mean assuming a heat transfer using an infinite number of reservoirs whose temperatures are infinitesimally different than each other would result in least change in the conditions of each reservoir after forward and reverse process the changes being so negligible that every part of the surroundings is observed with almost similar initial properties after the complete cycle?

 

[Chestermiller]

So you mean assuming a heat transfer using an infinite number of reservoirs whose temperatures are infinitesimally different than each other would result in least change in the conditions of each reservoir after forward and reverse process the changes being so negligible that every part of the surroundings is observed with almost similar initial properties after the complete cycle?

It's even more powerful than just that. In all these cases, by the end of the return path, only two of the reservoirs in the sequence would be affected in any way. These would be the reservoirs at the very beginning and at the very end of the sequence. All the other reservoirs would be exactly returned to their initial states. And, as far as the initial and final reservoirs are concerned, as we add more intermediate reservoirs to the sequence, their changes would become less and less.

Chet

 

[Soumalya]

I am all clear Chet.

Thank You

 

[rkmurtyp]

Hello,
It is said that any process accompanied by dissipating effects is an irreversible process.For instance, a piston cylinder arrangement undergoing an expansion of a gas with heat lost as friction between surfaces is an irreversible process.Here the system does work on the surroundings with heat being rejected to the surroundings as a result of friction.The process can be summed up as:

Work transfer for the system= -W
Work transfer to the surroundings=+W

Heat lost from the system=-Q
Heat gained by the surroundings=+Q

Then logically the process could be reversed at any stage if external work is done by the surroundings on the system equal to W and heat lost as friction be added to the system so that both system and surroundings are restored to their initial conditions.So why it's still called an irreversible process?

The irreversible process involving expansion of a gas with friction is more complicated than necessary. A simpler process to understand the concept of reversible and irreversible processes would be to consider either expansion with pressure of the surroundings lower than the pressure of the system OR compression of a gas with friction present. After understanding these you can certainly consider expansion with friction.

Let us take the expansion experiment.

Here we are conducting an irrversible expansion process (pressure of the surroundings lower than the pressure of the system) that takes the system from an equilibrium state A to another equilibrium state B, by say, an isothermal process (keeping the system in thermal contact with a heat reservoir (HR) at the temperature, T, the temperature of the system in state A, T_A (ofcourse it is also the temperature of the system in state B, T_B i.e.T_A =T_B = T)). The work W done by the gas is less than the reversible work W_rev (the maximum work) it would have done in going from A to B by a reversible process. The system absobs Q (<Q_rev) units of heat from the reservoir.

Thus the changes in the surroundings (when the system goes from state A to state B are: A mass in the surroundings is found at a higher level (this is a way of saying the system did work on the surroundings) and a HR at temperature T, loses Q (<Q_rev) units of heat.

Now let us take the system to its original state by a reversible isothermal process. This needs minimum expenditure of work of W_rev units of work (which is more than W). The HR at temperature T receives Q_rev units (>Q units it lost). Thus when the system reaches its original state we find a mass in the surroundings at a level lower than it was in the beginning of the process and the HR gained (Q_rev - Q) units of heat (=W_rev - W).

The net result of this cyclic process is transformation of (W_rev - W) units of work into (Q_rev -Q) units of heat.

Now we can consider the compression with friction involved.

We keep the system in thermal contact with HR at T and apply pressure P_ext (>P_sym)to overcome friction. (Note that, you can employ a higher pressure to compress even if there was no friction). The isothermal work done on the system W is greater than W_rev (the minimum work needed to take the system fromB to A). Heat Q (=W) units is supplied to the HR at T.

Now according to Kelvin's postulate of the second law of thermodynamics, it is impossible to absorb Q units of heat from HR at T and convert it into W (=Q) units of work in a cyclic process leaving no other changes elsewhere.

The best we can do to take the system from state A to state B is to employ a reversible isothermal expansion process. This aborbs Q_rev (<Q) units of heat and delivers W_rev (<W) units of work (raises a mass to a level lower than its original level). The net result at the end of the cycle is: the system reached its original state and, in the surroundings a mass is found at a lower level and HR gained an equivalent amount of heat.

To understand the concept of irreversibility in thermodyanmics we don't need the presence of friction and its dissipative effects. It is in dynamics that friction leads to dissipative effects and loss of efficiency of processes.
In thermodyanmics, we have this situation: Once an irreversible process occurs, it is impossible to reverse it with no other changes left elsewhere - even employing the most efficient revrsible prcess.

 

[rkmurtyp]

Hello,
It is said that any process accompanied by dissipating effects is an irreversible process.For instance, a piston cylinder arrangement undergoing an expansion of a gas with heat lost as friction between surfaces is an irreversible process.Here the system does work on the surroundings with heat being rejected to the surroundings as a result of friction.The process can be summed up as:

Work transfer for the system= -W
Work transfer to the surroundings=+W

Heat lost from the system=-Q
Heat gained by the surroundings=+Q

Then logically the process could be reversed at any stage if external work is done by the surroundings on the system equal to W and heat lost as friction be added to the system so that both system and surroundings are restored to their initial conditions.So why it's still called an irreversible process?

The irreversible process involving expansion of a gas with friction is more complicated than necessary. A simpler process to understand the concept of reversible and irreversible processes would be to consider either expansion with pressure of the surroundings lower than the pressure of the system OR compression of a gas with friction present. After understanding these you can certainly consider expansion with friction.

Let us take the expansion experiment.

Here we are conducting an irrversible expansion process (pressure of the surroundings lower than the pressure of the system) that takes the system from an equilibrium state A to another equilibrium state B, by say, an isothermal process (keeping the system in thermal contact with a heat reservoir (HR) at the temperature, T, the temperature of the system in state A, T_A (ofcourse it is also the temperature of the system in state B, T_B i.e.T_A =T_B = T)). The work W done by the gas is less than the reversible work W_rev (the maximum work) it would have done in going from A to B by a reversible process. The system absobs Q (<Q_rev) units of heat from the reservoir.

Thus the changes in the surroundings (when the system goes from state A to state B are: A mass in the surroundings is found at a higher level (this is a way of saying the system did work on the surroundings) and a HR at temperature T, loses Q (<Q_rev) units of heat.

Now let us take the system to its original state by a reversible isothermal process. This needs minimum expenditure of work of W_rev units of work (which is more than W). The HR at temperature T receives Q_rev units (>Q units it lost). Thus when the system reaches its original state we find a mass in the surroundings at a level lower than it was in the beginning of the process and the HR gained (Q_rev - Q) units of heat (=W_rev - W).

The net result of this cyclic process is transformation of (W_rev - W) units of work into (Q_rev -Q) units of heat.

Now we can consider the compression with friction involved.

We keep the system in thermal contact with HR at T and apply pressure P_ext (>P_sym)to overcome friction. (Note that, you can employ a higher pressure to compress even if there was no friction). The isothermal work done on the system W is greater than W_rev (the minimum work needed to take the system fromB to A). Heat Q (=W) units is supplied to the HR at T.

Now according to Kelvin's postulate of the second law of thermodynamics, it is impossible to absorb Q units of heat from HR at T and convert it into W (=Q) units of work in a cyclic process leaving no other changes elsewhere.

The best we can do to take the system from state A to state B is to employ a reversible isothermal expansion process. This aborbs Q_rev (<Q) units of heat and delivers W_rev (<W) units of work (raises a mass to a level lower than its original level). The net result at the end of the cycle is: the system reached its original state and, in the surroundings a mass is found at a lower level and HR gained an equivalent amount of heat.

To understand the concept of irreversibility in thermodyanmics we don't need the presence of friction and its dissipative effects. It is in dynamics that friction leads to dissipative effects and loss of efficiency of processes.
In thermodyanmics, we have this situation: Once an irreversible process occurs, it is impossible to reverse it with no other changes left elsewhere - even employing the most efficient revrsible prcess.

 

[Soumalya]

The irreversible process involving expansion of a gas with friction is more complicated than necessary. A simpler process to understand the concept of reversible and irreversible processes would be to consider either expansion with pressure of the surroundings lower than the pressure of the system OR compression of a gas with friction present. After understanding these you can certainly consider expansion with friction.

Let us take the expansion experiment.

Here we are conducting an irrversible expansion process (pressure of the surroundings lower than the pressure of the system) that takes the system from an equilibrium state A to another equilibrium state B, by say, an isothermal process (keeping the system in thermal contact with a heat reservoir (HR) at the temperature, T, the temperature of the system in state A, T_A (ofcourse it is also the temperature of the system in state B, T_B i.e.T_A =T_B = T)). The work W done by the gas is less than the reversible work W_rev (the maximum work) it would have done in going from A to B by a reversible process. The system absobs Q (<Q_rev) units of heat from the reservoir.

Thus the changes in the surroundings (when the system goes from state A to state B are: A mass in the surroundings is found at a higher level (this is a way of saying the system did work on the surroundings) and a HR at temperature T, loses Q (<Q_rev) units of heat.

Now let us take the system to its original state by a reversible isothermal process. This needs minimum expenditure of work of W_rev units of work (which is more than W). The HR at temperature T receives Q_rev units (>Q units it lost). Thus when the system reaches its original state we find a mass in the surroundings at a level lower than it was in the beginning of the process and the HR gained (Q_rev - Q) units of heat (=W_rev - W).

The net result of this cyclic process is transformation of (W_rev - W) units of work into (Q_rev -Q) units of heat.

Now we can consider the compression with friction involved.

We keep the system in thermal contact with HR at T and apply pressure P_ext (>P_sym)to overcome friction. (Note that, you can employ a higher pressure to compress even if there was no friction). The isothermal work done on the system W is greater than W_rev (the minimum work needed to take the system fromB to A). Heat Q (=W) units is supplied to the HR at T.

Now according to Kelvin's postulate of the second law of thermodynamics, it is impossible to absorb Q units of heat from HR at T and convert it into W (=Q) units of work in a cyclic process leaving no other changes elsewhere.

The best we can do to take the system from state A to state B is to employ a reversible isothermal expansion process. This aborbs Q_rev (<Q) units of heat and delivers W_rev (<W) units of work (raises a mass to a level lower than its original level). The net result at the end of the cycle is: the system reached its original state and, in the surroundings a mass is found at a lower level and HR gained an equivalent amount of heat.

To understand the concept of irreversibility in thermodyanmics we don't need the presence of friction and its dissipative effects. It is in dynamics that friction leads to dissipative effects and loss of efficiency of processes.
In thermodyanmics, we have this situation: Once an irreversible process occurs, it is impossible to reverse it with no other changes left elsewhere - even employing the most efficient revrsible prcess.

Very well explained

Thank You

 

[Chestermiller]

The irreversible process involving expansion of a gas with friction is more complicated than necessary. A simpler process to understand the concept of reversible and irreversible processes would be to consider either expansion with pressure of the surroundings lower than the pressure of the system OR compression of a gas with friction present. After understanding these you can certainly consider expansion with friction.

Let us take the expansion experiment.

Here we are conducting an irrversible expansion process (pressure of the surroundings lower than the pressure of the system) that takes the system from an equilibrium state A to another equilibrium state B, by say, an isothermal process (keeping the system in thermal contact with a heat reservoir (HR) at the temperature, T, the temperature of the system in state A, T_A (ofcourse it is also the temperature of the system in state B, T_B i.e.T_A =T_B = T)). The work W done by the gas is less than the reversible work W_rev (the maximum work) it would have done in going from A to B by a reversible process. The system absobs Q (<Q_rev) units of heat from the reservoir.

Thus the changes in the surroundings (when the system goes from state A to state B are: A mass in the surroundings is found at a higher level (this is a way of saying the system did work on the surroundings) and a HR at temperature T, loses Q (<Q_rev) units of heat.

Now let us take the system to its original state by a reversible isothermal process. This needs minimum expenditure of work of W_rev units of work (which is more than W). The HR at temperature T receives Q_rev units (>Q units it lost). Thus when the system reaches its original state we find a mass in the surroundings at a level lower than it was in the beginning of the process and the HR gained (Q_rev - Q) units of heat (=W_rev - W).

The net result of this cyclic process is transformation of (W_rev - W) units of work into (Q_rev -Q) units of heat.

Now we can consider the compression with friction involved.

We keep the system in thermal contact with HR at T and apply pressure P_ext (>P_sym)to overcome friction. (Note that, you can employ a higher pressure to compress even if there was no friction). The isothermal work done on the system W is greater than W_rev (the minimum work needed to take the system fromB to A). Heat Q (=W) units is supplied to the HR at T.

Now according to Kelvin's postulate of the second law of thermodynamics, it is impossible to absorb Q units of heat from HR at T and convert it into W (=Q) units of work in a cyclic process leaving no other changes elsewhere.

The best we can do to take the system from state A to state B is to employ a reversible isothermal expansion process. This aborbs Q_rev (<Q) units of heat and delivers W_rev (<W) units of work (raises a mass to a level lower than its original level). The net result at the end of the cycle is: the system reached its original state and, in the surroundings a mass is found at a lower level and HR gained an equivalent amount of heat.

To understand the concept of irreversibility in thermodyanmics we don't need the presence of friction and its dissipative effects. It is in dynamics that friction leads to dissipative effects and loss of efficiency of processes.
In thermodyanmics, we have this situation: Once an irreversible process occurs, it is impossible to reverse it with no other changes left elsewhere - even employing the most efficient revrsible prcess.

Hi rkmurtyp.

Thanks for the interesting analysis of reversibility/irreversibility for isothermal expansion without friction and for isothermal compression with friction. You may be interested in a thread that Red_CCF and I have been collaborating on quantifying the adiabatic quasi static compression and expansion of an ideal gas with friction between the piston and cylinder: https://www.physicsforums.com/showthread.php?t=750946
The first 12 posts of this thread focus on a fluid mechanics issue, but, starting with post #13, the thread switches to the thermo problem. We are currently focusing on determining the change in entropy, and confirming that it is greater than zero for this adiabatic irreversible process.

Chet

 

[rkmurtyp]

Yes, I will go to the thread you gave.