Understanding reversible and irreversible process

1. Apr 24, 2014

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?

2. Apr 24, 2014

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.

3. Apr 24, 2014

Soumalya

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

4. Apr 24, 2014

UltrafastPED

You are applying work to the system.

5. Apr 24, 2014

Soumalya

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

6. Apr 25, 2014

UltrafastPED

You can find these in your textbook.

7. Apr 28, 2014

Staff: Mentor

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

8. Apr 29, 2014

Andrew Mason

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

9. Apr 29, 2014

Soumalya

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?

10. Apr 29, 2014

Staff: Mentor

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/TI along the path (where TI 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

11. Apr 29, 2014

Soumalya

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????

12. Apr 29, 2014

Staff: Mentor

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

13. Apr 29, 2014

Soumalya

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.

14. Apr 29, 2014

Andrew Mason

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

15. Apr 29, 2014

Staff: Mentor

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

16. Apr 30, 2014

Soumalya

Yes yes I would be very relieved

But I do have some related doubts coming after this LOL

17. Apr 30, 2014

Staff: Mentor

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 = mCv(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

18. Apr 30, 2014

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"

19. Apr 30, 2014

Staff: Mentor

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

20. May 1, 2014

Soumalya

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.
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=Cv[T1-(T0+T1)/2]=Cv.(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=Cv(T1-T0)/2 again.

So the overall result is a net heat transfer of Cv(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.

You may use infinite number of reservoirs to reverse the process but the transfer of heat in the reverse process would the same as Cv(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

Last edited: May 1, 2014