First, I have attached a pdf with my calculations. In summary

T2 = (4/5)T1, and T3 = (7/5)T2 which leads to T3 = (28/25)T1

Second is the reason. The reason is because the work of a process that is not quasi-static is W = Pext*Delta_V. For a reversible process work is calculated using the integral (because infinitely small increments of change represent a continuous function). So, in this case going from state 1 to state 2, and then from state 2 to state 3, there are going to be two different external pressures. Pext1 = mg, Pext2 = 2mg, where P is pressure, m is mass, g is gravity, V is volume. For the same amount of work to be done in the two changes, obviously the change in volume must be different. So, you won't return to the same state.

Attached are the details.

I have to apologize. While I stand by my statement that the process you describe is not reversible, my "math" was poorly thought out and the entropy change could equal zero provided the ratio of volumes and temperatures were correct.

However, it is true that a process must not generate entropy if it is to be reversible, and my statement that your process would increase entropy unless your weight was infinitesimally small is still true.

Elliott and Lira describe this situation of expansion in a piston cylinder better than I can. (Elliot, J.R.; Lira, C.T. Introductory Chemical Engineering Thermodynamics, Prentice-Hall, 1999, first edition, pages 38 and 39)

In this situation, a pressure gradient is created inside the cylinder, and a new equilibrium is only reached once dissipative forces have caused the piston to stop moving.

A result of the dissipative forces, is that the temperature at the final state will not be as large as the temperature would be if the system didn't have dissipative forces, i.e. was reversible (Energy is lost to the dissipative forces).

Likewise, when you then put the weight back on, these dissipative forces, which allow the "flux" to come to zero, also cause the change in temperature to not be the same as if the change was reversible, and so, in the new state, the temperature will not be the same as in the original state.

Thus, unless your weight is infinitely small, there will be a pressure gradient, which will lead to dissipation forces, which will take away from the temperature change, and not allow you to return back to the original state at the same pressure, volume and temperature. Thus the very existence of a flux which stops the piston from moving after the addition or removal of your weights will not allow the process to be reversible. Elliott and Lira describe it much better than I have. And again, I apologize for the incorrect assumption in my first post.

I appreciate your reply

could u please provide me that explanation by elliot and lira, i can't find this book.

Hi,

I don't know how to legally get it to you :(

Here is a link for the online edition, which is actually the 2nd edition. If you have access to it great, if not, it is all I can do without violating copyright.

http://eds.b.ebscohost.com/eds/detail/detail?vid=1&sid=f23b5ee8-b2dc-485f-ba41-38d9141c4629%40sessionmgr114&hid=114&bdata=JnNpdGU9ZWRzLWxpdmUmc2NvcGU9c2l0ZQ%3d%3d#AN=alb.5922505&db=cat03710a

http://proquest.safaribooksonline.com/9780132788489?uicode=ualberta

Here is a link to a pdf you should be able to access. It doesn't directly answer your question, but it goes over the basic fundamentals in a very well written way.

http://puccini.che.pitt.edu/~karlj/Classes/CHE2101/Shell/Ch6_The_first_law_and_reversibility.pdf

Cheers,

Braden

If I put the weight back on the piston

then you have to do some work lifting the weight into position. So you are not analyzing an isolated system.

It's not reversible process though initial and final states are the same in ideal conditions (see below). The process is named reversible if it comes back through the same sequence of states. In Your case You remove suddanly one weight and the system went out of equilibrium. Sure the piston oscillates about new state of balance because of absence of dry friction until viscousity forces damp the oscillations and rise up a  temperature a little. The back process is completely unlike to the forward one because the oscillations take places around initial level, If You introduce dry friction between the piston and walls of canister the system won't return to the initial state. May be it will remain in new state at all though the the two weigts are on the piston.if mg

Assume the piston is filled with an ideal gas of N particles of volume V and temperature T. I.e., a pressure P such that PV = N kB T, and an internal energy U =  (3/2) NkBT. The equilibrium condition is that PA = 2Mg, where A is the cross-section of the piston and M is the mass of each weight.

When you remove one of the weights a lot of unpredictable things will happen, but in the end a new equilibrium condition will arise, with pressure P' such that P'A = Mg. In this process the piston have lifted the single weight an height h, such that the volume has increased to V' = V + hA, and the temperature has changed to T'. The weight has gained a potential energy Mgh, and the internal energy of the gas has changed to U' = (3/2) NkBT'.

By conservation of energy you must have U' + Mgh = U, and from the equation of state that P'V' = NkBT'. This gives two equations to solve for T' and h, leading to

h = NkBT/(PA + 4Mg/3) =  (3/10) * NkBT/(Mg),

T' = T * (PA + 2 Mg/3)/(PA + 4Mg/3) = (4/5) * T

Now put the second weight back. That increases the total energy by Mgh compared to the original system. Hence, the piston will not fall back to the original system, but to one with the two weights somewhat higher, and at a somewhat higher temperature. You can calculate it yourself. In general the details depends on (i) how the pressure varies with temperature and volume, and (ii) how the internal energy varies with temperature and volume.

Dear Raman: Many of the comments so far have already noted that the process you have in mind is an irreversible process even though the system has come back to the same equilibrium state. You need to remember an important fact. You can take a system between two equilibrium states by a variety of processes, some of which may be reversible and some irreversible. For a process to be reversible, all intermediate states must be in equilibrium; otherwise the process will be irreversible during which dissipation will occur and irreversible entropy would be generated. The latter is the hallmark of irreversibility. If you remove the weight gradually and very slowly, the syatem will remain in equilibrium and the process will be reversible. If you change the weight abruptly, the system is not in equilibrium during the process and the process is irreversible, even if you have the same weight at the end of the process.

I hope that these additional comments clarify the situation for you.

Puru

Gujrati> even though the system has come back to the same equilibrium state.

No, as I carefully explained, it does not come back to the original state! Because some extra energy is added from outside. (According to my interpretation of the process, which is somewhat vaguely formulated in places.) One may consider a reformulation such that the system return to the starting position. Then no extra energy should be added, but the total entropy of the universe (outside the system) has increased.

Raman, what is an adiabatic canister??

Dear Dr. Olaussen: My understanding of the situation is as follows: There is no heat exchange due to the adiabatic walls. Thus, the change in the energy U comes from the work done on the gas. Per your analysis, U=cT (c=3Nk/2) and PV=NkT with PA=2Mg. After the weight is halved, U'=U-Mgh and V'=V+hA in the new equilibrium. Once the removed weight is added, U'=U'+Mgh and P"A=2Mg=PA in the new equilibrium. Hence, P"=P and U"=U and we obtain the same initial equilibrium state.

You have said: "Now put the second weight back. That increases the total energy by Mgh compared to the original system." I do not see this. The additional weight increses U' by Mgh and not U by Mgh. This is where we differ.

Otherwise, we are in agreement.

Puru

Respected Dr. Kare by adiabatic canister i mean that we can't put in (or put out) any energy in the system via heat, only work is the only way to introduce energy in/out of the system.

now referring to your reply"

If I put the weight back on the piston

then you have to do some work lifting the weight into position. So you are not analyzing an isolated system"

Suppose before the process i already have a weight that i have kept at height h above the initial position so that now i don't have to do any work in bringing that (previously suddenly removed) weight upwards(when i say i put the weight back on) now how do you explain it?

Also suppose no dissipative forces are present.

Isn't viscosity a dissipative force??

Can somebody please elaborate upon the explanation of Dr. Viktor Vladimir Chistyakov , i am not getting him properly but i think he's delivering the explanation in the most general way??

Suppose before the process i already have a weight that i have kept at height h above the initial position so that now i don't have to do any work in bringing that (previously suddenly removed) weight upwards(when i say i put the weight back on) now how do you explain it?

It should not be necessary to explain. If you are unable to understand this on your own, no amount of explanation would help! Do you really believe that by buying a new car with money you had in the bank, you actually got it for free?

Also suppose no dissipative forces are present.

Thermodynamics does not care about what you suppose, if it is against the first and/or second laws.

Guyrati> After the weight is halved, U'=U-Mgh and V'=V+hA in the new equilibrium. Once the removed weight is added, U''=U'+Mgh and P"A=2Mg=PA in the new equilibrium.

No!!!

Start a height 0, then the initial energy is E = U (internal energy), and after the new equilibrium  E = U' + Mgh. With the new weight added, E' = U' + 2Mgh. When the last equilibrium is established, E' = U'' + 2 Mgh''.

Frankly, every 6-year old kid trading chocolate bars with you would manage to get this right (not computing U'' and h'' of course). Beware!  

Dear Dr. Olaussen: I look at it differently. The system I wish to look at consists of the gas with a piston and the two weights. The two weights originally are at the ground level (h=0) and the energy of the system is U. One of the weights is removed keeping at the ground level so no work is done in removing it. Thus, the energy of the system remains at U after the displacement of a weight. After equilibrium is achieved, U'=U-Mgh for the system in which one of the weights is at the ground level but the other one is at h>0. The displaced weight is now lifted to the height h>0 which requires the expenditure of work Mgh. This reduces the energy to U'-Mgh=U-2Mgh. Now after the equilibrium is reached, the system gains energy in the amount of 2Mgh as the weights come back to the ground level. The final energy after equilibration is U"=U-2Mgh+2Mgh=U as I had concluded.

You do not include the work required in lifting the displaced weight to the height h. This explains the difference. In effect, you are thinking of an additional weight, a third weight, at h and two weights at the ground level. What you do is to bring the third weight to the top of the piston at height h in the intermediate step. You can do that but then your final equilibrium state corresponds to all the weights at the ground level. This is not the initial state you started with. You now lift  a weight by h. If you do things carefully, you will come to the same conclusion as I have.

I hope that I have not made things too obscure.

Puru 

The displaced weight is now lifted to the height h>0 which requires the expenditure of work Mgh.

Which is taken from the reservoir, i.e from somewhere outside of the canister system, and added to the canister system.

This reduces the energy to U'-Mgh=U-2Mgh

Assume you have a 6-year old boy, giving you a chocolate bar, expecting to get another brand in return. And you tell him: Look boy  -- the number of chocolate bars in our family has not changed; why should I give you anything?

Well, if I were your 6-year boy I would first hit you on your nose (very deservingly), and start yelling while looking for something really solid to hit your bonehead with. 

I hope that I have not made things too obscure.

Only to yourself.

If the gas was real gas ( not ideal gas), it cools down on adiabatic expansion due to the work done in lifting the (remaining weight) weight irreversibly. The work done is less than the reversible work. When the removed weight is replaced, The gas is compressed, its temperature rises and a final equilibrium state is reached. But this new equilibrium sate is different from the initial equilibrium state (since the work interactions have different values for the irreversible expansion and compression processes). Therefore, the process is not a cyclic process. Again, it is neither quasi static nor reversible process.

Similar is the case with ideal gas; except that the numbers corresponding to the temperature, volume and pressure in the final state, the work interactions during expansion and compression are different. 

How is this process not quasi-static yet reversible?

Please, avoid to use term “adiabatic system”.  There is only adiabatic process and system you describe is closed.  It is not nickel and dimes but it is necessary in the first order to define type of thermodynamic system and then to define thermodynamic process. There are two types of closed systems with

·         Adiabatic boundary – not allowing any heat exchange: A thermally isolated system

·         Rigid boundary – not allowing exchange of work: A mechanically isolated system.

In an adiabatic process, energy is transferred only as work.

Thermodynamic system you suggested represents a closed, thermally isolated system that fits system described at: https://en.wikipedia.org/wiki/Adiabatic_process :

·         If the system walls are adiabatic (Q = 0), but not rigid (W ≠ 0), and, in a fictive idealized process, energy is added to the system in the form of frictionless, non-viscous pressure–volume work, and there is no phase change, the temperature of the system will rise. Such a process is called an isentropic process and is said to be "reversible". Fictively, if the process is reversed, the energy added as work can be recovered entirely as work done by the system. If the system contains a compressible gas and is reduced in volume, the uncertainty of the position of the gas is reduced, and seemingly would reduce the entropy of the system, but the temperature of the system will rise as the process is isentropic (ΔS = 0). Should the work be added in such a way that friction or viscous forces are operating within the system, then the process is not isentropic, and if there is no phase change, then the temperature of the system will rise, the process is said to be "irreversible", and the work added to the system is not entirely recoverable in the form of work.

Explanation:  The fact that the system you suggested is placed in a vacuum is irrelevant because work is done on the weight 2. If the weight 1 is removed and thus system is pushed out from equilibrium then the described system does work pdV expanding from volume V0 to volume V1 pushing the piston and the weight 2 to position  h1. h1>h0 increasing its potential energy…

First law for a closed thermally isolated system dU = - pdV leads to...

 In reversed process energy would be added in form potential energy of weight 1 and then transformed into kinetic energy of a piston doing work on the working substance (gas) changing U.

Obviously this process is reversible.

First, I have attached a pdf with my calculations. In summary

T2 = (4/5)T1, and T3 = (7/5)T2 which leads to T3 = (28/25)T1

Second is the reason. The reason is because the work of a process that is not quasi-static is W = Pext*Delta_V. For a reversible process work is calculated using the integral (because infinitely small increments of change represent a continuous function). So, in this case going from state 1 to state 2, and then from state 2 to state 3, there are going to be two different external pressures. Pext1 = mg, Pext2 = 2mg, where P is pressure, m is mass, g is gravity, V is volume. For the same amount of work to be done in the two changes, obviously the change in volume must be different. So, you won't return to the same state.

Attached are the details.

Kelly has given the correct mathematical analysis for the problem that started this thread. The analysis clearly shows that the final height of the two weights is higher than their initial height and that the final temperature is higher than the initial temperature. This was also mentioned by Olaussen, but I misunderstood his comments and was misled. Thanks to both of them for the correct answer.

Dear Raman, Dear Prof. Dr. Olaussen,

Raman: "Also suppose no dissipative forces are present."

Prof. Dr. Olaussen: "Thermodynamics does not care about what you suppose, if it is against the first and/or second laws".

I am sorry, but in fact the branch you call 'thermodynamics' does in no way deserve such an idenifier under the conditions you cite... 

The First Basic Law = The Energy Conservation Law (note: the TOTAL Energy Conservation!).

The Second Basic Law =  The Energy Transformation Law, the only physically consistent formulation of which was made by August Friedrich Horstmann relatively long ago, but this had been forgotten somehow.

To sum up, any realistic process to be triggered requires some Driving Force, or, as they used to call it long ago the Vis Viva (its correct translation from Latine: The Livening Force). Energetically (mechanically) seen, this ought to be the Kinetic Energy. Where should we then get the latter from ?

Here comes the First Important Energy Transformation: As the Total Energy is conserved, the First Law, we would need some 'Energy Stock' = 'The Potential Energy'. The latter should be somehow TRANSFORMED into the Kinetic Energy. Let us first assume the Hamiltonian Dynamical Representation: There are absolutely no obstacles/hindrances opposing the transormation involved. Then you get some amount of the Kinetic Energy EQUAL TO THE AVAILABLE AMOUNT OF THE Potential Energy, so that you are able to develop your Driving Force from now on...

Now we are capable of triggering our desired process - very nice! What the Driving Force will now be spent for?

Two things - the amount of the Driving Force ought to be enough for:

1. To neutralize the ubiquitous obstacles/hindrances.

2. To enable the desired process as it is.

A very simple living example: You have some money and you would like to buy a piece of bread.

To buy the bread you have to go to the bakery which is three tram stops away from your current spot. This is a natural OBSTACLE.

You need:

1.  To buy a ticket to cover the distance between you and the baker.

2. To buy the piece of bread you need.

If you have enough money for the BOTH ABOVE ACCOMPLISHMENTS, you will win!

This same happens with Thermodynamics:

1. You need to 'reconcile all the gaps available', to 'neutralize all the obstacles and hindrances available'.

2. You need to Drive your process to your Desired Aim.

Now, the true formulation of the Second Basic Law comes:

If you trigger nothing, you get no resistence (not only in the Electric sense, it is in General sense, Isaac Newton started formulating it as the Third Basic Law: any action creates an equal and oppositely directed reaction). The resistence increases as you incerase the driving force. And the resistence tends to come to its maximum - if you have the Driving Force enough to overcome the maximum of the resistence - you just need now only some final amount of the Driving Force to accomplish your desired goals.

This way you arrive at the Equilibruim between all the obstacles and your Will, you achieve your initial aim.

To sum up: The Increasing Entropy: the ubiquitous obstacles. What has been spent to overcome the latter is just the useless, the devaluated Energy. This is the Entropic Loss ... You might denote this as you like, but there is always a dialectic contradiction, a unity and struggle of the opposites, namely - between the Driving Force (say, Enthalpy) and the Hindrances (say, Entropy). Why this is a Struggle - it is clear from the above, and why it is a Unity: because all of this is related to the Energy notion...

...Thus, the true thermodynamic mechanism of any realistic process ought to include the Entropy-Enthalpy Compensation. The Helmholtz and Gibbs potentials might be analyzed within such a conceptual framework... 

... And the above ought to be the true classical thermodynamics.

To finalize, the terms like 'reversible', 'adiabatic', 'closed systems', 'quasi-static processes' etc. are in fact just misnomers having nothing to do with the actual physics. They have to do with the notorious attempts to 'physicalize' the ingenious cyclic widget by N. L. S. Carnot, which was nothing but just a skillfull theoretical tool to analyze the situations we have been discussing above...

Respectfully yours,

Evgeni B. Starikov

A reversible process is a process that can be reversed without leaving any trace on the surroundings. In this processm the system can return to its original state, but not the surroundings.

In the expansion process, the system does work on the surroundings (to against P0), To restore the system to its original state, we have to raise the weight to the height where the piston sits at the end of the expansion, thus leaves the trace on the surroundings.

Therefore, it is NOT the reversible process.

It won't be reversible as, reversibility is defined for a combination of system and surrounding considered together (or more generally, an isolated super-system of several coupled systems). In other words, this process (initiated by removal of the weight) would cause both system and surrounding to change their states. The surrounding here consists of atmosphere and the mass removing agent. Then, even if the system goes back to it's initial state after the mass is replaced, same cannot be said for the surrounding. And if somehow the surrounding is brought back to it's exact inital state, same can't be said for the system. Basically, reversible process is one for which, there is a reverse process in which initial states of both system and surrounding can be re-achieved simultanously.