Reversibility in thermodynamics

[PhysicsInNJ]

Conceptually, why does infinitesimally changing a system allow for a process to be reversible. For example, if we heat a system at temperature T1 to T2 by using a heat reservoir at T2, it is considered irreversible, but if we heat the system with many reservoirs at temperatures T1+dT, T1+2dT... the process becomes reversible.

I see that with the finite temperature difference, you would not be able to reverse the direction of heat flow without outside work, shouldn't the same problem exist for the "reversible" process, albeit with smaller magnitudes? You would not be able to spontaneously return a system at T1+dT to T1.

 

[Chestermiller]

This is a very good question. First let me say that the word "spontaneous" implies irreversible in thermodynamics. So I don't think you meant to use that word.

The question is: How can both the system and the sequence of reservoirs be returned to their original state without doing any work and without having more than a negligible effect on anything else? Here is a hint to how one could proceed: On the reverse path, slightly shift the sequence of reservoirs that the system is brought into contact with. Can you figure out from this hint how to do it?

Chet

 

[PhysicsInNJ]

@Chestermiller If we put the system while it is in a T1 + 2dT configuration back in contact with T1+dT we could revert the temperature to T1 +dT. But this requires work on our part (outside work) to move the system and reservoir in contact, right? Also this wouldn't be able to bring the system back to T1.

 

[Chestermiller]

@Chestermiller If we put the system while it is in a T1 + 2dT configuration back in contact with T1+dT we could revert the temperature to T1 +dT. But this requires work on our part (outside work) to move the system and reservoir in contact, right?

I don't see why it would require work.

Also this wouldn't be able to bring the system back to T1.

You put the system at temperature T+10dT into contact with the reservoir at T+9dT. Then you put the system at temperature T+9dT into contact with the reservoir at T+8dT. Then you put the system at temperature T+8dT into contact with the reservoir at T+7dT. etc.

 

[PhysicsInNJ]

I don't see why it would require work.

I was thinking that if we were hypothetically carrying out this process, we would have to move the reservoirs/systems requiring us to do physical work.

You put the system at temperature T+10dT into contact with the reservoir at T+9dT. Then you put the system at temperature T+9dT into contact with the reservoir at T+8dT. Then you put the system at temperature T+8dT into contact with the reservoir at T+7dT. etc.

But T+dT is the last reservoir we have, so the system would only be able to get down to T+dT, not return to T.

 

[Chestermiller]

I was thinking that if we were hypothetically carrying out this process, we would have to move the reservoirs/systems requiring us to do physical work.

We could reduce such work to zero by putting the system in a frictionless sheet of ice as we moved it into contact with the various reservoirs. This really wouldn't be what we consider thermodynamic work.

But T+dT is the last reservoir we have, so the system would only be able to get down to T+dT, not return to T.

Good call. Yes, we would have to add a reservoir at the one end, and not use the last reservoir at the other end. But, in the limit where dT really does approach zero, this would be a negligible effect.

Irrespective of the $\Delta T$ in each contact with a reservoir, the change in the entropy of the system would be zero for the complete cycle. The change in entropy of the reservoirs by doing this game plan for an arbitrary $\Delta T$ would be $$\Delta S=mC\Delta T\left(\frac{T_H-T_C}{T_HT_C}\right)$$ where $T_H$ is the high end temperature of the system, $T_C$ is the low end temperature of the system, and C is the heat capacity of the system. So, as $\Delta T$ gets smaller, the overall change in entropy for the combination of system and surroundings (reservoirs) would approach zero for the cycle.

 

[PhysicsInNJ]

We could reduce such work to zero by putting the system in a frictionless sheet of ice as we moved it into contact with the various reservoirs. This really wouldn't be what we consider thermodynamic work.

I see, I suppose I was being pedantic and not considering the overall picture.

This made sense, thank you Chet!