Solving the Thermodynamics Paradox with a Ball-Earth System

[Mausam]

I have learned that ΔG is the amount of available energy. So to understand it i used a simmple body system rather than chemical reaction,which is the sliding of a block on a fixed wedge(height h and inclination angleθ).
1)my system is ball -earth and rest surrounding both in thermal eqilibrium at temprature T
2)Now as per the definition H=U+PV, ΔH in this case would be -mgh.
3)If we consider that the wedge is frictionless the amount of energy that we could obtain when the ball reaches floor =mgh,hence the term TΔS=0.Hence the ΔS of both system and surroundings =0(because there is no transfer of heat even if we carry out this process reversibly) therefore ΔS of universe is 0. which contradicts the fact that this process was irreversible which must accompany increase in entropy of universe.

Can anyone please point out the mistake in this.I also had a contradictory situation when i assumed the wedge to have a friction coefficient.

 

[Chestermiller]

The process is not irreversible if the slide is frictionless. It is totally reversible, considering that the block can be allowed to slide up a second slope and regain all its potential energy. There is no mechanical energy dissipated in this process. Therefore, the changes in internal energy U, enthalpy H, entropy S, and free energy G are all zero. All the potential energy is converted to kinetic energy.

I don't understand why you say that there is a change in enthalpy.

 

[256bits]

Try looking for entropy of a pendulum and see if any success pops up.

 

[Mausam]

The process is not irreversible if the slide is frictionless. It is totally reversible, considering that the block can be allowed to slide up a second slope and regain all its potential energy. There is no mechanical energy dissipated in this process. Therefore, the changes in internal energy U, enthalpy H, entropy S, and free energy G are all zero. All the potential energy is converted to kinetic energy.

I don't understand why you say that there is a change in enthalpy.

Thank you for your inputs. I meant its spontaneous in the forward direction , so the entropy of the universe must increase.but the reverse process can't take place without external force working on it. And the kinetic energy is the available energy (gibbs free energy).
My point is that delta G is negative hence its an irreversible spontaneous process, but the entropy of the universe is not changing unless there is friction.
And this process can't be reversible because its not in equilibrium otherwise delta G would have been 0.

 

[Mausam]

Could u please elaborate 256bits

 

[Chestermiller]

Thank you for your inputs. I meant its spontaneous in the forward direction , so the entropy of the universe must increase.but the reverse process can't take place without external force working on it. And the kinetic energy is the available energy (gibbs free energy).
My point is that delta G is negative hence its an irreversible spontaneous process, but the entropy of the universe is not changing unless there is friction.
And this process can't be reversible because its not in equilibrium otherwise delta G would have been 0.

You seem very confused about all this. As long as there is no dissipation of mechanical energy, the system and the surroundings can be returned to their original state. There is no need for an external force for this system and surroundings to return to their original state. Just let the block slide down the slope, slide up an adjacent slope, slide back down the adjacent slope, and slide back up the original slope, and you are back at the original state for both the system and surroundings.

The kinetic energy is not gibbs free energy. What ever gave you that idea?

This is not considered a spontaneous process in the thermodynamic sense. It is just a conversion of one form of mechanical energy (potential energy) to another (kinetic energy). A thermodynaically spontaneous process must involve dissipation of mechanical "driving force", such as conversion of mechanical energy to heat or internal energy (via kinetic friction or fluid friction), spontaneous occurrence of chemical reaction, dissipation of temperature differences by conductive heat transfer, and dissipation of concentration differences by molecular diffusion. All these effects happen at the molecular scale, not the macroscopic scale.

As I said before, the changes in internal energy, enthalpy, entropy, and gibbs free energy for the process you describe are all zero. We know this because they are all physical properties (state functions) of the block material that depend only on its temperature, and its temperature has not changed.

 

[Mausam]

Yes sir actually i am too much confused with thermo ,either the high school books are not accurate or i am not good enough.I cleared lot of my doubts after reading your articles yesterday. i would be grateful if you could help me with this.I know i am wrong but i don't know where.
1)You said that entropy is temperature dependent,but in case of free expansion of a gas in vacuum ΔT=0 but ΔS_uni>0.Even in voltaic cell the temp remains constant but gibbs free energy is not zero.It also implies that to increase(or decrease) the randomness of a system T must change, but why?
2)And in case if there was friction between the wedge and the ball then,
ΔS=Qrev/T=(Work done by friction if the process is carried out reversibly)/T ,
but friction is a non conservative force ∴its work done depends on the path taken
Which contradicts with the fact that entropy is a state function.and even the gibbs free energy.
3)Sir you used the term Driving force for the dissipation of mechanical energy, but how could the dissipation of internal energy drive the process forward?

 

[Chestermiller]

Yes sir actually i am too much confused with thermo ,either the high school books are not accurate or i am not good enough.I cleared lot of my doubts after reading your articles yesterday. i would be grateful if you could help me with this.I know i am wrong but i don't know where.
1)You said that entropy is temperature dependent,but in case of free expansion of a gas in vacuum ΔT=0 but ΔS_uni>0.Even in voltaic cell the temp remains constant but gibbs free energy is not zero.It also implies that to increase(or decrease) the randomness of a system T must change, but why?

The entropy of a solid block is a function only of temperature. The entropy per unit mass of a gas is also a function specific volume or pressure. In a voltaic cell, the entropy is also a function of concentration.

2)And in case if there was friction between the wedge and the ball then,
ΔS=Qrev/T=(Work done by friction if the process is carried out reversibly)/T ,
but friction is a non conservative force ∴its work done depends on the path taken
Which contradicts with the fact that entropy is a state function.and even the gibbs free energy.

If there is frictional work done, the process is not reversible. You can only use the equation you wrote for a reversible path, which this isn't. So, you need to devise a reversible path between the same initial and final state of the block and determine the change in entropy for that path. The reversible path you devise does not have to bear any resemblance whatsoever to the actual path. Here is a Physics Forums Insights article I wrote which gives a cookbook recipe for determining the change in entropy for an irreversible path: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

3)Sir you used the term Driving force for the dissipation of mechanical energy, but how could the dissipation of internal energy drive the process forward?

Who said anything about dissipation of internal energy. By driving force, I mean a temperature gradient, a concentration gradient, a velocity gradient, or a chemical mixture being out of equilibrium.

 

[Stephen Tashi]

I meant its spontaneous in the forward direction , so the entropy of the universe must increase.but the reverse process can't take place without external force working on it.

(I don't see that the process is any more (or less) spontaneous than any other process. For, example, how did the block get on the wedge? Was it held there for a time and then released?)

There are conceptual difficulties in thinking about entropy - or even defining it in certain situations. Your example brings up the difficulty (and perhaps the impossibility) of defining "entropy" and "Gibbs free energy" in a certain situation. It doesn't reach the level of showing a paradox involving those quantities unless we can define them.

The definition of system undergoing a process that is "reversible" doesn't require that the process will eventually reverse itself. The definition of "reversible process" refers to only to the possibility that an earlier state of the process can be reached from a later state of the process by (theoretically) doing "infinitesimal changes" that never perturb the state out of an equilibrium state (which is a paradoxical statement in itself unless we allow for the magic of infinitesimals).

It isn't clear how "the entropy of the universe" is to be defined. The popularized meaning of "entropy of the universe" would only convey the intuitive notion of "disorder". It doesn't provide any technical definition of how entropy is measured. In thermodynamics , entropy is defined for systems in equilibrium. Do we consider the universe to be in equilibrium? - given that the distribution of matter is not homogeneous.

The current Wikipedia article on "Gibbs free energy" quotes Gibb's definition of it as:

the greatest amount of mechanical work which can be obtained from a given quantity of a certain substance in a given initial state, without increasing its total volume or allowing heat to pass to or from external bodies, except such as at the close of the processes are left in their initial condition.

So , in analyzing your example, what do we consider "the substance" to be? And what is the volume of that substance? Or are we attempting to create a more general definition of "Gibbs free energy" than Gibbs did?

Your example apparently intends to portray a deterministic system that is complete specified by the current position and velocity of a single mass. It isn't an "ensemble" of systems that have some nontrivial probability distribution of being in various microstates. At a given time t, your system is in a single state with probability 1. So, in your example, we can't usefully define "entropy" as the Shannon entropy measure of a probability distribution.

 

[Chestermiller]

The definition of system undergoing a process that is "reversible" doesn't require that the process will eventually reverse itself. The definition of "reversible process" refers to only to the possibility that an earlier state of the process can be reached from a later state of the process by (theoretically) doing "infinitesimal changes" that never perturb the state out of an equilibrium state (which is a paradoxical statement in itself unless we allow for the magic of infinitesimals).

In the block example that the OP gave, the changes are not infinitesimal, and mechanically, the system is always out of equilibrium. Yet the system could be returned to its original state without changing anything else. The key to reversibility is that there is no (or negligible) frictional dissipation of mechanical energy, no dissipation of temperature gradients at a finite rate, no dissipation of concentration gradients at a finite rate, and no chemical reaction at a finite rate.

 

[Stephen Tashi]

In the block example that the OP gave, the changes are not infinitesimal, and mechanically, the system is always out of equilibrium

I think you are pointing out that the process described in the OP can only be considered from the thermodynamic point of view if we define "the system" to be the material substances that make up the wedge, the block, and the the horizontal surface. The fact that the material in the block is in a volume that has a net relative motion to the volume of material in the wedge is not a thermodynamic consideration.

But let me make sure that's what you mean.

As the block is sliding down the wedge, I see that there are "unbalanced forces" so the system is out of equilibrium, and after it is moving horizontally, the forces are balanced. ( As to whether there is a net inflow or outflow to "the system", this depends on how the system is defined. If the "the system" has a boundary and the mass slides out of the boundary then I agree there would be a net outflow of energy.)

The thermodynamic definition of "reversible" doesn't require that the changes in the process as it goes in the "forward" direction be infinitesimal. It only requires that the process can be reversed (conceptually) by infinitesimal changes. (This doesn't preclude that the process could also be reversed by non-infinitesimal changes.) The thermodynamic definition of "reversible" does require that the infinitesimal changes move the system only through equilibrium states - so I agree the thermodynamic definition of "reversible" precludes moving the block back up the wedge through the states where the forces were unbalanced.

Expositions of "reversible" mechanical processes usually confine themselves to saying that , in real life, friction makes mechanical processes irreversible, but they fail to define precisely what a reversible mechanical process would be. One is left to conclude that the definition of a "reversible" mechanical process is that it would be a process where there is no friction. That definition-by-default doesn't deal with any concept of finding a path through equilibrium states. So the connection between the definition of a "reversible" mechanical process and "reversible" thermodynamic process is not explained.

Yet the system could be returned to its original state without changing anything else.

I don't understand what you mean by that. Suppose we (physically) add another wedge to the system and have the block slide up the wedge and then slide back down to reverse its direction. To me, we have changed the system. We added another wedge to it. On the other hand, I agree we could conceptually imagine another wedge in the system an imagine the process "reversing itself".

The example in the OP doesn't illustrate any sort of paradox about the concepts of "reversibility" and "Gibbs free energy". In my opinion, this is because it doesn't specify how to define the "Gibbs free energy" of the system in the example. I think your view is that the only things with a defined "Gibbs free energy" in the system are the pieces of material that make up the wedge, the block and the horizontal surface, all of which are at a constant state of thermal equilibrium (in the imagined absence of friction).

 

[Chestermiller]

I think you are pointing out that the process described in the OP can only be considered from the thermodynamic point of view if we define "the system" to be the material substances that make up the wedge, the block, and the the horizontal surface. The fact that the material in the block is in a volume that has a net relative motion to the volume of material in the wedge is not a thermodynamic consideration.

Actually, the system I was considering was just the block.

But let me make sure that's what you mean.

As the block is sliding down the wedge, I see that there are "unbalanced forces" so the system is out of equilibrium, and after it is moving horizontally, the forces are balanced. ( As to whether there is a net inflow or outflow to "the system", this depends on how the system is defined. If the "the system" has a boundary and the mass slides out of the boundary then I agree there would be a net outflow of energy.)

Again, the system I was considering was just the block.

The thermodynamic definition of "reversible" doesn't require that the changes in the process as it goes in the "forward" direction be infinitesimal. It only requires that the process can be reversed (conceptually) by infinitesimal changes. (This doesn't preclude that the process could also be reversed by non-infinitesimal changes.) The thermodynamic definition of "reversible" does require that the infinitesimal changes move the system only through equilibrium states - so I agree the thermodynamic definition of "reversible" precludes moving the block back up the wedge through the states where the forces were unbalanced.

Expositions of "reversible" mechanical processes usually confine themselves to saying that , in real life, friction makes mechanical processes irreversible, but they fail to define precisely what a reversible mechanical process would be. One is left to conclude that the definition of a "reversible" mechanical process is that it would be a process where there is no friction. That definition-by-default doesn't deal with any concept of finding a path through equilibrium states. So the connection between the definition of a "reversible" mechanical process and "reversible" thermodynamic process is not explained.

My understanding of a reversible processes for a system (sometimes referred to as an internally reversible process) is that the system passes through a continuous sequence of thermodynamic equilibrium states. My understanding of a reversible process for a system plus surroundings is that (a) both the system and the surroundings individually pass through a continuous sequence of thermodynamic equilibrium states and (b) that both system and surroundings can be returned to their original states (without affecting anything else in the universe).

I don't understand what you mean by that. Suppose we (physically) add another wedge to the system and have the block slide up the wedge and then slide back down to reverse its direction. To me, we have changed the system. We added another wedge to it. On the other hand, I agree we could conceptually imagine another wedge in the system an imagine the process "reversing itself".

If the system is the block and the surroundings are everything else, then, in the end both the system and the surroundings are returned to their original states.

The example in the OP doesn't illustrate any sort of paradox about the concepts of "reversibility" and "Gibbs free energy". In my opinion, this is because it doesn't specify how to define the "Gibbs free energy" of the system in the example. I think your view is that the only things with a defined "Gibbs free energy" in the system are the pieces of material that make up the wedge, the block and the horizontal surface, all of which are at a constant state of thermal equilibrium (in the imagined absence of friction).

Yes. I agree. I regard the thermodynamic functions as physical properties of the materials involved.