Question about Flow between Parallel Plates

[Chestermiller]

If the process was truly quasistatic, the system pressure = interface pressure = ideal gas pressure and thus pressure change is dependent on the volume change during compression. In reality, does the change in p(I) have little dependence on the volume change of the system and thus no longer quasistatic?

What happens in irreversible non-quasistatic compression is that gas immediately adjacent to the piston becomes compressed first, while the gas further away hasn't gotten compressed yet. This is what happens at short times. As time progresses, the compressed region of gas grows in spatial extent.

Does σ(I) immediately compensate for the drop in friction from static to kinetic (σ(I) + F_kin/A = P_ext +dP) such that the net force on the massless piston is zero?

Yes.

My thought is that since net force must equal 0 per Newton's Second Law, essentially 0 = 0*a, but now a is undefined since there's an infinite number of solutions, so how do we know the kinematics of the piston (displacement, velocity, acceleration) for every dP addition?

This is not correct. It should be F_net = 0*a = 0. Regarding the question about "how do we know the kinematics", the gas has mass/inertia, and you are using the piston to apply a force to it. To find out the time-dependent kinematics of the piston, you would have to solve the gas dynamics partial differential equations using Pext + dP as the time dependent stress boundary condition. Sometimes, this is what you would have to do to solve your problem, even if you are interested only in the initial and final equilibrium states.

Is absolute entropy used when finding ΔS (pre and post combustion), as the equation in my book seem to imply this by having the reference entropy on the right side?

I would have to see what your book has before I comment on this.

Is the reference state entropy is analogous/equivalent to the enthalpy of formation component of a specie's absolute enthalpy?

Yes. The enthalpy is usually referred to a reference state, rather than absolute zero.

How come it is independent pressure unlike the enthalpy of formation?

Neither the enthalpy of formation nor the entropy of formation is dependent on pressure, because they both refer to the pure specie at 1 atm.

Chet

 

[rkmurtyp]

If the process was truly quasistatic, the system pressure = interface pressure = ideal gas pressure and thus pressure change is dependent on the volume change during compression. In reality, does the change in p(I) have little dependence on the volume change of the system and thus no longer quasistatic? Does σ(I) immediately compensate for the drop in friction from static to kinetic (σ(I) + F_kin/A = P_ext +dP) such that the net force on the massless piston is zero?

Since you are also referring to chemical reactions as examples of processes under discussion, the following might help.

In the study of chemical reactions, besides thermodynamic considerations, we study kinetics of reactions to arrive at mechanism (Path followed by the reaction in going from a stste A to a state B) of reaction. Unlike the definite (unique) values for changes in the thermodynamic properties, the values of kinetic properties such as rate constants, do not have unique values. We can think of a number of (possible) mechanisms each of which satisfies some of the experimentally observed/measured properties for a given reaction.

In a similar fashion we can give a number of alternate explanations (for the rate of motion of pistion in the presence/absence of friction, whether we call it quasistatic process or irreversible process etc.

In short, uiqueness is lost in rates of processes - plurality of paths connecting two given states becomes possible, which is not the case with thermodynamics of the process. Thermodynamics gives 'yes' or 'no' type of answers ( a process is either reversible or irreversble - no plurality of answers!) with 100% certainity.

In essence, even if we arrive at a possible explanation of how the piston (whether massless or otherwise) moves, with some model for viscocity, pressure distribution etc of the gas, it will not be unique - there exists no way we can prove that it is the correct or unique path the process followed.

rkmurty

 

[Chestermiller]

In short, uiqueness is lost in rates of processes - plurality of paths connecting two given states becomes possible, which is not the case with thermodynamics of the process. Thermodynamics gives 'yes' or 'no' type of answers ( a process is either reversible or irreversble - no plurality of answers!) with 100% certainity.

In essence, even if we arrive at a possible explanation of how the piston (whether massless or otherwise) moves, with some model for viscocity, pressure distribution etc of the gas, it will not be unique - there exists no way we can prove that it is the correct or unique path the process followed.

If what you're saying here were correct, then we Chemical Engineers could never have done what we have been doing successfully for over a hundred years, which is confidently design and operate real world chemical plants involving equipment such as compressors, heat exchangers, distillation towers, absorption columns, chemical reactors, cooling towers, dryers, adsorption beds, ion exchange columns, evaporators, piping networks, filters, etc.

Chet

 

[Red_CCF]

What happens in irreversible non-quasistatic compression is that gas immediately adjacent to the piston becomes compressed first, while the gas further away hasn't gotten compressed yet. This is what happens at short times. As time progresses, the compressed region of gas grows in spatial extent.

In the example we were solving involving wall friction, in reality stress from gas viscosity must always be significant enough to render the process non-quasistatic to compensate for the static-kinetic friction change? Would this process approach quasistatic if wall frictional effects approaches 0?

This is not correct. It should be F_net = 0*a = 0. Regarding the question about "how do we know the kinematics", the gas has mass/inertia, and you are using the piston to apply a force to it. To find out the time-dependent kinematics of the piston, you would have to solve the gas dynamics partial differential equations using Pext + dP as the time dependent stress boundary condition. Sometimes, this is what you would have to do to solve your problem, even if you are interested only in the initial and final equilibrium states.

So a massless piston's acceleration can be any number including 0 per the gas dynamic equations?

Does the piston stop once σ(I) + Fkin/A = Pext +dP? If the piston continues to move, then σ(I) will continue to increase, and I can't wrap my head around what happens next.

I would have to see what your book has before I comment on this.

Yes. The enthalpy is usually referred to a reference state, rather than absolute zero.

Neither the enthalpy of formation nor the entropy of formation is dependent on pressure, because they both refer to the pure specie at 1 atm.

Chet

Unfortunately the book I have doesn't say much about the second law, just stuck in two equations in an example that computes product entropy (species and mixture) attached. The plot given was w.r.t. S of the mixture. It seems most books just state the enthalpy/entropy fundamental equations as facts without much explanation.

If both enthalpy/entropy of formation are independent of pressure, does this mean that if temperature is kept constant, the enthalpy/entropy of formation of any substance is constant w.r.t pressure (i.e. forming CO2 at 1atm or 100atm releases the same amount of energy)?

Thanks very much

 

[rkmurtyp]

If what you're saying here were correct, then we Chemical Engineers could never have done what we have been doing successfully for over a hundred years, which is confidently design and operate real world chemical plants involving equipment such as compressors, heat exchangers, distillation towers, absorption columns, chemical reactors, cooling towers, dryers, adsorption beds, ion exchange columns, evaporators, piping networks, filters, etc.

Chet

No doubt, we have been designing and operating equipment and processes in everyday practice. However, for each of those designs, we can have several alternate possible designs, the design takes several other factors (such as financial aspects) into consideration. Suppose we ask if a given design for a process is unique - do we get the answer, yes or no? We don't.

In contrast, thermodynamics of a process yeilds unique results/answers. For example, for questions such as: Can the system reach state B spontaneously from state A? we have unique answers, whereas, with regard to rate processes we can not get such unique answers.

rkmurty

 

[Chestermiller]

No doubt, we have been designing and operating equipment and processes in everyday practice. However, for each of those designs, we can have several alternate possible designs, the design takes several other factors (such as financial aspects) into consideration. Suppose we ask if a given design for a process is unique - do we get the answer, yes or no? We don't.

With all due respect, I totally disagree. If I design any of the pieces of equipment I mentioned in my previous post, I fully expect the model predictions to match the observed performance of the equipment with respect to temperatures, pressures, species concentrations, etc. entering and exiting the equipment, as well as throughout the equipment. This kind of predictive capability has validated observationally time and time again. If we engineers couldn't do these things, no one would be shelling out multi-billions of dollars to build chemical plants, and no one would be paying us engineers to big buck to design them.

Having an "all bets are off attitude" like this is highly detrimental. Before you go around telling us engineers what we are able and not able to do, maybe you should study engineering yourself and ascertain the true facts.
[/QUOTE]

 

[rkmurtyp]

With all due respect, I totally disagree. If I design any of the pieces of equipment I mentioned in my previous post, I fully expect the model predictions to match the observed performance of the equipment with respect to temperatures, pressures, species concentrations, etc. entering and exiting the equipment, as well as throughout the equipment. This kind of predictive capability has validated observationally time and time again. If we engineers couldn't do these things, no one would be shelling out multi-billions of dollars to build chemical plants, and no one would be paying us engineers to big buck to design them.

Having an "all bets are off attitude" like this is highly detrimental. Before you go around telling us engineers what we are able and not able to do, maybe you should study engineering yourself and ascertain the true facts.

[/QUOTE]

Thanks

 

[Red_CCF]

With all due respect, I totally disagree. If I design any of the pieces of equipment I mentioned in my previous post, I fully expect the model predictions to match the observed performance of the equipment with respect to temperatures, pressures, species concentrations, etc. entering and exiting the equipment, as well as throughout the equipment. This kind of predictive capability has validated observationally time and time again. If we engineers couldn't do these things, no one would be shelling out multi-billions of dollars to build chemical plants, and no one would be paying us engineers to big buck to design them.

Having an "all bets are off attitude" like this is highly detrimental. Before you go around telling us engineers what we are able and not able to do, maybe you should study engineering yourself and ascertain the true facts.

Hi Chet

When you get a chance, can you take a look at some of the questions I had in post #54 here

Thanks!

 

[Chestermiller]

Hi Chet

When you get a chance, can you take a look at some of the questions I had in post #54 here

Thanks!

Hi Red_CCF. Rest assured that I haven't forgotten #54. I just wanted to take a little more time than usual thinking about exactly how I wanted to answer these questions in an effective way.

Chet

 

[Chestermiller]

In the example we were solving involving wall friction, in reality stress from gas viscosity must always be significant enough to render the process non-quasistatic to compensate for the static-kinetic friction change?

The main contributor to the stress is the increased local pressure in the portion of the gas immediately adjacent to the piston. The viscous stress is secondary, but important long term. As the piston advances, the extent of the compressed gas region increases, and the newly accelerated gas allows the pressure at the piston face to be maintained. All this is happening at short times.

Would this process approach quasistatic if wall frictional effects approaches 0?
So a massless piston's acceleration can be any number including 0 per the gas dynamic equations? Does the piston stop once σ(I) + Fkin/A = Pext +dP? If the piston continues to move, then σ(I) will continue to increase, and I can't wrap my head around what happens next.

The answers to these questions depend on how we control the motion of the piston. We have total control on the external pressure P_ext(t) that we apply and/or the kinematics of the piston motion. Imagine that there is a push-rod attached to the top of the piston, and we control the motion of the rod by hand. When we feel the static friction give way, we can back off on the pressure we apply, so that the piston moves at whatever slow velocity we wish. Or we can try to hold the force we apply constant at the value that existed when the static friction gave way, in which case the gas immediately adjacent to the piston will start to accelerate. We can apply any motion we desire to the piston by controlling P_ext(t). When we do this, the non-uniform deformation within the gas will adjust itself in such a manner that σ(I) always satisfies the equation σ(I) + Fkin/A = Pext(t).

Unfortunately the book I have doesn't say much about the second law, just stuck in two equations in an example that computes product entropy (species and mixture) attached. The plot given was w.r.t. S of the mixture. It seems most books just state the enthalpy/entropy fundamental equations as facts without much explanation.

Sorry your book doesn't do a good job. I looked over the excerpt you sent, and it (at least) looks correct. What they are trying to do is calculate what the entropy of the reaction mixture would be at various conversions. The conversion that maximizes the entropy for this isolated reacting system is the one that corresponds to the equilibrium conversion.

If both enthalpy/entropy of formation are independent of pressure, does this mean that if temperature is kept constant, the enthalpy/entropy of formation of any substance is constant w.r.t pressure (i.e. forming CO2 at 1atm or 100atm releases the same amount of energy)?

It's not quite correct to say that the enthalpy/entropy of formation are "independent of pressure." Both the enthalpy of formation and the entropy of formation occur at a constant pressure of one atmosphere. For an ideal gas, the enthalpy is independent of pressure, but the entropy definitely depends on a pressure. After the material is formed from the elements at 1 atm. pressure, if the pressure of the species changes, its entropy changes.

Chet

 

[Red_CCF]

The main contributor to the stress is the increased local pressure in the portion of the gas immediately adjacent to the piston. The viscous stress is secondary, but important long term. As the piston advances, the extent of the compressed gas region increases, and the newly accelerated gas allows the pressure at the piston face to be maintained. All this is happening at short times.

What did you mean by long term?

The answers to these questions depend on how we control the motion of the piston. We have total control on the external pressure P_ext(t) that we apply and/or the kinematics of the piston motion. Imagine that there is a push-rod attached to the top of the piston, and we control the motion of the rod by hand. When we feel the static friction give way, we can back off on the pressure we apply, so that the piston moves at whatever slow velocity we wish. Or we can try to hold the force we apply constant at the value that existed when the static friction gave way, in which case the gas immediately adjacent to the piston will start to accelerate. We can apply any motion we desire to the piston by controlling P_ext(t). When we do this, the non-uniform deformation within the gas will adjust itself in such a manner that σ(I) always satisfies the equation σ(I) + Fkin/A = Pext(t).

Let's say the system is initially at P_o + Fstat/A = Pext and some force is added such that static friction begins to give away to kinetic friction. Assuming that we can back off this force fast enough such that the gas only sees an addition of dP to the external pressure such that now P + Fkin/A = Pext + dP.

1. Can we say at this point that the process so far is quasistatic (P = ideal gas pressure) as we are adding an infinitesimal dP as opposed to an finite F_stat-F_kin which is significant enough for to local pressure increase and viscous stresses?

2. I would assume that regardless of whether the piston is massless or not, at the point in which the gas pressure is P where P + Fkin/A = Pext + dP, the piston probably still be moving. If the piston continues to move, then P + Fkin/A > Pext + dP and the piston would decelerate and push back (if not massless) or just stop (if massless) to maintain force balance. To maintain a positive forward motion without stoppage, the only way I see is to add dP continuously. At the end of the compression, would a massless piston just stop without trouble (and maintain quasistatic) but a piston with mass will now oscillate barring no changes in Pext?

Sorry your book doesn't do a good job. I looked over the excerpt you sent, and it (at least) looks correct. What they are trying to do is calculate what the entropy of the reaction mixture would be at various conversions. The conversion that maximizes the entropy for this isolated reacting system is the one that corresponds to the equilibrium conversion.

Was the equation used for the plot the absolute entropy, since the reference entropy is on the right side of the equation?

It's not quite correct to say that the enthalpy/entropy of formation are "independent of pressure." Both the enthalpy of formation and the entropy of formation occur at a constant pressure of one atmosphere. For an ideal gas, the enthalpy is independent of pressure, but the entropy definitely depends on a pressure. After the material is formed from the elements at 1 atm. pressure, if the pressure of the species changes, its entropy changes.

Chet

For ideal gases, both its enthalpy and enthalpy of formation is pressure independent but the entropy of formation is pressure dependent? Most examples I found conveniently had reactions at 1atm, but if a reaction is at some other pressure can I use the same enthalpy/entropy of formation?

I'm also curious of the way absolute enthalpy is defined. Given absolute enthalpy = enthalpy of formation at (T_ref, P_ref or 298K and 1atm) + sensible enthalpy. I noticed that enthalpy of formation at (T_f, P_ref ) is much lower than the absolute enthalpy at (T_f,P_ref) from using the proper equation. Is there a reason why we assume the formation occurs at some standard state (seems a bit arbitrary) and heated to the final temperature as opposed to forming directly at T_f,P_ref, which gave the wrong result?

Thanks very much

 

[Chestermiller]

What did you mean by long term?

Well, you remember that with a frictionless piston, there can be oscillation. We would have to solve the gas dynamics equations to figure out what is happening after the early part of the deformation, as when, for example, the piston reaches its forward extent for the first time and may want to move backward. The inertia of the gas can also contribute to this picture, since parts of the gas may still be moving when the piston reaches the end of its first stroke.

Let's say the system is initially at P_o + Fstat/A = Pext and some force is added such that static friction begins to give away to kinetic friction. Assuming that we can back off this force fast enough such that the gas only sees an addition of dP to the external pressure such that now P + Fkin/A = Pext + dP.

1. Can we say at this point that the process so far is quasistatic (P = ideal gas pressure) as we are adding an infinitesimal dP as opposed to an finite F_stat-F_kin which is significant enough for to local pressure increase and viscous stresses?

Yes. However, if the process is quasistatic like this, the viscous stresses in the gas are negligible, and the gas pressure is uniform throughout the cylinder. This is the problem we have have already analyzed.

2. I would assume that regardless of whether the piston is massless or not, at the point in which the gas pressure is P where P + Fkin/A = Pext + dP, the piston probably still be moving. If the piston continues to move, then P + Fkin/A > Pext + dP and the piston would decelerate and push back (if not massless) or just stop (if massless) to maintain force balance. To maintain a positive forward motion without stoppage, the only way I see is to add dP continuously.

Yes. That's what we found when we analyzed the problem with friction previously. The gas pressure within the cylinder under these quasistatic conditions will increase, and Pext will have to increase. The gas pressure will be uniform in the cylinder, and the viscous stresses will be negligible.

At the end of the compression, would a massless piston just stop without trouble (and maintain quasistatic) but a piston with mass will now oscillate barring no changes in Pext?

This question confuses me. If the process is quasistatic, there will be no oscillation. But, Pext will have to increase as the pressure of the gas increases.

Was the equation used for the plot the absolute entropy, since the reference entropy is on the right side of the equation?

When I hear the term "absolute entropy," I think of a reference state at absolute zero temperature. I don't think this is what you mean. The entropy on the left side of the equation is the entropy of the mixture, relative to the combined entropy of the pure components (separated) and each in its reference state of 1 atm and 25C.

For ideal gases, both its enthalpy and enthalpy of formation is pressure independent but the entropy of formation is pressure dependent?

The entropy of formation describes the change:
Equilibrium State 1: Pure elements in stoichiometric proportions at 25 C and 1 atm
Equilibrium State 2: Pure compound at 25C and 1 atm
As with any system, to determine the entropy change, you would have to identify a reversible path between state 1 and state 2, and then determine the Q for the path.

Most examples I found conveniently had reactions at 1atm, but if a reaction is at some other pressure can I use the same enthalpy/entropy of formation?

Yes. In the examples, the total pressure was constant, but not the partial pressures of the various species. The term involving partial pressure in your equations for the individual species partial molar entropies takes into account the difference between the total pressure and the partial pressure of the species.

I'm also curious of the way absolute enthalpy is defined. Given absolute enthalpy = enthalpy of formation at (T_ref, P_ref or 298K and 1atm) + sensible enthalpy. I noticed that enthalpy of formation at (T_f, P_ref ) is much lower than the absolute enthalpy at (T_f,P_ref) from using the proper equation. Is there a reason why we assume the formation occurs at some standard state (seems a bit arbitrary) and heated to the final temperature as opposed to forming directly at T_f,P_ref, which gave the wrong result?

I don't quite understand this question. The reason for using a reference state is so we don't have to tabulate the enthalpies of the species at every temperature. All we need to do is provide the enthalpy of formation at one temperature (the reference temperature), and provide the heat capacity of the species. I'm sure you already knew this.

Chet

 

[Red_CCF]

This question confuses me. If the process is quasistatic, there will be no oscillation. But, Pext will have to increase as the pressure of the gas increases.

In the process I was imagining, if the piston (with mass) is still moving when P + Fkin/A = Pext + dP, any additional motion by the piston would mean that P + Fkin/A > Pext + dP and the piston will decelerate and move backwards where forces are balanced which is not desirable. I thought a solution would be to add dP as soon as the forces balance continuously, which will accelerate the piston by small amounts each time. At the end of the compression, I would still end up with positive momentum for the piston when P + Fkin/A = Pext + ∫dP so if the external forces are maintained constant, I see an oscillation occurring if the piston moves further down (increasing P).

If the piston is massless at the point where P + Fkin/A = Pext + dP, would it stop instantly as the case where P + Fkin/A > Pext + dP is not physically possible?

Yes. That's what we found when we analyzed the problem with friction previously. The gas pressure within the cylinder under these quasistatic conditions will increase, and Pext will have to increase. The gas pressure will be uniform in the cylinder, and the viscous stresses will be negligible.

Is the quasistatic nature/viscous stresses affected by the actual speed of the piston or simply how much we are disturbing it from equilibrium? In the scenario I mentioned above I am continuously accelerating the piston so I would get a continuously higher velocities.

When I hear the term "absolute entropy," I think of a reference state at absolute zero temperature. I don't think this is what you mean. The entropy on the left side of the equation is the entropy of the mixture, relative to the combined entropy of the pure components (separated) and each in its reference state of 1 atm and 25C.

With regards to the equation:
s_i = s_i^o (Tref) + ∫c_p\frac{dT}{T} - R_uln(\frac{P_i}{P^o})

I took the latter two terms to represent entropy change from the reference state to the final state, and the first to be the formation entropy. I also assumed that the first term to be an absolute entropy (from absolute zero) such that s_i is now relative to absolute zero, is this correct?

The entropy of formation describes the change:
Equilibrium State 1: Pure elements in stoichiometric proportions at 25 C and 1 atm
Equilibrium State 2: Pure compound at 25C and 1 atm
As with any system, to determine the entropy change, you would have to identify a reversible path between state 1 and state 2, and then determine the Q for the path.

Is it possible even theoretically for a reaction be reversible?

I don't quite understand this question. The reason for using a reference state is so we don't have to tabulate the enthalpies of the species at every temperature. All we need to do is provide the enthalpy of formation at one temperature (the reference temperature), and provide the heat capacity of the species. I'm sure you already knew this.

Chet

My book tabulates both the enthalpy of formation as a function of temperature as well as the sensible enthalpy change from reference (labelled as absolute entropy - enthapy of formation @ 298K) all at 1atm. What I had thought is that, for say CO2, the enthalpy of formation at say 1000K should be equal to enthalpy of formation @ 298K + Sensible Enthalpy (298K to 1000K). I found the two to be nowhere close to equal, so I'm confused on why we have to use formation enthalpy at a reference state always even if the reaction takes place at much higher temperatures such that species would be forming at higher temperature. I have the same question about entropy as well.

Thank you

 

[Chestermiller]

In the process I was imagining, if the piston (with mass) is still moving when P + Fkin/A = Pext + dP, any additional motion by the piston would mean that P + Fkin/A > Pext + dP and the piston will decelerate and move backwards where forces are balanced which is not desirable. I thought a solution would be to add dP as soon as the forces balance continuously, which will accelerate the piston by small amounts each time. At the end of the compression, I would still end up with positive momentum for the piston when P + Fkin/A = Pext + ∫dP so if the external forces are maintained constant, I see an oscillation occurring if the piston moves further down (increasing P).

You are increasing Pext gradually in a quasistatic process. You can do this as slowly as you wish, such that the piston momentum is negligible, even if it has mass. Also, as soon at the piston tries to reverse direction, static friction will kick in again. This whole question is not really a thermodynamics question. It is strictly a mechanics question. I'd like you to try modelling it yourself. Model the gas as a Hooke's law spring with an initial compression. Increase the force Pext a tiny incremental, and see if, with friction, the piston indeed does oscillate.

If the piston is massless at the point where P + Fkin/A = Pext + dP, would it stop instantly as the case where P + Fkin/A > Pext + dP is not physically possible?

I don't understand this question. Solving the modeling exercise above will probably answer this question too.

Is the quasistatic nature/viscous stresses affected by the actual speed of the piston or simply how much we are disturbing it from equilibrium? In the scenario I mentioned above I am continuously accelerating the piston so I would get a continuously higher velocities.

Viscous stresses are caused by velocity gradients in the fluid. If there are no significant velocity gradients, then the viscous stresses are negligible. The absolute magnitude of the deformation is not important (as far as viscous stresses are concerned). If you want to get an idea of how viscous stresses come into play, include a damping element in parallel with the spring in the model I discussed above.

With regards to the equation:
s_i = s_i^o (Tref) + ∫c_p\frac{dT}{T} - R_uln(\frac{P_i}{P^o})

I took the latter two terms to represent entropy change from the reference state to the final state, and the first to be the formation entropy. I also assumed that the first term to be an absolute entropy (from absolute zero) such that s_i is now relative to absolute zero, is this correct?

It doesn't matter, as long as everything is done consistently. You can take s_i as the entropy of the pure species relative to absolute zero, or you can take the entropy of pure elements at Tref as zero. Either way, you get the same answer for changes in entropy.

Is it possible even theoretically for a reaction be reversible?

Yes. Look up van't Hoff equilibrium box in your book or google

My book tabulates both the enthalpy of formation as a function of temperature as well as the sensible enthalpy change from reference (labelled as absolute entropy - enthapy of formation @ 298K) all at 1atm. What I had thought is that, for say CO2, the enthalpy of formation at say 1000K should be equal to enthalpy of formation @ 298K + Sensible Enthalpy (298K to 1000K). I found the two to be nowhere close to equal, so I'm confused on why we have to use formation enthalpy at a reference state always even if the reaction takes place at much higher temperatures such that species would be forming at higher temperature. I have the same question about entropy as well.

You have to use Hess' law, and subtract the sensible heats of carbon and oxygen (weighted stoichiometrically).

Chet

 

[rkmurtyp]

Is it possible even theoretically for a reaction be reversible?

Thank you

If a chemical reaction involves (theoritically) transformation of chemical energy into any form of energy other than heat (electrical energy, for example), such reaction is deemed to be reversible.

Therefore, the question we are left with is about reactions involving energy transformations involving heat.

We can consider the following situations:

1. Initial state (A) is just the reactants (or just the products) and final state (B) is a
stiochiometric mixture of reactants and products.

2. Initial state is equilibrium mixture of reactants and products in state A, and final state is
equilibrium mixture of reactants and products in state B.

In either case, we take the system from state B to state A by a reversible process, so that the system completes a cycle. The only changes left now, are in the surroundings.

We sum up the quantities Q_i/T_i, that is, calculate ∑(Q_i/T_i). If we get a positive number we say the process AB is irreversible; if we get zero, we say the process AB is reversible.

It is, therefore, possible for a chemical reaction to be reversible in principle (theoritically).

rkmurty

 

[Red_CCF]

You are increasing Pext gradually in a quasistatic process. You can do this as slowly as you wish, such that the piston momentum is negligible, even if it has mass. Also, as soon at the piston tries to reverse direction, static friction will kick in again. This whole question is not really a thermodynamics question. It is strictly a mechanics question. I'd like you to try modelling it yourself. Model the gas as a Hooke's law spring with an initial compression. Increase the force Pext a tiny incremental, and see if, with friction, the piston indeed does oscillate.


I don't understand this question. Solving the modeling exercise above will probably answer this question too.

In the situation I was thinking, the piston wouldn't be stopping and going at every dP addition. I was trying to avoid the constant back and forth switch between kinetic and static friction, which means that the piston is constantly accelerating up to the end of the compression. The force balance I came up with was:

APext + AdP + mg - k|(x-x₀)| - Fkin = ma

I am not sure if the above is correct or how I would solve this to get the velocity at the end of the compression.

Viscous stresses are caused by velocity gradients in the fluid. If there are no significant velocity gradients, then the viscous stresses are negligible. The absolute magnitude of the deformation is not important (as far as viscous stresses are concerned). If you want to get an idea of how viscous stresses come into play, include a damping element in parallel with the spring in the model I discussed above.

Would the speed of the piston have an effect of on the velocity gradient in the fluid? I am thinking that the faster I compress, the less time the fluid further away from the piston face to "react" to the change, and the further I deviate from a quasistatic process.

It doesn't matter, as long as everything is done consistently. You can take s_i as the entropy of the pure species relative to absolute zero, or you can take the entropy of pure elements at Tref as zero. Either way, you get the same answer for changes in entropy.

The s^o(T_ref) from the tables is different for every species, if I were to subtract s^o(T_ref) from s_i such that the entropy is relative to a specie's reference state and then using that to calculate ΔS, would this be the same as ΔS calculated using absolute entropy?

Thanks very much

 

[Chestermiller]

In this response, I will be referring to the attachment below:
https://www.physicsforums.com/attachment.php?attachmentid=70091&stc=1&d=1401192844

In the situation I was thinking, the piston wouldn't be stopping and going at every dP addition. I was trying to avoid the constant back and forth switch between kinetic and static friction, which means that the piston is constantly accelerating up to the end of the compression. The force balance I came up with was:

APext + AdP + mg - k|(x-x₀)| - Fkin = ma

I am not sure if the above is correct or how I would solve this to get the velocity at the end of the compression.

In the attachment, I have included two "models" of the gas behavior to help is quantify our discussion. The first model has one degree of freedom, and features a spring in parallel with a damper (the combination is commonly referred to as a Voigt element). The spring simulates the local p-v-t behavior of the gas, and the damper simulates the viscous dissipation. The parameter x1 represents the displacement of the piston.

The second model has two Voigt elements, with a mass in-between (simulating the mass of the gas). This model has two degrees of freedom. In this lumped model, x2 represents the displacement of the piston, and x1 represents the displacement half-way through the gas. The stiffness of the springs and the damper constants in this model are not the same as for the single degree of freedom model, but are appropriately adjusted to give equivalent behavior.

In my responses in this posting, we will be focusing exclusively on the first model. Later we can get to the second model, if necessary.

For the single degree of freedom model, the force balance just before static friction releases would read :
APext+ mg + kx₁₀ - Fstat = 0
where x₁₀ is negative to represent initial compression of the gas.

I'm going to stop here for now, and give you a chance to digest the gas models, and ask questions.

Chet

 

[Chestermiller]

Would the speed of the piston have an effect of on the velocity gradient in the fluid? I am thinking that the faster I compress, the less time the fluid further away from the piston face to "react" to the change, and the further I deviate from a quasistatic process.

Yes.

The s^o(T_ref) from the tables is different for every species, if I were to subtract s^o(T_ref) from s_i such that the entropy is relative to a specie's reference state and then using that to calculate ΔS, would this be the same as ΔS calculated using absolute entropy?

I'm a little confused over this question. Are you referring to ΔS for a single species or ΔS for a mixture of species?

Chet

 

[Red_CCF]

In this response, I will be referring to the attachment below:
https://www.physicsforums.com/attachment.php?attachmentid=70091&stc=1&d=1401192844

In the attachment, I have included two "models" of the gas behavior to help is quantify our discussion. The first model has one degree of freedom, and features a spring in parallel with a damper (the combination is commonly referred to as a Voigt element). The spring simulates the local p-v-t behavior of the gas, and the damper simulates the viscous dissipation. The parameter x1 represents the displacement of the piston.

The second model has two Voigt elements, with a mass in-between (simulating the mass of the gas). This model has two degrees of freedom. In this lumped model, x2 represents the displacement of the piston, and x1 represents the displacement half-way through the gas. The stiffness of the springs and the damper constants in this model are not the same as for the single degree of freedom model, but are appropriately adjusted to give equivalent behavior.

In my responses in this posting, we will be focusing exclusively on the first model. Later we can get to the second model, if necessary.

For the single degree of freedom model, the force balance just before static friction releases would read :
APext+ mg + kx₁₀ - Fstat = 0
where x₁₀ is negative to represent initial compression of the gas.

I'm going to stop here for now, and give you a chance to digest the gas models, and ask questions.

Chet

Is there a reason why we are adding a damper? If we assume that there is always a dP difference between Pext and Friction + P_I (the system is able to pull back force once static gives away to kinetic friction), the process should be quasistatic so would the "damping" effect from viscous stresses would be zero and are we just doing a general case where we can set the damping to coefficient to zero for the quasistatic model?

If the piston was massless, can we just set m = 0 of the general solution or does it affect the fundamental setup in some way?

I'm a little confused over this question. Are you referring to ΔS for a single species or ΔS for a mixture of species?

Chet

I was referring to ΔS of the mixture. For the formation of water H₂ + 1/2O₂ -> H₂O, assuming the process goes to completion with no dissociation and initial and final equilibrium temperatures are the same at 298K (and constant pressure), the entropy of formation s^o(T_ref) for the reactants is 130.595 + 205.043*0.5 = 232.52kJ/molK but for the product (just water) it is 188.715kJ/molK and the two doesn't cancel. In this case the absolute entropy = entropy of formation so had I used the entropy of each species as relative to its reference state entropy, I should have gotten zero for both (and ΔS_mix = 0)?

Thanks very much

 

[Chestermiller]

Is there a reason why we are adding a damper? If we assume that there is always a dP difference between Pext and Friction + P_I (the system is able to pull back force once static gives away to kinetic friction), the process should be quasistatic so would the "damping" effect from viscous stresses would be zero and are we just doing a general case where we can set the damping to coefficient to zero for the quasistatic model?

If the piston was massless, can we just set m = 0 of the general solution or does it affect the fundamental setup in some way?

In recent posts, you were referring to the effects of viscous stresses, and to pistons with mass that accelerate. So to try to address these all at one time (rather than bouncing from one to the other), I thought it would be better if we got a little more general. We will be solving several cases. Is that OK with you?

I was referring to ΔS of the mixture. For the formation of water H₂ + 1/2O₂ -> H₂O, assuming the process goes to completion with no dissociation and initial and final equilibrium temperatures are the same at 298K (and constant pressure), the entropy of formation s^o(T_ref) for the reactants is 130.595 + 205.043*0.5 = 232.52kJ/molK but for the product (just water) it is 188.715kJ/molK and the two doesn't cancel. In this case the absolute entropy = entropy of formation so had I used the entropy of each species as relative to its reference state entropy, I should have gotten zero for both (and ΔS_mix = 0)?

These entropies are absolute entropies. The difference between the 188.715kJ/molK and the 232.52kJ/molK is the change in entropy as a result of reaction, in going from pure oxygen and pure hydrogen at 298K and 1 atm to pure water at 298K and 1 atm.

Chet

 

[Chestermiller]

Continuing the analysis, the force balance on the piston after static friction releases is given by:

AP_{ext}(t)+mg-F_{kin}+\left(kx_1+C\frac{dx_1}{dt}\right)=-m\frac{d^2x_1}{dt^2}

where C is the (viscous) damping constant and where the piston displacement x₁ is measured upward (so, for compression, x₁ is negative).

If we subtract the force balance equation just before static friction releases from this equation, we obtain:

AΔP_{ext}(t)+(F_{stat}-F_{kin})+\left(kΔx_1+C\frac{d(Δx_1)}{dt}\right)=-m\frac{d^2(Δx_1)}{dt^2}
where ΔP_{ext}=P_{ext}-P_{ext,static} and Δx_1=x_1-x_{10}.

Is this formulation acceptable to you so far?

Chet

 

[Red_CCF]

These entropies are absolute entropies. The difference between the 188.715kJ/molK and the 232.52kJ/molK is the change in entropy as a result of reaction, in going from pure oxygen and pure hydrogen at 298K and 1 atm to pure water at 298K and 1 atm.

Chet

So does this mean that the absolute entropy must always be used, and we must never use entropy that is w.r.t. the reference state/entropy of formation (298K, 1atm) as the ΔS of the reaction becomes 0 in the above case?

Continuing the analysis, the force balance on the piston after static friction releases is given by:

AP_{ext}(t)+mg-F_{kin}+\left(kx_1+C\frac{dx_1}{dt}\right)=-m\frac{d^2x_1}{dt^2}

where C is the (viscous) damping constant and where the piston displacement x₁ is measured upward (so, for compression, x₁ is negative).

If we subtract the force balance equation just before static friction releases from this equation, we obtain:

AΔP_{ext}(t)+(F_{stat}-F_{kin})+\left(kΔx_1+C\frac{d(Δx_1)}{dt}\right)=-m\frac{d^2(Δx_1)}{dt^2}
where ΔP_{ext}=P_{ext}-P_{ext,static} and Δx_1=x_1-x_{10}.

Is this formulation acceptable to you so far?

Chet

In the first equation, at the instant the friction switches to kinetic, is the LHS = dP such that dP = -m\frac{dx^2}{dt^2} (difference between applied pressure and sum of system forces)?

Thanks very much

 

[Chestermiller]

So does this mean that the absolute entropy must always be used, and we must never use entropy that is w.r.t. the reference state/entropy of formation (298K, 1atm) as the ΔS of the reaction becomes 0 in the above case?

No. We could have taken the entropies of formation of the elemental species to be zero at 298 and 1atm, in which case the entropy of formation of water in the reference state would have been 189-233 = -44. In this approach, the free energy of formation of the water (from the elements) in the reference state would have been -241826+(298)(44), where the -241826 is the heat of formation of the water (g) in the reference state. If we are working with the equilibrium of chemical reactions, it is important to know the free energies of formation of the species in the reference state, and the enthalpy of formation in the reference state (the latter to get the temperature dependence of the equilibrium constant). The entropy of formation is secondary to this exercise. However, if we are using the entropy of formation to calculate the free energy of formation, the reference state for entropy and for free energy of formation must be consistent.

In the first equation, at the instant the friction switches to kinetic, is the LHS = dP such that dP = -m\frac{dx^2}{dt^2} (difference between applied pressure and sum of system forces)?

No. A differential can not be equal to something that is not differential.

Try to think of it this way. We have numerous options regarding Pext, since we have complete control over it. So we need to decide what we want to do. Each option results in a different deformation history. Here are some options:

1. We instantly back off on the Pext, so that equilibrium is re-established, such that Pext decreases by exactly F_stat-T_kin. This re-establishes static friction, and the piston doesn't move.
2. We move the piston in whatever way necessary such that Pext does not change once friction switches to kinetic. This results in an irreversible deformation.
3. We instantly back off on Pext, so that Pext decreases by F_stat-T_kin-dP. In this case, the deformation is quasistatic. Under these circumstances, the gas deforms reversibly, but the adiabatic process for the combination of gas plus piston is irreversible (because of the friction, as we already showed). Here, we have to decide what overall pressure change we want to impose.
4. We apply an impulse force to the piston at time zero, so it starts moving with an initial velocity, and then let friction and viscous dissipation slow it down.


I think the interesting options are 2 and 4. Can you think of others? What do you want to consider?

Chet

 

[Red_CCF]

No. We could have taken the entropies of formation of the elemental species to be zero at 298 and 1atm, in which case the entropy of formation of water in the reference state would have been 189-233 = -44. In this approach, the free energy of formation of the water (from the elements) in the reference state would have been -241826+(298)(44), where the -241826 is the heat of formation of the water (g) in the reference state. If we are working with the equilibrium of chemical reactions, it is important to know the free energies of formation of the species in the reference state, and the enthalpy of formation in the reference state (the latter to get the temperature dependence of the equilibrium constant). The entropy of formation is secondary to this exercise. However, if we are using the entropy of formation to calculate the free energy of formation, the reference state for entropy and for free energy of formation must be consistent.

I think I just have a logic issue. My thought is that, if I set the entropy of each species to be relative to its absolute entropy at reference state (298K, 1atm), which is essentially its entropy of formation, then in the formation of water example I had, the entropy of each species would be 0 since the entropy is measured at the reference state relative to the reference state. So if I do Δs_reaction = s_prod - s_reactant, I'd end up with 0 = 0 - 0 when it is supposed to be -44 regardless of how I define the reference.

No. A differential can not be equal to something that is not differential.

Try to think of it this way. We have numerous options regarding Pext, since we have complete control over it. So we need to decide what we want to do. Each option results in a different deformation history. Here are some options:

1. We instantly back off on the Pext, so that equilibrium is re-established, such that Pext decreases by exactly F_stat-T_kin. This re-establishes static friction, and the piston doesn't move.
2. We move the piston in whatever way necessary such that Pext does not change once friction switches to kinetic. This results in an irreversible deformation.
3. We instantly back off on Pext, so that Pext decreases by F_stat-T_kin-dP. In this case, the deformation is quasistatic. Under these circumstances, the gas deforms reversibly, but the adiabatic process for the combination of gas plus piston is irreversible (because of the friction, as we already showed). Here, we have to decide what overall pressure change we want to impose.
4. We apply an impulse force to the piston at time zero, so it starts moving with an initial velocity, and then let friction and viscous dissipation slow it down.


I think the interesting options are 2 and 4. Can you think of others? What do you want to consider?

Chet

For option 3, how come we can say that Pext decreases by F_stat-T_kin-dP but yet cannot set an equality involving dP?

I think that option 2 is closer to practical applications and is probably similar to option 3; those two are the ones I am most interested in looking at.

Thank you very much

 

[Chestermiller]

I think I just have a logic issue. My thought is that, if I set the entropy of each species to be relative to its absolute entropy at reference state (298K, 1atm), which is essentially its entropy of formation, then in the formation of water example I had, the entropy of each species would be 0 since the entropy is measured at the reference state relative to the reference state. So if I do Δs_reaction = s_prod - s_reactant, I'd end up with 0 = 0 - 0 when it is supposed to be -44 regardless of how I define the reference.

Only the entropies of formation of the elemental species are taken to be zero. The entropies of formation of compounds are not.

For option 3, how come we can say that Pext decreases by F_stat-T_kin-dP but yet cannot set an equality involving dP?

I don't understand this question.

I think that option 2 is closer to practical applications and is probably similar to option 3; those two are the ones I am most interested in looking at.

Let's do option 2. We will first consider the case where the piston mass m = 0, and then do the case where the piston has mass. Please take a shot at writing the differential equation for the case of no mass (in terms of the Δ's). If you are comfortable with your differential equation, please solve it for Δx₁ as a function to time t.

Chet

 

[Red_CCF]

Only the entropies of formation of the elemental species are taken to be zero. The entropies of formation of compounds are not.

I notice that all element enthalpy of formation is zero at reference state so I assume this is from the same reasoning? How come this is true? For entropy, I assumed that if everything was relative to the reference state, each specie's entropy would be defined as s_absolute - s_{reference state}; how come this only holds for elements?

I don't understand this question.

I meant, how come we can express in words that Pext is decreased by F_stat-T_kin-dP but we can't set a force balance like P_ext = P_ext,static - (F_stat-T_kin-dP)

Let's do option 2. We will first consider the case where the piston mass m = 0, and then do the case where the piston has mass. Please take a shot at writing the differential equation for the case of no mass (in terms of the Δ's). If you are comfortable with your differential equation, please solve it for Δx₁ as a function to time t.

Chet

\frac{d(Δx_1)}{dt}+\frac{k}{C}Δx_1= -\frac{A}{C}ΔP_{ext}(t)-\frac{1}{C}(F_{stat}-F_{kin})

Δx_1 = -\frac{1}{e^\frac{kt}{C}}∫e^\frac{kt}{C}\frac{A}{C}ΔPext(t)dt - \frac{Fstat - Fkin}{k}+\frac{CONST}{e^\frac{kt}{C}}

I imagine that Pext(t) is the forcing function and its properties be known. I am unsure how to evaluate the arbitrary constant without knowledge of the first integral though.

I'm also wondering, what is the difference between the equation with and without Δ?

Thanks

 

[Chestermiller]

I notice that all element enthalpy of formation is zero at reference state so I assume this is from the same reasoning? How come this is true? For entropy, I assumed that if everything was relative to the reference state, each specie's entropy would be defined as s_absolute - s_{reference state}; how come this only holds for elements?

What we are trying to do here is determine the changes in entropy as a result of changes in pressure, temperature, and chemical reaction between two thermodynamic equilibrium states, say A and B. To do this, we need to establish a proper mathematical framework for the analysis. It doesn't matter whether we choose the absolute entropy of a pure species as the zero point for entropy (in the case of the elements), or whether we choose the entropy to be zero for the pure elemental species at some other state such as 298 and 1 atm. Either way, when you calculate the change in entropy from state A to state B, you will get the same answer. You need to convince yourself of that. I don't really know the details of how they calculate absolute entropies of species, but this doesn't really matter to me. If I know that this is the basis for the entropy values in a table, I can use them. Similarly, if I know that the entropies in the table are relative to the pure elemental species in their equilibrium state at 298 and 1 atm, I can use them. These are the two different bases that are commonly used in various tables.

I meant, how come we can express in words that Pext is decreased by F_stat-T_kin-dP but we can't set a force balance like P_ext = P_ext,static - (F_stat-T_kin-dP)

I don't have any problem with this, but what does dP represent? Also, what is this "force balance" being applied to?

\frac{d(Δx_1)}{dt}+\frac{k}{C}Δx_1= -\frac{A}{C}ΔP_{ext}(t)-\frac{1}{C}(F_{stat}-F_{kin})

Δx_1 = -\frac{1}{e^\frac{kt}{C}}∫e^\frac{kt}{C}\frac{A}{C}ΔPext(t)dt - \frac{Fstat - Fkin}{k}+\frac{CONST}{e^\frac{kt}{C}}

I imagine that Pext(t) is the forcing function and its properties be known. I am unsure how to evaluate the arbitrary constant without knowledge of the first integral though.

In the description of Case 2, what do the words "Pext does not change" mean to you with regard to ΔP_ext? This should give you enough information to determine CONST.

I'm also wondering, what is the difference between the equation with and without Δ?

It just seems easier to me to work in terms of the equation with the Δ's; it eliminates the x₀ and the mg. Mathematically, of course, there's no difference.

Chet

 

[Red_CCF]

What we are trying to do here is determine the changes in entropy as a result of changes in pressure, temperature, and chemical reaction between two thermodynamic equilibrium states, say A and B. To do this, we need to establish a proper mathematical framework for the analysis. It doesn't matter whether we choose the absolute entropy of a pure species as the zero point for entropy (in the case of the elements), or whether we choose the entropy to be zero for the pure elemental species at some other state such as 298 and 1 atm. Either way, when you calculate the change in entropy from state A to state B, you will get the same answer. You need to convince yourself of that. I don't really know the details of how they calculate absolute entropies of species, but this doesn't really matter to me. If I know that this is the basis for the entropy values in a table, I can use them. Similarly, if I know that the entropies in the table are relative to the pure elemental species in their equilibrium state at 298 and 1 atm, I can use them. These are the two different bases that are commonly used in various tables.

What I am confused by is, why are pure species treated differently than compounds when setting s = 0 in reference state? How come I can't subtract off, from the absolute entropy at some T and p, a substances' entropy at its reference state such that any substance has zero entropy at its reference state, which I am imagining as a "origin/zero point"?

I don't have any problem with this, but what does dP represent? Also, what is this "force balance" being applied to?

I was thinking in terms of the force balance for the quasistatic case at the instant that friction switches from static to kinetic. In such a case, would P_ext(t = 0⁺) = P_{ext, static} - (Fstatic/A - Fkinetic/A - dP)? This would result in P_ext(t = 0⁺)A + mg - P_IA - Fkinetic = dP = ma.

In the description of Case 2, what do the words "Pext does not change" mean to you with regard to ΔP_ext? This should give you enough information to determine CONST.

It just seems easier to me to work in terms of the equation with the Δ's; it eliminates the x₀ and the mg. Mathematically, of course, there's no difference.

Chet

If ΔP_ext(0) = 0 and Δx1(t) = 0

CONST = e^\frac{kt}{C}(\frac{Fstat - Fkin}{k}) and
Δx_1 = -\frac{1}{e^\frac{kt}{C}}∫e^\frac{kt}{C}\frac{A}{C}ΔPext(t)dt

Thank you

 

[Chestermiller]

What I am confused by is, why are pure species treated differently than compounds when setting s = 0 in reference state? How come I can't subtract off, from the absolute entropy at some T and p, a substances' entropy at its reference state such that any substance has zero entropy at its reference state, which I am imagining as a "origin/zero point"?

If we take S = 0 for the pure elements in the reference state, then the entropy of the compound formed from these elements at the reference temperature and pressure can't be zero. Let's consider your specific example:

State A: 1/2 mole of pure oxygen(g) and 1 mole of pure hydrogen(g) (in separate containers) each at the reference temperature and pressure.

State B: 1 mole of pure liquid water at the reference temperature and pressure (in a third container).

The change in entropy for a process that goes from State A to State B is not equal to zero. The "entropy of formation" of liquid water at the reference temperature and pressure is defined as the change in entropy ΔS for this specific process.

Now, the real challenge is to dream up a reversible process that goes from State A to State B (so that the change in entropy can be measured). Can you think of such a process? It might help to think in terms of a vant Hoff equilibrium box, and semipermeable membranes.

I was thinking in terms of the force balance for the quasistatic case at the instant that friction switches from static to kinetic. In such a case, would P_ext(t = 0⁺) = P_{ext, static} - (Fstatic/A - Fkinetic/A - dP)? This would result in P_ext(t = 0⁺)A + mg - P_IA - Fkinetic = dP = ma.

Sorry. I still don't follow. Is dP supposed to be the change in pressure of the gas, or is it supposed to represent a slight incremental increase in the external pressure?

If ΔP_ext(0) = 0 and Δx1(t) = 0

CONST = e^\frac{kt}{C}(\frac{Fstat - Fkin}{k}) and
Δx_1 = -\frac{1}{e^\frac{kt}{C}}∫e^\frac{kt}{C}\frac{A}{C}ΔPext(t)dt

Case 2 states that "P_ext does not change" means that ΔP_ext is not just zero at time t = 0. It means that ΔP_ext is held equal to zero for all times. In other words, P_ext is held constant throughout the deformation at its (high) value that existed just before static friction released. So:
Δx_1 = -\frac{(F_{stat} - F_{kin})}{k}\left(1-e^{-\frac{kt}{C}}\right)
This is how the piston has to be moved to keep P_ext constant at its value that existed just before static friction released.

Chet

 

[Red_CCF]

If we take S = 0 for the pure elements in the reference state, then the entropy of the compound formed from these elements at the reference temperature and pressure can't be zero. Let's consider your specific example:

State A: 1/2 mole of pure oxygen(g) and 1 mole of pure hydrogen(g) (in separate containers) each at the reference temperature and pressure.

State B: 1 mole of pure liquid water at the reference temperature and pressure (in a third container).

The change in entropy for a process that goes from State A to State B is not equal to zero. The "entropy of formation" of liquid water at the reference temperature and pressure is defined as the change in entropy ΔS for this specific process.

Now, the real challenge is to dream up a reversible process that goes from State A to State B (so that the change in entropy can be measured). Can you think of such a process? It might help to think in terms of a vant Hoff equilibrium box, and semipermeable membranes.

From my limited understanding of Van't Hoff's box, I believe there would be 3 piston cylinder assemblies (each containing one of O2, H2, H2O) each separated from a reaction chamber by a semipermeable membrane. Assuming the reaction chamber containing O2, H2, H2O is initially at equilibrium, adding infinitessimal amounts of O2 and H2 produces an infinitessimal amount of H2O that is immediately extracted at equilibrium pressure and I assume that the heat generated is also reversibly rejected. I know ΔS=0 for the reaction chamber but not sure how to prove that ΔS of the three piston-cylinder combined plus heat loss is also 0.

I was imagining the term reference like an origin and that if absolute zero was the reference, formation entropy would be zero for every species so moving this reference to another state (298K, 1atm) should mean that the entropy of every species (compound or not) starts at zero from that reference, even though I understand that thermodynamically this gives invalid results.

From what I have read, am I correct to understand reference state as some "arbitrary" T and p (conventionally taken as 298K, 1atm) where it is defined that only element formation properties (entropy, enthalpy, free energy) are equal to zero? Also, I read that elements refers to elements in their "natural state". I understand qualitatively what it means (O2 but not O, S8 but not S), but is there a formal definition on what natural state is and would it vary with T and p?

Sorry. I still don't follow. Is dP supposed to be the change in pressure of the gas, or is it supposed to represent a slight incremental increase in the external pressure?

I meant dP to mean an incremental increase in external pressure.

Case 2 states that "P_ext does not change" means that ΔP_ext is not just zero at time t = 0. It means that ΔP_ext is held equal to zero for all times. In other words, P_ext is held constant throughout the deformation at its (high) value that existed just before static friction released. So:
Δx_1 = -\frac{(F_{stat} - F_{kin})}{k}\left(1-e^{-\frac{kt}{C}}\right)
This is how the piston has to be moved to keep P_ext constant at its value that existed just before static friction released.

Chet

So in this example, the P-v curve would be a horizontal line? Since the piston is massless, the force balance at the system boundary at any instance in time, but the second derivative of Δx is equal to k²/C²e^-kt/C which isn't zero at any point in time implying that the massless piston is always accelerating, what could be causing this?

Thanks very much

 

[Chestermiller]

From my limited understanding of Van't Hoff's box, I believe there would be 3 piston cylinder assemblies (each containing one of O2, H2, H2O) each separated from a reaction chamber by a semipermeable membrane. Assuming the reaction chamber containing O2, H2, H2O is initially at equilibrium, adding infinitessimal amounts of O2 and H2 produces an infinitessimal amount of H2O that is immediately extracted at equilibrium pressure and I assume that the heat generated is also reversibly rejected. I know ΔS=0 for the reaction chamber but not sure how to prove that ΔS of the three piston-cylinder combined plus heat loss is also 0.

Very close. You have the right idea; there are only a few steps missing. Initially you have 1/2 mole of oxygen in the oxygen cylinder, 1 mole of hydrogen in the hydrogen cylinder, and 0 moles of water in the water cylinder. You first expand the oxygen and the hydrogen within their individual cylinders isothermally and reversibly from their initial pressures of 1 atm until their pressures are equal to their equilibrium partial pressures within the box. You then inject them isothermally and reversibly through their semipermeable membranes into the equilibrium box, while at the same time removing water vapor isothermally and reversibly at its equilibrium partial pressure through its semipermeable membrane (into the water cylinder which is held at a pressure equal to the water vapor partial pressure within the equilibrium box). After all the oxygen and hydrogen have been injected into the equilibrium box and the water vapor produced by the reaction has been removed into its cylinder, there will be one mole of water vapor in the water cylinder at the water partial pressure within the equilibrium box. The final step is to compress the water within its cylinder isothermally and reversibly from its equilibrium partial pressure in the box to a pressure of 1 atm., including condensing the water to liquid water at 1 atm.

You are correct in noting that, for the contents of the equilibrium box, there is no entropy change as a result of this process. However, the ΔS of the three piston-cylinder combined plus heat loss is not zero. If you had measured the reversible heat added when expanding the oxygen and hydrogen, the reversible heat removed from the equilibrium box as a result of the reaction, and the reversible heat added when the compressing the water to 1 atm, you would find that the net reversible heat would not be zero. If we divided this net heat by the temperature of the system, we would get the entropy of formation of liquid water.

I was imagining the term reference like an origin and that if absolute zero was the reference, formation entropy would be zero for every species so moving this reference to another state (298K, 1atm) should mean that the entropy of every species (compound or not) starts at zero from that reference, even though I understand that thermodynamically this gives invalid results.

I'm not quite sure what you're saying, but, even if absolute zero were the reference, formation entropy for compounds would not be zero, since the reaction to produce the compound from the elements would involve a finite entropy change, even at absolute zero.

From what I have read, am I correct to understand reference state as some "arbitrary" T and p (conventionally taken as 298K, 1atm) where it is defined that only element formation properties (entropy, enthalpy, free energy) are equal to zero?

Yes. This is the basis for one of the two different approaches. The other is that the entropies of the elements are zero at absolute zero, and, at 298 and 1 atm, the entropies of the elements are their absolute entropies at their natural state at 298 and 1 atm. With either way of doing it, you get the same result for the entropy change for any and all reactions.

Also, I read that elements refers to elements in their "natural state". I understand qualitatively what it means (O2 but not O, S8 but not S), but is there a formal definition on what natural state is and would it vary with T and p?

The natural state is usually understood as you described it without separately specifying it for each species. The natural state also applies to compounds. It doesn't vary with T and p, because T and p for the reference state are 298 and 1 atm, period.

So in this example, the P-v curve would be a horizontal line?

Sure. The P_ext-v curve would be a horizontal line.

Since the piston is massless, the force balance at the system boundary at any instance in time, but the second derivative of Δx is equal to k²/C²e^-kt/C which isn't zero at any point in time implying that the massless piston is always accelerating, what could be causing this?

The net force on a massless piston that is accelerating is zero. So what. What is the big surprise?

Anyway, next we are going to be considering the case where the piston has mass. Then you'll be able to see what the difference is. So, for Case 2, what is the differential equation for the piston motion Δx₁ in the case where the piston has mass? What do you get for the solution to this differential equation?

Here is a summary so far for the piston motion:

1. No viscous dissipation, no piston mass, no gas mass: Δx₁=ΔF/k

2. With viscous dissipation, no piston mass, no gas mass: Δx_1=\frac{ΔF}{k}(1-e^{-\frac{kt}{C}})

(Note that, in this situation, the piston does not oscillate.)

Chet

 

[Chestermiller]

I was thinking in terms of the force balance for the quasistatic case at the instant that friction switches from static to kinetic. In such a case, would P_ext(t = 0⁺) = P_{ext, static} - (Fstatic/A - Fkinetic/A - dP)? This would result in P_ext(t = 0⁺)A + mg - P_IA - Fkinetic = dP = ma.

Hi. It finally dawned on me what you are asking here. For this case,
P_{ext}(t)=P_{ext,static}-\frac{(F_{static}-F_{kinetic})}{A}+δP_{ext}(t)
This is the same as:
ΔP_{ext}(t)=-\frac{(F_{static}-F_{kinetic})}{A}+δP_{ext}(t)
If we substitute this into our differential equation, we obtain:
AδP_{ext}(t)=-kΔx_1-C\frac{dΔx_1}{dt}-m\frac{d^2Δx_1}{dt^2}
Now, let's suppose that we force the piston to undergo a very small constant acceleration of a. The question is, in what way would δP have to be varied as a function of time to accomplish this? With constant acceleration, we would have
Δx_1=\frac{1}{2}at^2
and
t=\sqrt{\frac{2Δx_1}{a}}
If we substitute these relationships into the differential equation, we obtain:
AδP_{ext}(t)=-(kΔx_1+C\sqrt{2aΔx_1}+ma)
Now, if the acceleration is very small, but we wait a very long time for the displacement to reach a desired value of Δx₁, this equation reduces to:
AδP_{ext}(t)=-kΔx_1
Do you see that, in this quasistatic case, the acceleration term and the viscous dissipation term become negligible?

Chet

 

[Red_CCF]

Very close. You have the right idea; there are only a few steps missing. Initially you have 1/2 mole of oxygen in the oxygen cylinder, 1 mole of hydrogen in the hydrogen cylinder, and 0 moles of water in the water cylinder. You first expand the oxygen and the hydrogen within their individual cylinders isothermally and reversibly from their initial pressures of 1 atm until their pressures are equal to their equilibrium partial pressures within the box. You then inject them isothermally and reversibly through their semipermeable membranes into the equilibrium box, while at the same time removing water vapor isothermally and reversibly at its equilibrium partial pressure through its semipermeable membrane (into the water cylinder which is held at a pressure equal to the water vapor partial pressure within the equilibrium box). After all the oxygen and hydrogen have been injected into the equilibrium box and the water vapor produced by the reaction has been removed into its cylinder, there will be one mole of water vapor in the water cylinder at the water partial pressure within the equilibrium box. The final step is to compress the water within its cylinder isothermally and reversibly from its equilibrium partial pressure in the box to a pressure of 1 atm., including condensing the water to liquid water at 1 atm.

You are correct in noting that, for the contents of the equilibrium box, there is no entropy change as a result of this process. However, the ΔS of the three piston-cylinder combined plus heat loss is not zero. If you had measured the reversible heat added when expanding the oxygen and hydrogen, the reversible heat removed from the equilibrium box as a result of the reaction, and the reversible heat added when the compressing the water to 1 atm, you would find that the net reversible heat would not be zero. If we divided this net heat by the temperature of the system, we would get the entropy of formation of liquid water.

Why does the initial state of the reactants and the end state of the products have to be at 1atm? Would the potential sources of irreversibilities for this process would include that from heating, piston work, and deviation from equilibrium of the reaction chamber?

I'm not quite sure what you're saying, but, even if absolute zero were the reference, formation entropy for compounds would not be zero, since the reaction to produce the compound from the elements would involve a finite entropy change, even at absolute zero.

So for compounds, the absolute entropy of say H2O wouldn't be zero at absolute zero?

Yes. This is the basis for one of the two different approaches. The other is that the entropies of the elements are zero at absolute zero, and, at 298 and 1 atm, the entropies of the elements are their absolute entropies at their natural state at 298 and 1 atm. With either way of doing it, you get the same result for the entropy change for any and all reactions.

In the formation of H2O at 298 and 1atm, what is the difference between the -44kJ/kmolK and the 188.7kJ/kmolK values? The table I have has formation enthalpy/free energy relative to reference state (element h_f/g_f = 0), but it seems that formation entropy isn't (element s(Tref) ≠ 0 and varies with temperature, also they didn't use a subscript _f for s like with the others which may be significant). If I were to remake this table w.r.t. reference state for H2O, would I put the -44 value under its entropy?

The natural state is usually understood as you described it without separately specifying it for each species. The natural state also applies to compounds. It doesn't vary with T and p, because T and p for the reference state are 298 and 1 atm, period.

Is there no formal definition on what a species'/element natural state is for species' where it won't be intuitively obvious, or is it just the most common form of the species that we encounter?

The net force on a massless piston that is accelerating is zero. So what. What is the big surprise?

I was thinking that a net force is needed to produce acceleration, now I am thinking that as m -> 0, would F -> 0 to produce some finite acceleration?

Anyway, next we are going to be considering the case where the piston has mass. Then you'll be able to see what the difference is. So, for Case 2, what is the differential equation for the piston motion Δx₁ in the case where the piston has mass? What do you get for the solution to this differential equation?

Here is a summary so far for the piston motion:

1. No viscous dissipation, no piston mass, no gas mass: Δx₁=ΔF/k

2. With viscous dissipation, no piston mass, no gas mass: Δx_1=\frac{ΔF}{k}(1-e^{-\frac{kt}{C}})

(Note that, in this situation, the piston does not oscillate.)

Chet

For the case with mass, the ODE becomes:
m\frac{d^2(Δx_1)}{dt^2}+C\frac{d(Δx_1)}{dt}+kΔx_1=(F_{kin}-F_{stat})
The solution will depend on the actual values of m,C, and k
r = \frac{-C\pm\sqrt{C^2-4mk}}{2m}
assuming that the roots are real the homogeneous solution would be
Δx_1 = c_1e^{\frac{-C+\sqrt{C^2-4mk}}{2m}t} + c_2e^{\frac{-C-\sqrt{C^2-4mk}}{2m}t}+(F_{kin}-F_{stat})/k
if the roots are imaginary, the homogeneous solution would be
Δx_1 = e^{\frac{-C}{2m}t}(c_1sin(\frac{\sqrt{4mk-C^2}}{2m}) + c_2cos(\frac{\sqrt{4mk-C^2}}{2m})+(F_{kin}-F_{stat})/k
which gives a decaying oscillation. I assume that the first peak would take the piston past equilibrium BDC and as t -> ∞ we converge at the actual BDC which happen to be (F_kin-F_stat)/k.

Thanks very much

 

[Red_CCF]

Hi. It finally dawned on me what you are asking here. For this case,
P_{ext}(t)=P_{ext,static}-\frac{(F_{static}-F_{kinetic})}{A}+δP_{ext}(t)
This is the same as:
ΔP_{ext}(t)=-\frac{(F_{static}-F_{kinetic})}{A}+δP_{ext}(t)
If we substitute this into our differential equation, we obtain:
AδP_{ext}(t)=-kΔx_1-C\frac{dΔx_1}{dt}-m\frac{d^2Δx_1}{dt^2}
Now, let's suppose that we force the piston to undergo a very small constant acceleration of a. The question is, in what way would δP have to be varied as a function of time to accomplish this? With constant acceleration, we would have
Δx_1=\frac{1}{2}at^2
and
t=\sqrt{\frac{2Δx_1}{a}}
If we substitute these relationships into the differential equation, we obtain:
AδP_{ext}(t)=-(kΔx_1+C\sqrt{2aΔx_1}+ma)
Now, if the acceleration is very small, but we wait a very long time for the displacement to reach a desired value of Δx₁, this equation reduces to:
AδP_{ext}(t)=-kΔx_1
Do you see that, in this quasistatic case, the acceleration term and the viscous dissipation term become negligible?

Chet

Is there a reason why δP is used here instead of dP?

I've always thought that an infinitessimal quantity is already the smallest possible unit (and essentially a "constant"), but here we have to vary this to maintain a certain acceleration small enough for the approximation to hold. Are we allowed to vary this (and to what degree) and still call it an infintessimal disturbance as it appears that δP increases with time as Δx increases with time?

Thanks very much

 

[Chestermiller]

Why does the initial state of the reactants and the end state of the products have to be at 1atm?

That's just the convention that's used. It's assumed that 1 atm is low enough for ideal gas behavior to apply.

Would the potential sources of irreversibilities for this process would include that from heating, piston work, and deviation from equilibrium of the reaction chamber?

I don't quite get what you are driving at here. In the description we discussed, the entire transition from the initial equilibrium state to the final equilibrium state was reversible. In irreversible processes, irreversibilities result from finite heat conduction, finite viscous dissipation, and finite chemical reaction rates.

So for compounds, the absolute entropy of say H2O wouldn't be zero at absolute zero?

No, it wouldn't.

In the formation of H2O at 298 and 1atm, what is the difference between the -44kJ/kmolK and the 188.7kJ/kmolK values? The table I have has formation enthalpy/free energy relative to reference state (element h_f/g_f = 0), but it seems that formation entropy isn't (element s(Tref) ≠ 0 and varies with temperature, also they didn't use a subscript _f for s like with the others which may be significant). If I were to remake this table w.r.t. reference state for H2O, would I put the -44 value under its entropy?

I'm puzzled by all this. The free energy, enthalpy, and entropy at the reference state have to be related by:

g_ref=h_ref-T_refs_ref

If g_ref and h_ref of the elements are zero at the reference state, then s_ref also has to be zero. The g, h, and s of formation of the elements vary with temperature, but at the reference state, they are taken as zero. The g, h, and s of formation of compounds also vary with the temperature, but at the reference state, they are not zero. I wish I could see your table so I could understand better.

Is there no formal definition on what a species'/element natural state is for species' where it won't be intuitively obvious, or is it just the most common form of the species that we encounter?

Typically, the latter. But, in the case of any ambiguity, the state of the species has to be specified precisely.

I was thinking that a net force is needed to produce acceleration, now I am thinking that as m -> 0, would F -> 0 to produce some finite acceleration?

Yes. That's what I've been trying to say. We'll see how this plays out in our analyses.

For the case with mass, the ODE becomes:
m\frac{d^2(Δx_1)}{dt^2}+C\frac{d(Δx_1)}{dt}+kΔx_1=(F_{kin}-F_{stat})
The solution will depend on the actual values of m,C, and k
r = \frac{-C\pm\sqrt{C^2-4mk}}{2m}
assuming that the roots are real the homogeneous solution would be
Δx_1 = c_1e^{\frac{-C+\sqrt{C^2-4mk}}{2m}t} + c_2e^{\frac{-C-\sqrt{C^2-4mk}}{2m}t}+(F_{kin}-F_{stat})/k

Yes. This is correct. Now, before we talk about the cases where the mass is large enough for the roots to be imaginary, let's talk a little more about the case of real roots. I would like to re-express the roots in a slightly different form:
r_1 = \frac{-C+\sqrt{C^2-4mk}}{2m}=\frac{-2k}{C+\sqrt{C^2-4mk}}=\frac{-2k/C}{1+\sqrt{1+\left(\frac{4mk}{C^2}\right)}}
r_2 = \frac{-C-\sqrt{C^2-4mk}}{2m}=\frac{-2k}{C-\sqrt{C^2-4mk}}=\frac{-2k/C}{1-\sqrt{1-\left(\frac{4mk}{C^2}\right)}}
For small values of the piston mass m, do you know how to linearize these roots with respect to $\frac{4mk}{C^2}$? If so, please do. Also, from the initial conditions, do you know how to determine the constants of integration c₁ and c₂? If so, please do.

Chet

 

[Chestermiller]

Is there a reason why δP is used here instead of dP?

I've always thought that an infinitessimal quantity is already the smallest possible unit (and essentially a "constant"), but here we have to vary this to maintain a certain acceleration small enough for the approximation to hold. Are we allowed to vary this (and to what degree) and still call it an infintessimal disturbance as it appears that δP increases with time as Δx increases with time?

Compressing a gas quasistatically by, say, a factor of 2 requires more than just a single differential change in the compressive pressure on the piston. It requires a sequence of differential pressure changes applied gradually over time so that they sum to a finite pressure change δP. The more the gas is compressed, the higher the cumulative pressure increase δP has to be.

Chet

 

[Red_CCF]

I don't quite get what you are driving at here. In the description we discussed, the entire transition from the initial equilibrium state to the final equilibrium state was reversible. In irreversible processes, irreversibilities result from finite heat conduction, finite viscous dissipation, and finite chemical reaction rates.

I didn't word that question very well. I was really wondering how adding the reactants too quickly or not removing the products fast enough from the reaction chamber will lead to irreversibilities?

I'm puzzled by all this. The free energy, enthalpy, and entropy at the reference state have to be related by:

g_ref=h_ref-T_refs_ref

If g_ref and h_ref of the elements are zero at the reference state, then s_ref also has to be zero. The g, h, and s of formation of the elements vary with temperature, but at the reference state, they are taken as zero. The g, h, and s of formation of compounds also vary with the temperature, but at the reference state, they are not zero. I wish I could see your table so I could understand better.

I have attached part of the table for H₂. I am probably missing/mis-interpreting something. I believe that the entropy give is absolute while the free energy and enthalpy of formation are relative to the reference state; however it seems the book on purposely chose not to term the entropy as "formation" (with subscript f). I believe that the s^o(T) in the table is the same as that in the equation in the second attachment.

Yes. This is correct. Now, before we talk about the cases where the mass is large enough for the roots to be imaginary, let's talk a little more about the case of real roots. I would like to re-express the roots in a slightly different form:
r_1 = \frac{-C+\sqrt{C^2-4mk}}{2m}=\frac{-2k}{C+\sqrt{C^2-4mk}}=\frac{-2k/C}{1+\sqrt{1+\left(\frac{4mk}{C^2}\right)}}
r_2 = \frac{-C-\sqrt{C^2-4mk}}{2m}=\frac{-2k}{C-\sqrt{C^2-4mk}}=\frac{-2k/C}{1-\sqrt{1-\left(\frac{4mk}{C^2}\right)}}
For small values of the piston mass m, do you know how to linearize these roots with respect to $\frac{4mk}{C^2}$? If so, please do. Also, from the initial conditions, do you know how to determine the constants of integration c₁ and c₂? If so, please do.

Chet

From assuming Δx = 0 and d(Δx)/dt = 0 at t = 0, for the constants I got:

c_1 = \frac{(F_{kin} - F_{stat})(1-\sqrt{1-\frac{4mk}{C^2}})}{2k\sqrt{1+\frac{4mk}{C^2}}}
c_2 = \frac{(F_{stat} - F_{kin})(1+\sqrt{1+\frac{4mk}{C^2}})}{2k\sqrt{1+\frac{4mk}{C^2}}}

I'm not sure if I'm doing this correctly, but the square root would be linearized, for f(x) = \sqrt{1\pm x} :

L_1(x) = \sqrt{1-\frac{4mk}{C^2}} - 1/2(x-\frac{4mk}{C^2})(1-\frac{4mk}{C^2})^{-0.5} for \sqrt{1-\frac{4mk}{C^2}}

and

L_2(x) = \sqrt{1+\frac{4mk}{C^2}} + 1/2(x-\frac{4mk}{C^2})(1+\frac{4mk}{C^2})^{-0.5} for \sqrt{1+\frac{4mk}{C^2}}

Compressing a gas quasistatically by, say, a factor of 2 requires more than just a single differential change in the compressive pressure on the piston. It requires a sequence of differential pressure changes applied gradually over time so that they sum to a finite pressure change δP. The more the gas is compressed, the higher the cumulative pressure increase δP has to be.

Chet

To maintain a small but constant acceleration, are we basically applying δP (via series of dP) to let the piston accelerate initially and wait for the piston to decelerate by a certain extent before applying δP again?

Thanks very much

 

[Red_CCF]

The attachments didn't attach for some reason.

 

[Chestermiller]

I didn't word that question very well. I was really wondering how adding the reactants too quickly or not removing the products fast enough from the reaction chamber will lead to irreversibilities?

The irreversibility really comes in in allowing the reaction to proceed at a finite rate. If you allow the reaction to proceed at a finite rate, then you can't return the system to the initial state (temperature, pressure, composition) without causing a significant change in the surroundings.

I have attached part of the table for H₂. I am probably missing/mis-interpreting something. I believe that the entropy give is absolute while the free energy and enthalpy of formation are relative to the reference state; however it seems the book on purposely chose not to term the entropy as "formation" (with subscript f). I believe that the s^o(T) in the table is the same as that in the equation in the second attachment.

I'm not familiar with what they are doing in these tables, and need more context. What they're doing seems very confusing to me, and apparently it has confused you also. Here's what several texts that I have access to do:

Smith and Van Ness, Introduction to Chemical Engineering thermodynamics:
The standard heat of formation of a species is equal to the change in enthalpy in going from the elements to the species at 1 atm and temperature T.
The standard heat of formation of a species in the reference state is the standard heat of formation at 298.
The standard heat of formation of the elements is equal to zero in the reference state.
The standard free energy of formation of a species is equal to the change in free energy in going from the elements to the species at 1 atm and temperature T.
The standard free energy of formation of a species in the reference state is the standard free energy of formation at 298.
The standard free energy of formation of the elements is equal to zero in the reference state.

Perry and Chilton, Chemical Engineers' Handbook
Same as for Smith and Van Ness

Hougan, Watson, and Ragatz, Chemical Process Principles
Heats of formation handled the same as previous two references.
Standard Molar Absolute Entropies used to get entropy change for a reaction in the reference state (298, 1 atm).
Standard Molar Enthalpies and Entropies used to get change in free energy for a reaction in the reference state.
Standard free energy changes (1atm) for reactions at other temperatures calculated from \frac{d(ΔG^o/RT)}{d(1/T)}=\frac{ΔH^o}{R}

From assuming Δx = 0 and d(Δx)/dt = 0 at t = 0, for the constants I got:

c_1 = \frac{(F_{kin} - F_{stat})(1-\sqrt{1-\frac{4mk}{C^2}})}{2k\sqrt{1+\frac{4mk}{C^2}}}
c_2 = \frac{(F_{stat} - F_{kin})(1+\sqrt{1+\frac{4mk}{C^2}})}{2k\sqrt{1+\frac{4mk}{C^2}}}

I'm not sure if I'm doing this correctly, but the square root would be linearized, for f(x) = \sqrt{1\pm x} :

L_1(x) = \sqrt{1-\frac{4mk}{C^2}} - 1/2(x-\frac{4mk}{C^2})(1-\frac{4mk}{C^2})^{-0.5} for \sqrt{1-\frac{4mk}{C^2}}

and

L_2(x) = \sqrt{1+\frac{4mk}{C^2}} + 1/2(x-\frac{4mk}{C^2})(1+\frac{4mk}{C^2})^{-0.5} for \sqrt{1+\frac{4mk}{C^2}}

I'm going to come back to this problem tomorrow.

To maintain a small but constant acceleration, are we basically applying δP (via series of dP) to let the piston accelerate initially and wait for the piston to decelerate by a certain extent before applying δP again?

No. δP is a function of time. It is the time-dependent pressure variation we need to apply over the entire deformation to hold the acceleration constant at a very small value. When we reach the desired final deformation, we stop increasing δP and, since it then only results in a negligible net load on the piston, the deformation ceases.
AδP=\left(\frac{1}{2}at^2\right)k+(at)C+ma
where a is the downward acceleration.

Chet

 

[Red_CCF]

The irreversibility really comes in in allowing the reaction to proceed at a finite rate. If you allow the reaction to proceed at a finite rate, then you can't return the system to the initial state (temperature, pressure, composition) without causing a significant change in the surroundings.

For the H2O example what kind of effect would a finite rate have on the properties inside reaction chamber?

I'm not familiar with what they are doing in these tables, and need more context. What they're doing seems very confusing to me, and apparently it has confused you also. Here's what several texts that I have access to do:

Smith and Van Ness, Introduction to Chemical Engineering thermodynamics:
The standard heat of formation of a species is equal to the change in enthalpy in going from the elements to the species at 1 atm and temperature T.
The standard heat of formation of a species in the reference state is the standard heat of formation at 298.
The standard heat of formation of the elements is equal to zero in the reference state.
The standard free energy of formation of a species is equal to the change in free energy in going from the elements to the species at 1 atm and temperature T.
The standard free energy of formation of a species in the reference state is the standard free energy of formation at 298.
The standard free energy of formation of the elements is equal to zero in the reference state.

Perry and Chilton, Chemical Engineers' Handbook
Same as for Smith and Van Ness

Hougan, Watson, and Ragatz, Chemical Process Principles
Heats of formation handled the same as previous two references.
Standard Molar Absolute Entropies used to get entropy change for a reaction in the reference state (298, 1 atm).
Standard Molar Enthalpies and Entropies used to get change in free energy for a reaction in the reference state.
Standard free energy changes (1atm) for reactions at other temperatures calculated from \frac{d(ΔG^o/RT)}{d(1/T)}=\frac{ΔH^o}{R}

I'm not sure if they help but I have attached the cover sheet of the appendix which outlines the source and calculation method that is slightly different than the above. It appears that the entropy value is absolute and at least close to some other ones on the internet; however I am not sure why it is varying with temperature while the enthalpy/free energies do not.

No. δP is a function of time. It is the time-dependent pressure variation we need to apply over the entire deformation to hold the acceleration constant at a very small value. When we reach the desired final deformation, we stop increasing δP and, since it then only results in a negligible net load on the piston, the deformation ceases.
AδP=\left(\frac{1}{2}at^2\right)k+(at)C+ma
where a is the downward acceleration.

Chet

How do we maintain a constant acceleration while adding different increments of δP, as it is essentially a net force that will always add to the acceleration?

Thanks very much

 

[Red_CCF]

Forgot about the attachment again...

 

[Chestermiller]

From assuming Δx = 0 and d(Δx)/dt = 0 at t = 0, for the constants I got:

c_1 = \frac{(F_{kin} - F_{stat})(1-\sqrt{1-\frac{4mk}{C^2}})}{2k\sqrt{1-\frac{4mk}{C^2}}}
c_2 = \frac{(F_{stat} - F_{kin})(1+\sqrt{1-\frac{4mk}{C^2}})}{2k\sqrt{1-\frac{4mk}{C^2}}}

I'm not sure if I'm doing this correctly, but the square root would be linearized, for f(x) = \sqrt{1\pm x} :

L_1(x) = \sqrt{1-\frac{4mk}{C^2}} - 1/2(x-\frac{4mk}{C^2})(1-\frac{4mk}{C^2})^{-0.5} for \sqrt{1-\frac{4mk}{C^2}}

and

L_2(x) = \sqrt{1+\frac{4mk}{C^2}} + 1/2(x-\frac{4mk}{C^2})(1+\frac{4mk}{C^2})^{-0.5} for \sqrt{1+\frac{4mk}{C^2}}

I've corrected some sign errors in the equations for c1 and c2 (the discriminant should have a minus sign; I also made this same typo in one of my previous responses).

Linearizing the equations with respect to \frac{4mk}{C^2} involves making use of the following linearization approximations:
\sqrt{1-x}≈1-\frac{x}{2}
\frac{1}{1-x}≈1+x
After applying these approximations, I obtain:

Δx_1≈-\frac{(F_{stat}-F_{kin})}{k}\left[\left(1+\frac{mk}{C^2}\right)(1-e^{r_1t})-\left(\frac{mk}{C^2}\right)(1-e^{r_2t})\right]
where, in this linearized approximation,
r_1≈-\frac{k}{C}\left(1+\frac{mk}{C^2}\right)
r_2≈-\frac{C}{m}\left(1-\frac{mk}{C^2}\right)
Note the similarity between this solution, and the solution obtained when m = 0. In both cases, the piston displacement at long times Δx₁ is exactly the same value. Note also that for small values of \frac{4mk}{C^2}, the piston does not oscillate, but its displacement simply increases monotonically and asymptotically to the final value.

Chet

 

[Chestermiller]

For the H2O example what kind of effect would a finite rate have on the properties inside reaction chamber?

I want to think about this question some more.

I'm not sure if they help but I have attached the cover sheet of the appendix which outlines the source and calculation method that is slightly different than the above. It appears that the entropy value is absolute and at least close to some other ones on the internet; however I am not sure why it is varying with temperature while the enthalpy/free energies do not.

The entropy, enthalpy, and free energy of a pure substance at constant pressure all increase with temperature. I looked over your Tables A1 and A2 description, and can see what they are doing.

My understanding of the variables is as follows:

\bar{h_i^o}(T)=enthalpy of pure species i at 1 atm and temperature T
\bar{h_i^o}(T_{ref})=enthalpy of pure species i at 1 atm and temperature T_ref
\bar{h_i^o}(T) and \bar{h_i^o}(T_{ref}) are related by:
\bar{h_i^o}(T)=\bar{h_i^o}(T_{ref})+\int_{T_{ref}}^T{\bar{c}_p(T)dT}
These equations apply both to the elements and to compounds. However, for the elements at T_ref, \bar{h_i^o}(T_{ref}) is taken as zero.
\bar{h_{f,i}^o}(T)=heat of formation of pure species i at 1 atm and temperature T from the elements. If

\bar{s_i^o}(T)=absolute entropy of pure species i at 1 atm and temperature T
\bar{s_i^o}(T_{ref})=absolute entropy of pure species i at 1 atm and temperature T_ref
\bar{s_i^o}(T) and \bar{s_i^o}(T_{ref}) are related by:
\bar{s_i^o}(T)=\bar{s_i^o}(T_{ref})+\int_{T_{ref}}^T{\frac{\bar{c}_p(T)}{T}dT}

\bar{g_i^o}(T)=free energy of pure species i at 1 atm and temperature T
\bar{g_i^o}(T_{ref})=absolute entropy of pure species i at 1 atm and temperature T_ref
\bar{g_i^o}(T)=\bar{h_i^o}(T)-T\bar{s_i^o}(T)
This all looks pretty straightforward. Does this help answer your questions?

How do we maintain a constant acceleration while adding different increments of δP, as it is essentially a net force that will always add to the acceleration?

You're thinking about it backwards. We are applying whatever force necessary to maintain the acceleration of the piston constant (at a very small value). Are you saying that you can't figure out how to compel the piston to exhibit a desired acceleration?
Chet

 

[Red_CCF]

I've corrected some sign errors in the equations for c1 and c2 (the discriminant should have a minus sign; I also made this same typo in one of my previous responses).

Linearizing the equations with respect to \frac{4mk}{C^2} involves making use of the following linearization approximations:
\sqrt{1-x}≈1-\frac{x}{2}
\frac{1}{1-x}≈1+x
After applying these approximations, I obtain:

Δx_1≈-\frac{(F_{stat}-F_{kin})}{k}\left[\left(1+\frac{mk}{C^2}\right)(1-e^{r_1t})-\left(\frac{mk}{C^2}\right)(1-e^{r_2t})\right]
where, in this linearized approximation,
r_1≈-\frac{k}{C}\left(1+\frac{mk}{C^2}\right)
r_2≈-\frac{C}{m}\left(1-\frac{mk}{C^2}\right)
Note the similarity between this solution, and the solution obtained when m = 0. In both cases, the piston displacement at long times Δx₁ is exactly the same value. Note also that for small values of \frac{4mk}{C^2}, the piston does not oscillate, but its displacement simply increases monotonically and asymptotically to the final value.

Chet

Is there a restriction somewhere that mk/C^2 be less than 1?

The entropy, enthalpy, and free energy of a pure substance at constant pressure all increase with temperature. I looked over your Tables A1 and A2 description, and can see what they are doing.

My understanding of the variables is as follows:

\bar{h_i^o}(T)=enthalpy of pure species i at 1 atm and temperature T
\bar{h_i^o}(T_{ref})=enthalpy of pure species i at 1 atm and temperature T_ref
\bar{h_i^o}(T) and \bar{h_i^o}(T_{ref}) are related by:
\bar{h_i^o}(T)=\bar{h_i^o}(T_{ref})+\int_{T_{ref}}^T{\bar{c}_p(T)dT}
These equations apply both to the elements and to compounds. However, for the elements at T_ref, \bar{h_i^o}(T_{ref}) is taken as zero.
\bar{h_{f,i}^o}(T)=heat of formation of pure species i at 1 atm and temperature T from the elements. If

\bar{s_i^o}(T)=absolute entropy of pure species i at 1 atm and temperature T
\bar{s_i^o}(T_{ref})=absolute entropy of pure species i at 1 atm and temperature T_ref
\bar{s_i^o}(T) and \bar{s_i^o}(T_{ref}) are related by:
\bar{s_i^o}(T)=\bar{s_i^o}(T_{ref})+\int_{T_{ref}}^T{\frac{\bar{c}_p(T)}{T}dT}

\bar{g_i^o}(T)=free energy of pure species i at 1 atm and temperature T
\bar{g_i^o}(T_{ref})=absolute entropy of pure species i at 1 atm and temperature T_ref
\bar{g_i^o}(T)=\bar{h_i^o}(T)-T\bar{s_i^o}(T)
This all looks pretty straightforward. Does this help answer your questions?

With regards to the entropy equation, is there supposed to be a pressure component on the right hand side as well?

My main confusion is that they apparently put absolute entropy values next to enthalpy and free energy that is w.r.t the reference state. Is there a general relationship between the absolute entropy in the table and the entropy of formation, as I expect that for elements that 0 = 0 - T*0 (G_f = H_f - TS_f)?

How come that entropy here is temperature dependent but G_f and H_f are not? Are the formation entropy, enthalpy, free energy values in general dependent where we set the reference state (T,p)?

You're thinking about it backwards. We are applying whatever force necessary to maintain the acceleration of the piston constant (at a very small value). Are you saying that you can't figure out how to compel the piston to exhibit a desired acceleration?
Chet

Are we trying to "match" the P_I increase with δP increase such that there's always a constant (small) net force difference on the piston? Is it theoretically possible to have the piston maintained at a = 0 (outside some starting acceleration to get it moving) and is the premise of a small acceleration that velocity of the piston is small enough to not induce viscous stresses?

Thanks

 

[Chestermiller]

Is there a restriction somewhere that mk/C^2 be less than 1?

No, but we are just trying to address some the doubts you had regarding the transition from small finite piston mass to the limit of zero piston mass, such as:

1. How can you have an accelerating piston if the piston mass is zero?
2. Does the piston oscillate if its mass is very small, but not zero?
3. Is the transition smooth from small finite piston mass to zero piston mass?
This solution answers all these questions.

We can continue the solution to larger mass, if you'd like. If mk/C^2 is equal to 1, the system is critically damped. If it is greater than 1, then the system may oscillate, depending on whether static friction is re-established at the moment that the piston reaches its maximum compression (for the first time).

With regards to the entropy equation, is there supposed to be a pressure component on the right hand side as well?

Those superscript o's refer specifically to the "standard state" where the pressure is 1 atm.

My main confusion is that they apparently put absolute entropy values next to enthalpy and free energy that is w.r.t the reference state. Is there a general relationship between the absolute entropy in the table and the entropy of formation, as I expect that for elements that 0 = 0 - T*0 (G_f = H_f - TS_f)?

How come that entropy here is temperature dependent but G_f and H_f are not? Are the formation entropy, enthalpy, free energy values in general dependent where we set the reference state (T,p)?

The approach they've described in your book is very confusing, as evidenced by the fact that it has even confused a smart guy like you. I even find it a little confusing myself, and would never work with the kind of notation that they have used. Here is that approach that I learned.

1. Let h(T,p), s(T,p), and g(T,p) represent the enthalpy, entropy, and free energy of a pure species at temperature T and p.
2. Let h°(T), s°(T), and g°(T) represent the values of h(T,p), s(T,p), and g(T,p) at p = 1 atm
3. Let h°(298), s°(298), and g°(298) represent the enthalpy, entropy, and free energy of formation of the pure species from its elements at 298 K and 1 atm. So, for an elemental species, h°(298), s°(298), and g°(298) are equal to zero, and for a compound, they are not. In the case of the entropy of a compound, for example, s°(298) = (absolute entropy of compound at 298 and 1 atm) - (absolute entropies of elements comprising the compound at 298 and 1 atm).
We also have:
h°(T)=h°(298)+\int_{298}^T{C_p(T')dT'}
s°(T)=s°(298)+\int_{298}^T{\frac{C_p(T')}{T'}dT'}
g°(T)=h°-Ts°
For a species in an ideal gas mixture (or a pure species), if p is the partial pressure of a species, then,
h(T,p)=h°(T)=h°(298)+\int_{298}^T{C_p(T')dT'}
s(T,p)=s°(T)-R\ln(p)=s°(298)+\int_{298}^T{\frac{C_p(T')}{T'}dT'}-R\ln(p)
g(T,p)=h(T,p)-Ts(T,p)

The standard heat of a reaction ΔH°(T,1atm)=(sum of h°(T) of the pure reactants)-(sum of h°(T) of the pure products)
The standard free energy of reaction ΔG°(T,1atm)=(sum of g°(T) of the pure reactants)-(sum of g°(T) of the pure products)

Are we trying to "match" the P_I increase with δP increase such that there's always a constant (small) net force difference on the piston?

Not exactly. We are trying "match" the δP to the the P_I increase such that there's always a constant(small) net force difference on the piston.

Is it theoretically possible to have the piston maintained at a = 0 (outside some starting acceleration to get it moving) and is the premise of a small acceleration that velocity of the piston is small enough to not induce viscous stresses?

We won't be able get the pressure increase we want by maintaining the net force exactly equal to zero after the piston starts moving, since the piston will slow down and stop. But, yes, the idea is to make the velocity of the piston small enough for viscous stresses to be negligible, and we can do this by maintaining a small constant velocity until we want the piston to stop. We have total control over the kinematics of the piston motion (i.e., its displacement vs time). We can do that by driving the piston with a great big gigantic motor that is insensitive to the back pressure of the gas, and provides whatever displacement history we desire to the piston.

Chet

 

[Red_CCF]

Those superscript o's refer specifically to the "standard state" where the pressure is 1 atm.

The approach they've described in your book is very confusing, as evidenced by the fact that it has even confused a smart guy like you. I even find it a little confusing myself, and would never work with the kind of notation that they have used. Here is that approach that I learned.

1. Let h(T,p), s(T,p), and g(T,p) represent the enthalpy, entropy, and free energy of a pure species at temperature T and p.
2. Let h°(T), s°(T), and g°(T) represent the values of h(T,p), s(T,p), and g(T,p) at p = 1 atm
3. Let h°(298), s°(298), and g°(298) represent the enthalpy, entropy, and free energy of formation of the pure species from its elements at 298 K and 1 atm. So, for an elemental species, h°(298), s°(298), and g°(298) are equal to zero, and for a compound, they are not. In the case of the entropy of a compound, for example, s°(298) = (absolute entropy of compound at 298 and 1 atm) - (absolute entropies of elements comprising the compound at 298 and 1 atm).
We also have:
h°(T)=h°(298)+\int_{298}^T{C_p(T')dT'}
s°(T)=s°(298)+\int_{298}^T{\frac{C_p(T')}{T'}dT'}
g°(T)=h°-Ts°
For a species in an ideal gas mixture (or a pure species), if p is the partial pressure of a species, then,
h(T,p)=h°(T)=h°(298)+\int_{298}^T{C_p(T')dT'}
s(T,p)=s°(T)-R\ln(p)=s°(298)+\int_{298}^T{\frac{C_p(T')}{T'}dT'}-R\ln(p)
g(T,p)=h(T,p)-Ts(T,p)

The standard heat of a reaction ΔH°(T,1atm)=(sum of h°(T) of the pure reactants)-(sum of h°(T) of the pure products)
The standard free energy of reaction ΔG°(T,1atm)=(sum of g°(T) of the pure reactants)-(sum of g°(T) of the pure products)

My current understanding is, there are three sets of properties to consider, absolute h, g, s (relative to absolute zero), relative h^o,g^o,s^o (relative to standard state 298K, 1 atm), and also formation h_f, g_f, s_f, which is found from the formation reaction of a compound. If this is correct, then is it true that:

1. The formation h,g,s are independent of the reference state and I should get the same value whether absolute or relative h, g, s are used (and for water was calculated as -44kJ/kmolK) as it appears to be the difference of reactant and product h, g, or s?

2. Assuming absolute properties are used, Δh_f = ∑n_ih_prod - ∑n_ih_reac (and the same for Δs_f and Δg_f ) and h,g,s here can be either relative to absolute zero or reference state and reactants are elements only

3. I take the standard state h^o = 0 (formation) + h_sensible for elements in their natural state and using this as reference, h^o = h_f + h_sensible for compounds, where sensible heat = 0 at 298K (and same for g^o and s^o). For converting absolute and standard state properties, for enthalpy would it be h_abs = h^o + c_p(298-0).

4. For elements at absolute zero, I assume that absolute h, g, s are zero, but does this hold for compounds?

5. How come in the entropy equation there is a -Rln(P) as opposed to a ratio of P_i/P_o as in my book?

Not exactly. We are trying "match" the δP to the the P_I increase such that there's always a constant(small) net force difference on the piston.

We won't be able get the pressure increase we want by maintaining the net force exactly equal to zero after the piston starts moving, since the piston will slow down and stop. But, yes, the idea is to make the velocity of the piston small enough for viscous stresses to be negligible, and we can do this by maintaining a small constant velocity until we want the piston to stop. We have total control over the kinematics of the piston motion (i.e., its displacement vs time). We can do that by driving the piston with a great big gigantic motor that is insensitive to the back pressure of the gas, and provides whatever displacement history we desire to the piston.

Chet

If the piston is initially accelerated to some small velocity, and we control from this point on the external pressure such that the net force is always zero, what would slow the piston down?

Thanks very much

 

[Chestermiller]

My current understanding is, there are three sets of properties to consider, absolute h, g, s (relative to absolute zero), relative h^o,g^o,s^o (relative to standard state 298K, 1 atm), and also formation h_f, g_f, s_f, which is found from the formation reaction of a compound. If this is correct, then is it true that:

1. The formation h,g,s are independent of the reference state and I should get the same value whether absolute or relative h, g, s are used (and for water was calculated as -44kJ/kmolK) as it appears to be the difference of reactant and product h, g, or s?

2. Assuming absolute properties are used, Δh_f = ∑n_ih_prod - ∑n_ih_reac (and the same for Δs_f and Δg_f ) and h,g,s here can be either relative to absolute zero or reference state and reactants are elements only

3. I take the standard state h^o = 0 (formation) + h_sensible for elements in their natural state and using this as reference, h^o = h_f + h_sensible for compounds, where sensible heat = 0 at 298K (and same for g^o and s^o). For converting absolute and standard state properties, for enthalpy would it be h_abs = h^o + c_p(298-0).

I can't follow all the notation, so I'm going to take another shot at explaining it in the (simpler) way that I understand it. I hope this works for you.

Let's first focus an a compound, rather than an element. Let's talk about two equilibrium states:

State 1: The separate pure elements comprising the compound in stoichiometric proportions (to produce 1 mole of the compound) in their natural states at 298 and 1 atm. Let's call this the standard reference state.
State 2: The compound at temperature T and pressure p.

Let h(T,p) represent the change in enthalpy between State 1 and State 2. From now on, we will call this simply the enthalpy of the compound. The same goes for s(T,p) and g(T,p).

Now for an element:

State 1: The pure element in its natural state at 298 and 1 atm.
State 2: The pure element at temperature T and pressure p.

Note that
h(298,1 atm)= 0
s(298,1 atm) = 0
g(298,1 atm) = 0

In addition, we define the following for states at 1 atm (we call states at 1 atm standard states, and use a superscript o to identify them):

h(298,1 atm) = h°(298) = standard enthalpy of compound or element in the reference state = standard enthalpy of formation of compound or element in the reference state

h(T, 1 atm) = h°(T) = standard enthalpy of compound or element at temperature T.

∑n_ih°(T)_prod - ∑n_ih°(T)_reac= standard heat of formation of compound or element at temperature T

The same definitions apply to s and g.

4. For elements at absolute zero, I assume that absolute h, g, s are zero, but does this hold for compounds?

No. There is an entropy change involved in forming the compound.

5. How come in the entropy equation there is a -Rln(P) as opposed to a ratio of P_i/P_o as in my book?

I was assuming that the pressure was expressed in atm, and that P_o was 1 atm.

If the piston is initially accelerated to some small velocity, and we control from this point on the external pressure such that the net force is always zero, what would slow the piston down?

You apply a small deceleration at the end just as you applied a small acceleration at the beginning.

Chet

Thanks very much[/QUOTE]

 

[Red_CCF]

I can't follow all the notation, so I'm going to take another shot at explaining it in the (simpler) way that I understand it. I hope this works for you.

Let's first focus an a compound, rather than an element. Let's talk about two equilibrium states:

State 1: The separate pure elements comprising the compound in stoichiometric proportions (to produce 1 mole of the compound) in their natural states at 298 and 1 atm. Let's call this the standard reference state.
State 2: The compound at temperature T and pressure p.

Let h(T,p) represent the change in enthalpy between State 1 and State 2. From now on, we will call this simply the enthalpy of the compound. The same goes for s(T,p) and g(T,p).

Now for an element:

State 1: The pure element in its natural state at 298 and 1 atm.
State 2: The pure element at temperature T and pressure p.

Note that
h(298,1 atm)= 0
s(298,1 atm) = 0
g(298,1 atm) = 0

In addition, we define the following for states at 1 atm (we call states at 1 atm standard states, and use a superscript o to identify them):

h(298,1 atm) = h°(298) = standard enthalpy of compound or element in the reference state = standard enthalpy of formation of compound or element in the reference state

h(T, 1 atm) = h°(T) = standard enthalpy of compound or element at temperature T.

∑n_ih°(T)_prod - ∑n_ih°(T)_reac= standard heat of formation of compound or element at temperature T

Using the above notation, from what I understand:
h(T,1atm) = h^o(T) = h_formation + h_sensible = ∑n_ih°(T)_prod - ∑n_ih°(T)_reac + c_p(T-298)

So h(T,p) from the above notation is not absolute enthalpy of a compound or element, its reference is still set at standard state, but is the difference here that h^o essentially a subset of h(T,p) and equivalent to h(T,p = 1atm)?

If I were to define an absolute enthalpy, h_abs(T,p) reference at absolute zero, can I expect ∑n_ih_abs(T)_prod - ∑n_ih_abs(T)_reac to be the equal to the heat of formation in the last line above? If so, does this mean that h_abs(T,1atm)= h°(T) + ∫c_pdT (integral from 0 to 298K)?

No. There is an entropy change involved in forming the compound.

Is the absolute enthalpy (and therefore free energy) and formation enthalpy also equal to zero for elements and compounds?

You apply a small deceleration at the end just as you applied a small acceleration at the beginning.

Chet

Outside of the beginning and the end, why is it necessary to apply a constant small acceleration through the process vs. zero acceleration?

Thanks very much

 

[Chestermiller]

Using the above notation, from what I understand:
h(T,1atm) = h^o(T) = h_formation + h_sensible = ∑n_ih°(T)_prod - ∑n_ih°(T)_reac + c_p(T-298)

No. It's
h(T,1atm) = h^o(T) = h_formation(298) + h_sensible = h°_compound(298) - ∑n_ih°_elements(298) + c_p(T-298)=h°_compound(298)+ c_p(T-298)

So h(T,p) from the above notation is not absolute enthalpy of a compound or element, its reference is still set at standard state, but is the difference here that h^o essentially a subset of h(T,p) and equivalent to h(T,p = 1atm)?

Yes.

If I were to define an absolute enthalpy, h_abs(T,p) reference at absolute zero, can I expect ∑n_ih_abs(T)_prod - ∑n_ih_abs(T)_reac to be the equal to the heat of formation in the last line above? If so, does this mean that h_abs(T,1atm)= h°(T) + ∫c_pdT (integral from 0 to 298K)?

Well, since the last line above was not correct, and it's not a heat of formation, I don't know how to address these questions.

Is the absolute enthalpy (and therefore free energy) and formation enthalpy also equal to zero for elements and compounds?

I don't know exactly how to answer this question because I don't understand it. But, in the framework that I've described, the enthalpy of formation, entropy of formation, and free energy of formation of the elements at 298 and 1 atm are zero. The enthalpy of formation, entropy of formation, and free energy of formation of compounds at 298 and 1 atm are not.

Outside of the beginning and the end, why is it necessary to apply a constant small acceleration through the process vs. zero acceleration?

It's not. This was just an example that I conceived of which I thought would help you understand things better. But, if it works better to understand it in terms of your constant velocity example, that's fine with me.

Chet

 

[Chestermiller]

I think I have another way of explaining it in a way that you can relate to.

Let H(T,p), S(T,p), and G(T,p) represent the absolute enthalpy, absolute entropy, and absolute free energy of a pure species at temperature T and pressure p (assuming that such things can be determined). Let H°(T), S°(T), and G°(T) represent the absolute enthalpy, absolute entropy, and absolute free energy of a pure species at temperature T and pressure 1 atm. Then

h°(298)=H°(298)-∑n_iH°_i(298)
s°(298)=S°(298)-∑n_iS°_i(298)
g°(298)=G°(298)-∑n_iG°_i(298)

where the summation on the RHS is over the stoichiometric number of moles of the elements comprising the species. From this, it follows that, for the elements,

h°(298)=0
s°(298)=0
g°(298)=0

We know that

H°(T)=H°(298)+\int_{298}^T{C_p}dT
S°(T)=H°(298)+\int_{298}^T{\frac{C_p}{T}}dT
G°(T)=H°(298)-TS°(T)

From this, it follows that

h°(T)=h°(298)+\int_{298}^T{C_p}dT
s°(T)=s°(298)+\int_{298}^T{\frac{C_p}{T}}dT
g°(T)=h°(298)-Ts°(T)

Formation Properties:

At 298:

h_f°(298)=H°(298)-∑n_iH°_i(298)=h°(298)
s_f°(298)=S°(298)-∑n_iS°_i(298)=s°(298)
g_f°(298)=G°(298)-∑n_iG°_i(298)=g°(298)

At temperature T:

h_f°(T)=H°(T)-∑n_iH°_i(T)=h°(T)-∑n_ih°_i(T)
s_f°(T)=S°(T)-∑n_iS°_i(T)=s°(T)-∑n_is°_i(T)
g_f°(T)=G°(T)-∑n_iG°_i(T)=g°(T)-∑n_ig°_i(T)

Hope this helps.

Chet