Show that entropy is a state function

[Philip Koeck]

In a (reversible) Carnot cycle the entropy increase of the system during isothermal expansion at temperature T_H is the same as its decrease during isothermal compression at T_C. We can conclude that the entropy change of the system is zero after a complete Carnot cycle.
The mentioned textbook now states that any reversible cyclic process can be constructed from Carnot cycles (towards the end of chapter 20 in the 14th global edition of Young and Freedman.)

The conclusion is that any reversible cyclic process has zero entropy change for the system.

As I see it, this does not show that entropy is a state function.
One would also have to show that the entropy change of the system is zero in any irreversible cyclic process.

Is it okay to simply generalise the argument in the textbook and say that any reversible or irreversible cyclic process can be approximated by (reversible) Carnot processes and therefore the entropy of the system is always unchanged after one complete cycle, no matter whether the process is reversible or irreversible?

 

[vanhees71]

If the process is not reversible, you look at an open system, but entropy (and the other thermodynamical variables) are only state functions for closed systems!

 

[Lord Jestocost]

Summary:: I'm trying to show that entropy is a state function based on an analysis of the Carnot cycle and without using advanced mathematics. I'm not satisfied with the presentation in the textbook "University Physics" by Young and Freedman and would like some feedback.

Maybe, chapter "B. The Entroy" in Richard Becker's book "The Theory of Heat" might be of help:
https://books.google.de/books?id=wS...de&source=gbs_toc_r&cad=3#v=onepage&q&f=false

 

[Philip Koeck]

Maybe, chapter "B. The Entroy" in Richard Becker's book "The Theory of Heat" might be of help:
https://books.google.de/books?id=wS...de&source=gbs_toc_r&cad=3#v=onepage&q&f=false

Thanks for the reference.
I'm trying to work only with Young and Freedman at the moment since that is what we use in our courses.
Unfortunately I'm missing some step that would make the presentation logically consistent, at least for me.
I'll discuss it more in my reply to vanHees.

 

[Philip Koeck]

If the process is not reversible, you look at an open system, but entropy (and the other thermodynamical variables) are only state functions for closed systems!

In that case I have a serious, didactic problem later on in the same chapter.
Young and Freedman proceed to calculating entropy changes for irreversible processes such as free expansion and irreversible heat transfer.
The argument they offer is that for such calculations, any irreversible process can be replaced by a reversible process between the same initial and final state, because entropy is a state function.
Now, if entropy is not a state function for irreversible processes, how do I know that I will get the correct value for its change after making the replacement.
Do you see my problem?

 

[Chestermiller]

If an irreversible process is cyclic, this means that it starts and ends in the same state. It is possible to devise an infinite number of alternate reversible processes that also start and end in this same state. The change in entropy for all these reversible processes is zero. Therefore the change in entropy for the irreversible process is zero.

 

[Andy Resnick]

Summary:: I'm trying to show that entropy is a state function based on an analysis of the Carnot cycle and without using advanced mathematics. I'm not satisfied with the presentation in the textbook "University Physics" by Young and Freedman and would like some feedback.

Thinking about this question sent me down a rabbit hole... not sure if I have a good response, but here goes:

First, it's important to note the difference between the entropy of a system S and *changes* to the system entropy ΔS that occurs during a process.

Regarding 'S', since S is a property of a system, the entropy value associated with a particular *equilibrium* state can be expressed in terms of other state variables such as P, V, and T. Then, the entropy can be simply considered as another variable specifying the state and the state can be specified in terms of, for example, S and V or T and S, etc. Thus, S is a state variable. [There is an unstated assumption that S is uniquely determined in any particular equilibrium state, not sure if we need to 'prove' that.]

I got hung up (and honestly, and still uncertain) on both the difference between S and ΔS and the requirement for the state to be in equilibrium. For example:

Consider the copy of Young and Freedman on your desk- put the book into a box. Your system is the book. What is S? I honestly don't know how to calculate it and I believe it's actually an open question:

Xiong W, Faes L, Ivanov PC. Entropy measures, entropy estimators, and their performance in quantifying complex dynamics: Effects of artifacts, nonstationarity, and long-range correlations. Phys Rev E. 2017;95(6-1):062114. doi:10.1103/PhysRevE.95.062114 , available at https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6117159/

So, in frustration, I open the container and rip up the book. Or, maybe I open the box and burn the book. Maybe I fill the container with sulfuric acid and destroy the book. The thing is, I can 'easily' calculate ΔS for each of those three irreversible processes!

For an irreversible process, as long as the start and end states are equilibrium states ('thermostatics'), I can calculate ΔS in terms of any contrived reversible process that connects the two states. It's also possible to compute ΔS if the two states are *stationary* rather than equilibrium ('thermokinetics'), but a full-on nonequilibrium thermodynamic calculation is (AFAIK) not possible at this time.

Not sure if this helps... clarification is welcome!

 

[Philip Koeck]

If an irreversible process is cyclic, this means that it starts and ends in the same state. It is possible to devise an infinite number of alternate reversible processes that also start and end in this same state. The change in entropy for all these reversible processes is zero. Therefore the change in entropy for the irreversible process is zero.

Yes, that is very helpful.
So essentially I have to correct the textbook a little.
Any cyclic process, both reversible and irreversible, can be approximated by a series of Carnot processes.

An interesting example is a process that's almost identical to the Carnot-cycle, but the reversible, isothermal expansion is replaced by a free expansion.
It can be replaced by a single Carnot-cycle for calculating the entropy change of the system.

 

[vanhees71]

If an irreversible process is cyclic, this means that it starts and ends in the same state. It is possible to devise an infinite number of alternate reversible processes that also start and end in this same state. The change in entropy for all these reversible processes is zero. Therefore the change in entropy for the irreversible process is zero.

Have you now proven that there are no irreversible processes?

Of course not. The reason is that in general when you consider time-dependent situations of many-body systems they are usually not described by local thermal equilibrium at any instance of time. That's only the case if the changes on the macroscopic relevant observables are slow compared to the thermalization time of the system. Only then the entropy stays constant over all time. In other words for general non-equilibrium processes the entropy is not constant but increasing.

 

[Chestermiller]

Yes, that is very helpful.
So essentially I have to correct the textbook a little.
Any cyclic process, both reversible and irreversible, can be approximated by a series of Carnot processes.

Absolutely not. To determine the entropy change for an irreversible process, we need to devise an alternative reversible process between the same two thermodynamic end states, and calculate the integral of dq/T for that reversible process. The alternative reversible process does not need to bear any resemblance whatsoever to the actual irreversible process, as long as it passes through the same two end states. And there are an infinite number of reversible processes that will do this.

 

[Philip Koeck]

Absolutely not. To determine the entropy change for an irreversible process, we need to devise an alternative reversible process between the same two thermodynamic end states, and calculate the integral of dq/T for that reversible process. The alternative reversible process does not need to bear any resemblance whatsoever to the actual irreversible process, as long as it passes through the same two end states. And there are an infinite number of reversible processes that will do this.

Then I wonder whether I have gained anything. I want to show that S is a state function.
In your recipe you use the fact that S is a state function.
The problem I have with our textbook is that it shows that S is a state function only for reversible processes, but then it uses this finding even for irreversible processes.
I'm missing one step in the argument, so to say.

 

[Chestermiller]

I really don't know how to answer this question. I guess we just accepted it when we learned ti. My main strength lies more in applying the fundaments to solve specific problems.

 

[vanhees71]

A general change of state is described by off-equilibrium many-body theory, which can be addressed on various levels: At the most fundamental level you can use, e.g., non-equilibrium quantum-field theory methods (relativistic or non-relativistic), e.g., the Schwinger-Keldysh real-time-contour formalism, which leads to the Kadanoff-Baym equations in the 2PI (kown under the various names of Luttinger-Ward, Kadanoff-Baym, or Jackiw-Tomboulis) formalism. This is a very sophisticated approach and the solution of the equations is usually pretty difficult. It has been done mostly for toy models like the $\phi^4$, linear $\sigma$ model, often in lower space-time dimensions.

The next level of description are Boltzmann-type (quantum) transport equations, which is found formally by using a gradient expansion or equivalently a formal expansion in powers of $\hbar$ leading to a Markovian description, usually taking into account only $2 \rightarrow 2$ scattering processes but also higher-order inelastic processes like $2 \leftrightarrow 3$ scattering etc.

Through the assumption of "molecular chaos" to cut the typical "BBGKY hierarchy" one introduces a "thermodynamical arrow of time" which is by construction identical with the "causal arrow of time" underlying all physics. The principle of detailed balance, guaranteed by the unitarity of the S-matrix, ensures that the (coarse-grained) entropy does not decrease ("H-theorem"), and that in the long-time limit thermal equilibrium is reached.

This final equilibrium state is uniquely determined by the temperature, chemical potential(s) of conserved quantitities, and external paramaters (like volume, presence of external fields, etc.), and for this equilibrium state the entropy is maximal and a "state function", which means that the equilibrium state, reached under the applied constraints is the state of "minimal information", and you cannot say from the knowledge of this state, how the system came into this state, i.e., you have no information about the "history" of the system before it reached this equilibrium state.

 

[Lord Jestocost]

The problem I have with our textbook is that it shows that S is a state function only for reversible processes,

When you once have proven that the entropy of a system is a state function which depends only on the values of some state variables of the system, it doesn’t matter how you bring the system into the state where it has these values of these state variables.

 

[vanhees71]

Yes, you only have to wait until everything is in thermal equilibrium. On its way to thermal equilibrium entropy is never decreasing (H-theorem).

 

[Philip Koeck]

I think I'm getting there.
What the book really shows is that every reversible cyclic process has entropy change zero.
That obviously means that the entropy change between two states can be calculated using any reversible path between these states and the result will be the same. So far everything is clear.

The step I need is that the only explanation for the above is that the system's entropy is a state function.
So it doesn't really matter that only reversible paths are considered.
Does that sound right?

With this step I can conclude that an irreversible process between the same two states will also have the same entropy change.

 

[Chestermiller]

I think I'm getting there.
What the book really shows is that every reversible cyclic process has entropy change zero.
That obviously means that the entropy change between two states can be calculated using any reversible path between these states and the result will be the same. So far everything is clear.

The step I need is that the only explanation for the above is that the system's entropy is a state function.
So it doesn't really matter that only reversible paths are considered.
Does that sound right?

With this step I can conclude that an irreversible process between the same two states will also have the same entropy change.

How are you defining the entropy change of a process?

 

[Philip Koeck]

How are you defining the entropy change of a process?

The whole discussion in the book starts with the Carnot cycle, so the entropy change during a reversible isotherm is Q/T, whereas for the adiabatic processes it's zero.

 

[Chestermiller]

The whole discussion in the book starts with the Carnot cycle, so the entropy change during a reversible isotherm is Q/T, whereas for the adiabatic processes it's zero.

How do you define it for an arbitrary process?

 

[Stephen Tashi]

With this step I can conclude that an irreversible process between the same two states will also have the same entropy change.

A technicality about "irreversible process":
Is a "process" defined as physical phenomena that can be represented as a path on a P-V diagram? If so, then is "free expansion" a process? If a gas is confined to one side of a cylinder by a partition and the partition is removed then as the gas expands without being in equilibrium, does the gas have a defined volume and pressure?

 

[Stephen Tashi]

A quotation from "Extended Irreversible Thermodynamics" by Jou, Casas-Vazquez and Lebon

Using "CIT" to mean "classical irreversible thermodynamics":

The fundamental hypothesis underlying CIT is that of local equilibrium. It postulates that the local and instantaneous relations between the thermal and mechanical properties of a physical system are the same as for a uniform system at equilibrium. It assumes that the system under study can be mentally split into a series of cells sufficiently large to allow them to be treated as macroscopic thermodynamic systems, but sufficiently small that equilibrium is very close to being realized in each cell.

 

[Philip Koeck]

How do you define it for an arbitrary process?

As I see it, dS is always dQ/T for a reversible process. For an irreversible process it's not defined or at least can't be calculated directly.

 

[Philip Koeck]

A technicality about "irreversible process":
Is a "process" defined as physical phenomena that can be represented as a path on a P-V diagram? If so, then is "free expansion" a process? If a gas is confined to one side of a cylinder by a partition and the partition is removed then as the gas expands without being in equilibrium, does the gas have a defined volume and pressure?

Yes, I see that problem with processes where p and V are not defined, because they happen too quickly for example. They have two end-points on the pV-diagram but nothing in between so to say. That should mean they can't be approximated by reversible processes.

 

[Chestermiller]

As I see it, dS is always dQ/T for a reversible process. For an irreversible process it's not defined or at least can't be calculated directly.

It can be calculated directly for an irreversible process, but it isn't as simple a calculation as one might think. It involves solving the complicated partial differential equations in space and time within the system that include the transport processes of viscous fluid dynamics and heat conduction. This enables one to calculate the local rate of entropy generation within the system and to integrate that over the volume of the system. For the basics of how this is done, see Chapter 11, Transport Phenomena, Bird, Stewart, and Lightfoot, Problem 11D.1.

For pure irreversible heat conduction problems, say involving steady state conduction in a rod or even transient heat conduction in a rod, the calculation is much more straightforward to apply in practice, and I can show how it is done, if anyone is interested.

 

[hilbert2]

Every time you predict the direction of a constant-pressure chemical reaction or other process with Gibbs free energy arguments, you assume that entropy is a state function. Otherwise the system could have more than one value of S and G in the same apparent macroscopic state.

 

[Philip Koeck]

For pure irreversible heat conduction problems, say involving steady state conduction in a rod or even transient heat conduction in a rod, the calculation is much more straightforward to apply in practice, and I can show how it is done, if anyone is interested.

Yes, that would be interesting.
In basic textbooks irreversible heat exchange is often treated as if it was reversible without further explanation.

 

[Chestermiller]

Consider the irreversible process of steady state heat conduction occurring along a rod in contact with a reservoir of temperature $T_H$ at x = 0 and a second reservoir of temperature $T_C$ at x = L. The temperature profile along the rod is linear, and given by $$T=T_H-(T_H-T_C)\frac{x}{L}$$and the heat flux q along the rod is constant, and given by: $$q=-k\frac{dT}{dx}=k\frac{(T_H-T_C)}{L}$$ where k is the thermal conductivity of the rod metal.

The temperature profile has been determined by satisfying the differential heat balance equation, given by:
$$k\frac{d^2T}{dx^2}=0$$
The differential entropy balance equation on the rod can be obtained by dividing the differential heat balance equation by the absolute temperature, to yield:
$$\frac{k}{T}\frac{d^2T}{dx^2}=\frac{d}{dx}\left(\frac{k}{T}\frac{dT}{dx}\right)+\frac{k}{T^2}\left(\frac{dT}{dx}\right)^2=0$$or, equivalently $$-\frac{d}{dx}\left(\frac{q}{T}\right)+\frac{q^2}{kT^2}=0$$The second term in this equation represents physically the local rate of entropy generation per unit volume within the rod. If we integrate this equation between x = 0 and x = L, we obtain: $$\frac{q_H}{T_H}-\frac{q_C}{T_C}+\frac{q^2}{kT_HT_C}L=0\tag{1}$$where $q_H=q_C=q$. The first term on the LHS represents the rate of entropy entering the rod per unit area at x = 0, and the second term represents the rate of entropy exiting the rod per unit area at x = L. The coefficient of L in the third term is positive definite, and represents the average rate of entropy generation per unit volume within the rod. Since the rod is operating at steady state, we know that the entropy of the rod is not changing with time. So $L\frac{dS}{dt}=0$, where dS/dt is the average rate of change of entropy per unit volume of the rod. So we can write:
$$0=L\frac{dS}{dt}\gt \frac{q_H}{T_H}-\frac{q_C}{T_C}=-\frac{q^2}{kT_HT_C}L\tag{2}$$Eqn. 2 is equivalent to the Clausius inequality applied to the irreversible heat conduction process in the rod.

Note that, in this very simple case, we have been able to precisely determine the rate of entropy generation within the system. Note also that, Eqn. 1 tells us that the rate of entropy exiting the rod at x = L is just equal to the rate of entropy entering the rod at x = 0, plus the total rate of entropy generation within the rod.

 

[Philip Koeck]

Consider the irreversible process of steady state heat conduction occurring along a rod in contact with a reservoir of temperature $T_H$ at x = 0 and a second reservoir of temperature $T_C$ at x = L. The temperature profile along the rod is linear, and given by $$T=T_H-(T_H-T_C)\frac{x}{L}$$and the heat flux q along the rod is constant, and given by: $$q=-k\frac{dT}{dx}=k\frac{(T_H-T_C)}{L}$$ where k is the thermal conductivity of the rod metal.

The temperature profile has been determined by satisfying the differential heat balance equation, given by:
$$k\frac{d^2T}{dx^2}=0$$
The differential entropy balance equation on the rod can be obtained by dividing the differential heat balance equation by the absolute temperature, to yield:
$$\frac{k}{T}\frac{d^2T}{dx^2}=\frac{d}{dx}\left(\frac{k}{T}\frac{dT}{dx}\right)+\frac{k}{T^2}\left(\frac{dT}{dx}\right)^2=0$$or, equivalently $$-\frac{d}{dx}\left(\frac{q}{T}\right)+\frac{q^2}{kT^2}=0$$The second term in this equation represents physically the local rate of entropy generation per unit volume within the rod. If we integrate this equation between x = 0 and x = L, we obtain: $$\frac{q_H}{T_H}-\frac{q_C}{T_C}+\frac{q^2}{kT_HT_C}L=0\tag{1}$$where $q_H=q_C=q$. The first term on the LHS represents the rate of entropy entering the rod per unit area at x = 0, and the second term represents the rate of entropy exiting the rod per unit area at x = L. The coefficient of L in the third term is positive definite, and represents the average rate of entropy generation per unit volume within the rod.

Just some comments and questions to start with.

Your equation (1) can be rewritten as $$T_C = T_H - \frac{q L}{k}$$
This also follows directly by combining the first two equations in your text and then letting x = L.
Isn't equation (1) just another way of writing the temperature distribution then?
But maybe that's as it should be.

When you interpret equation (1) you say that the first term is the entropy entering the rod at x=0, for example.
Q is not transported into the rod in a reversible process in this case, so I don't understand how you can make this interpretation.

 

[Chestermiller]

Just some comments and questions to start with.

Your equation (1) can be rewritten as $$T_C = T_H - \frac{q L}{k}$$
This also follows directly by combining the first two equations in your text and then letting x = L.
Isn't equation (1) just another way of writing the temperature distribution then?
But maybe that's as it should be.

Yes, this equation is correct. So?

When you interpret equation (1) you say that the first term is the entropy entering the rod at x=0, for example.
Q is not transported into the rod in a reversible process in this case, so I don't understand how you can make this interpretation.

The difference between reversible and irreversible is not how entropy is transported into- and out of the system at its boundary with its surroundings. That transport is always described by the same equation (q/T). The difference between reversible and irreversible is that, in an irreversible process, entropy is also generated within the system and in a reversible process, it is not.

 

[Philip Koeck]

The difference between reversible and irreversible is that, in an irreversible process, entropy is also generated within the system and in a reversible process, it is not.

I'll try to apply this idea to the free expansion of a gas, versus the reversible isothermal expansion.

In a free expansion the entropy change of the system is exactly the same as for a reversible isothermal expansion at the same temperature. The difference between the two processes seems to lie in what happens in the surroundings. This is what you see in summary, so to say.

In fact entropy is generated inside the system during the free expansion, whereas entropy flows from the surroundings to the system in a reversible isothermal expansion.
So, although the entropy change in the system has the same value for both processes they are fundamentally different.

Does that sound about right?

 

[Chestermiller]

I'll try to apply this idea to the free expansion of a gas, versus the reversible isothermal expansion.

In a free expansion the entropy change of the system is exactly the same as for a reversible isothermal expansion at the same temperature. The difference between the two processes seems to lie in what happens in the surroundings. This is what you see in summary, so to say.

In fact entropy is generated inside the system during the free expansion, whereas entropy flows from the surroundings to the system in a reversible isothermal expansion.
So, although the entropy change in the system has the same value for both processes they are fundamentally different.

Does that sound about right?

It has nothing to do with the surroundings and everything to do with the system. In the alternative reversible process, you need to add heat reversibly and isothermally to the system, so $$Q_{rev}=nRT\ln{(V_2/V_1)}$$and$$\Delta S=\frac{Q_{rev}}{T}=nR\ln{(V_2/V_1)}$$
In the irreversible free expansion process between the same two states, $$Q=0$$and $$\Delta S=nR\ln{(V_2/V_1)}=\sigma+\frac{Q}{T}=\sigma$$where $\sigma$ is the entropy generated within the system during the irreversible free expansion as a result of viscous dissipation.

To give you a feel for how viscous dissipation comes into play in generating entropy, consider the following homework problem I've dreamed up for you: You have an incompressible viscous Newtonian fluid situated between two infinite insulated parallel plates in which the bottom plate is stationary and the upper plate is moving tangentially with constant velocity V. The velocity profile within the fluid between the plates is given by $$v=V\frac{y}{h}$$where h is the distance between the plates. The transient thermal energy balance on the fluid is given by$$\frac{dU}{dt}=\dot{W}$$ (where $\dot{W}$ is the rate at which the surroundings do work on the system), or, per unit volume, $$\rho C \frac{dT}{dt}=\mu\left(\frac{V}{h}\right)^2$$where $\rho$ is the fluid density, C is its heat capacity, and $\mu$ is its viscosity. If we assume that the fluid density, heat capacity, and viscosity are independent of temperature, and that the initial temperature (at time t = 0) is $T_0$, what is the temperature at time T?

If we divide this heat balance equation by the temperature T, we obtain the entropy balance equation on the fluid. What is that entropy balance equation (in terms of the entropy per unit volume), and, by integrating that equation, what is the change in entropy per unit volume of the fluid between time t = 0 and time t?

I'll continue after you complete this.

Notice that we are doing all this without even looking at an alternative reversible path yet.

 

[Philip Koeck]

To give you a feel for how viscous dissipation comes into play in generating entropy, consider the following homework problem I've dreamed up for you: You have an incompressible viscous Newtonian fluid situated between two infinite insulated parallel plates in which the bottom plate is stationary and the upper plate is moving tangentially with constant velocity V. The velocity profile within the fluid between the plates is given by $$v=V\frac{y}{h}$$where h is the distance between the plates. The transient thermal energy balance on the fluid is given by$$\frac{dU}{dt}=\dot{W}$$ (where $\dot{W}$ is the rate at which the surroundings do work on the system), or, per unit volume, $$\rho C \frac{dT}{dt}=\mu\left(\frac{V}{h}\right)^2$$where $\rho$ is the fluid density, C is its heat capacity, and $\mu$ is its viscosity. If we assume that the fluid density, heat capacity, and viscosity are independent of temperature, and that the initial temperature (at time t = 0) is $T_0$, what is the temperature at time T?

If we divide this heat balance equation by the temperature T, we obtain the entropy balance equation on the fluid. What is that entropy balance equation (in terms of the entropy per unit volume), and, by integrating that equation, what is the change in entropy per unit volume of the fluid between time t = 0 and time t?

For the temperature I just get a linear increase: $$T(t)=T_0+\frac{μV^2t}{ρCh^2}$$

For the entropy increase per unit volume I would integrate dU/T from T₀ to T(t), which gives me:
$$ΔS=ρC⋅ln\frac{T(t)}{T_0}$$ with the above expression for T(t).
Does that look right?

 

[Chestermiller]

For the temperature I just get a linear increase: $$T(t)=T_0+\frac{μV^2t}{ρCh^2}$$

For the entropy increase per unit volume I would integrate dU/T from T₀ to T(t), which gives me:
$$ΔS=ρC⋅ln\frac{T(t)}{T_0}$$ with the above expression for T(t).
Does that look right?

These results are both correct. I was also looking for the equation for the instantaneous rate of change of entropy per unit volume which is $$\frac{dS}{dt}=\rho C \frac{d\ln{T}}{dt}=\frac{\mu}{T}\left(\frac{V}{h}\right)^2$$This equation tells us that the instantaneous rate of change of entropy per unit volume during this process is just equal to the rate of entropy generation (since there is no entropy transfer across the boundaries of the system), which, in turn, is equal to the rate of viscous dissipation divided by the absolute temperature.

If we eliminate the product of density and heat capacity ($\rho C$) from your two equations, we obtain the average rate of entropy change (and average rate of entropy generation) between time zero and time t:: $$\frac{\Delta S}{t}=\frac{\mu}{T_{lm}}\left(\frac{V}{h}\right)^2$$where $T_{lm}$ is called the logarithmic mean temperature of. the system over the time interval: $$T_{lm}=\frac{(T-T_0)}{\ln{(T/T_0)}}$$It turns out mathematically that the log-mean temperature is very close to the arithmetic mean of the temperatures for all but very large differences in the temperatures. We know this from engineering experience with temperature differences in the design of heat exchangers.

These results illustrate the intimate relationship between the rate of viscous dissipation and the rate of entropy generation within a system experiencing an irreversible process. This same concept applies directly to irreversible process of free expansion, although, in free expansion, the gas velocities and viscous dissipation rates are much more complicated to calculate (and typically require the use of computational fluid dynamics). Fortunately, in the case of free expansion, we can easily establish the final state from macroscopic thermodynamics considerations, which allow us to circumvent the detailed fluid mechanics calculations. This, of course, is, in most cases, not possible.

Comments?

I have another example to offer after this one, which involves entropy generation from both viscous dissipation and finite heat conduction simultaneously.

 

[Philip Koeck]

I have another example to offer after this one, which involves entropy generation from both viscous dissipation and finite heat conduction simultaneously.

Yes, that would be interesting.

 

[Philip Koeck]

These results are both correct. I was also looking for the equation for the instantaneous rate of change of entropy per unit volume which is $$\frac{dS}{dt}=\rho C \frac{d\ln{T}}{dt}=\frac{\mu}{T}\left(\frac{V}{h}\right)^2$$This equation tells us that the instantaneous rate of change of entropy per unit volume during this process is just equal to the rate of entropy generation (since there is no entropy transfer across the boundaries of the system), which, in turn, is equal to the rate of viscous dissipation divided by the absolute temperature.

If we eliminate the product of density and heat capacity ($\rho C$) from your two equations, we obtain the average rate of entropy change (and average rate of entropy generation) between time zero and time t:: $$\frac{\Delta S}{t}=\frac{\mu}{T_{lm}}\left(\frac{V}{h}\right)^2$$where $T_{lm}$ is called the logarithmic mean temperature of. the system over the time interval: $$T_{lm}=\frac{(T-T_0)}{\ln{(T/T_0)}}$$It turns out mathematically that the log-mean temperature is very close to the arithmetic mean of the temperatures for all but very large differences in the temperatures. We know this from engineering experience with temperature differences in the design of heat exchangers.

These results illustrate the intimate relationship between the rate of viscous dissipation and the rate of entropy generation within a system experiencing an irreversible process. This same concept applies directly to irreversible process of free expansion, although, in free expansion, the gas velocities and viscous dissipation rates are much more complicated to calculate (and typically require the use of computational fluid dynamics). Fortunately, in the case of free expansion, we can easily establish the final state from macroscopic thermodynamics considerations, which allow us to circumvent the detailed fluid mechanics calculations. This, of course, is, in most cases, not possible.

Comments?

I've started thinking about the mechanism of entropy generation in your example and also for free expansion.
In your example the upper plate is moved against the friction due to the viscosity of the liquid. The work that's done is converted to inner energy and increases the temperature of the liquid and generates entropy.

In free expansion there's no work being done from the outside, no heat is transferred.
The average kinetic energy of these molecules is unchanged since the inner energy is constant (at least for an ideal gas).
When the gas expands it does no work on the surroundings.
The only thing that really happens is that molecules fly around chaotically and move further from each other.
For a real gas there's also viscous friction between the molecules, is that right?

For a real gas: The expansion should increase the potential energy of the molecules since there are attractive forces between them. I'm wondering about the energy balance in that case. Which energy decreases?
That should mean that the temperature has to decrease a bit during the free expansion of a real gas.

For an ideal gas: There is no potential energy and also no friction between the molecules. The only reason I can see for the entropy increase is the expansion.

 

[Chestermiller]

I've started thinking about the mechanism of entropy generation in your example and also for free expansion.
In your example the upper plate is moved against the friction due to the viscosity of the liquid. The work that's done is converted to inner energy and increases the temperature of the liquid and generates entropy.

This is correct. There is a mechanistic difference between free expansion of an ideal (compressible) gas and the example I gave of doing work on an incompressible liquid. But viscous dissipation plays a role in both cases.

In free expansion there's no work being done from the outside, no heat is transferred.
The average kinetic energy of these molecules is unchanged since the inner energy is constant (at least for an ideal gas).
When the gas expands it does no work on the surroundings.
The only thing that really happens is that molecules fly around chaotically and move further from each other.
For a real gas there's also viscous friction between the molecules, is that right?

An ideal gas is the limiting case of a real gas at low density, and, even at low density, there are significant numbers of collisions between molecules. (For example, check out the mean free path of air molecules at 1 bar, clearly in the ideal gas region). It is collisions and associated energy transfer between molecules that are responsible for viscous effects. So even an ideal gas exhibits viscous dissipation and is not inviscid. See Bird, Stewart, and Lightfoot for a graph showing the viscosity of gases in the ideal gas limit.

For an ideal gas: There is no potential energy and also no friction between the molecules. The only reason I can see for the entropy increase is the expansion.

As I said, even for an ideal gas there is viscous friction. In the case of free expansion of an ideal gas, even though the gas does no work on its container, the parcels of gas still expand against each other, and this would ordinarily result in expansion cooling. However, because of viscous dissipation in the gas, the expansion cooling is exactly canceled by viscous heating, so there is no net effect on the internal energy of the gas and its temperature. And the viscous heating is ultimately responsible for the entropy increase (since no entropy enters the gas through the insulated boundary of the container).

 

[Chestermiller]

Yes, that would be interesting.

We have the same incompressible viscous fluid as before, being sheared between two infinite parallel plates, but, in this case, rather than both plates being insulated, the lower plate at y = 0 is held at constant temperature To. The system attains a steady state temperature profile and the entropy of the liquid becomes constant. However, entropy is being generated within the fluid both by viscous dissipation and by heat conduction, and this generated entropy is being removed at the lower plate by heat transfer. We are going to solve for the rate of entropy generation and show that it is equal to the rate of entropy removal.

The differential thermal energy balance for this system is described by $$0=k\frac{d^2T}{dy^2}+\mu\left(\frac{V}{h}\right)^2$$To obtain the differential entropy balance equation for the system, we divide this equation by the local absolute temperature T (at arbitrary y), and then do exactly the same mathematical manipulation with the heat conduction term that I did in post #27. What do you obtain when you do this?

 

[Chestermiller]

The only reason I can see for the entropy increase is the expansion.

This is looking at the wrong side of the equation. The volume expansion term $R\ln{(V_2/V_1)}$ is part of the entropy change effect, not the cause of the entropy change. For a closed system, there are only two physical mechanisms that can result in an entropy change during a process: 1. entropy transport across the boundary of the system from heat flow and 2. entropy generation within the confines of the system as a result of viscous dissipation and finite internal heat fluxes. For a reversible process, only the first mechanism is present.

 

[Philip Koeck]

This is looking at the wrong side of the equation. The volume expansion term $R\ln{(V_2/V_1)}$ is part of the entropy change effect, not the cause of the entropy change. For a closed system, there are only two physical mechanisms that can result in an entropy change during a process: 1. entropy transport across the boundary of the system from heat flow and 2. entropy generation within the confines of the system as a result of viscous dissipation and finite internal heat fluxes. For a reversible process, only the first mechanism is present.

I'm not sure what to think about that.
There's the textbook example of calculating entropy change in a free expansion using Boltzmann's formula.
To me that looks like the volume increase, which is proportional to the increase in microstates, is treated as the cause of entropy increase.
Another example that comes to mind is the mixing of two different gases at the same pressure and temperature, which leads to an increase in entropy.
On the other hand, if the gases are the same the entropy doesn't increase although at a molecular level the same thing happens: Molecules fly around collide with each other.
Shouldn't an increase in available space be considered a cause of entropy increase?

 

[Chestermiller]

I'm not sure what to think about that.
There's the textbook example of calculating entropy change in a free expansion using Boltzmann's formula.
To me that looks like the volume increase, which is proportional to the increase in microstates, is treated as the cause of entropy increase.
Another example that comes to mind is the mixing of two different gases at the same pressure and temperature, which leads to an increase in entropy.
On the other hand, if the gases are the same the entropy doesn't increase although at a molecular level the same thing happens: Molecules fly around collide with each other.
Shouldn't an increase in available space be considered a cause of entropy increase?

I'm not familiar with the analysis of free expansion using the Boltzmann formula. But somehow that approach must be consistent with the continuum approach I presented based on viscous dissipation. That probably happens indirectly through the application of gas kinetic theory to derive the gas viscosity.

The mixing of two different gases at the same pressure and temperature involves another entropy generation mechanism that I didn't yet mention (to keep things simple). That mechanism is molecular diffusion, in which the rate of entropy generation is proportional to the square of the concentration gradients of the gases. This is also addressed in Transport Phenomena, but in a later chapter.

As far as an increase in available space being considered a cause of entropy increase, that is not the case if the expansion takes place adiabatically and reversibly (where there is no entropy change). In addition, none of what you are indicating could be considered a cause of entropy change for an irreversible path, since they all exclusively involve looking at alternate reversible paths (for which no entropy generation occurs). What I'm talking about is determining the entropy change directly for an irreversible process without indirectly invoking reversible paths. After all, the original question was related to entropy change as a state property even if the change occurs irreversibly. To understand how that can happen, one first needs to understand how entropy generation plays a role in irreversible processes, and how this entropy generation contributes in such a way that the overall entropy change only depends on the initial and final states. The development in Transport Phenomena clearly lays this out. However, in the present thread, what I've been trying to do is given you a better feel for how entropy generation comes into play.

 

[Philip Koeck]

As far as an increase in available space being considered a cause of entropy increase, that is not the case if the expansion takes place adiabatically and reversibly (where there is no entropy change).

One could argue that in a reversible adiabatic expansion the entropy increase due to volume increase is exactly balanced by the entropy decrease due to cooling.

 

[Philip Koeck]

We have the same incompressible viscous fluid as before, being sheared between two infinite parallel plates, but, in this case, rather than both plates being insulated, the lower plate at y = 0 is held at constant temperature To. The system attains a steady state temperature profile and the entropy of the liquid becomes constant. However, entropy is being generated within the fluid both by viscous dissipation and by heat conduction, and this generated entropy is being removed at the lower plate by heat transfer. We are going to solve for the rate of entropy generation and show that it is equal to the rate of entropy removal.

The differential thermal energy balance for this system is described by $$0=k\frac{d^2T}{dy^2}+\mu\left(\frac{V}{h}\right)^2$$To obtain the differential entropy balance equation for the system, we divide this equation by the local absolute temperature T (at arbitrary y), and then do exactly the same mathematical manipulation with the heat conduction term that I did in post #27. What do you obtain when you do this?

Dividing by the absolute temperature I get:
$$\frac{k}{T}\frac{d^2T}{dy^2}+\frac{\mu}{T}\left(\frac{V}{h}\right)^2=\frac{d}{dy}\left(\frac{k}{T}\frac{dT}{dy}\right)+\frac{k}{T^2}\left(\frac{dT}{dy}\right)^2+\frac{\mu}{T}\left(\frac{V}{h}\right)^2=0$$or, equivalently $$-\frac{d}{dy}\left(\frac{q}{T}\right)+\frac{q^2}{kT^2}+\frac{\mu}{T}\left(\frac{V}{h}\right)^2=0$$

 

[Chestermiller]

One could argue that in a reversible adiabatic expansion the entropy increase due to volume increase is exactly balanced by the entropy decrease due to cooling.

with respect, this makes no sense to me. Bird, et al have identified the only two mechanisms for entropy change in a system as

1. entropy transfer by heat flow across the boundaries of the system

2. entropy generation from irreversible transport processes occurring within the system

In a reversible process, only the first mechanism is significant.

Entropy is a material property of a substance, and is a unique function of its temperature and specific volume, independent of any process path. So, of course, if specific volume changes, it follows that entropy may change. But this has nothing to do with the fundamental mechanism for the change.

Saying that volume change is a cause of entropy change is analogous to saying that change in elevation is a cause of gravitational potential energy change. It is a force acting through a distance that is the fundamental cause (mechanism) of potential energy change.

 

[Chestermiller]

Dividing by the absolute temperature I get:
$$\frac{k}{T}\frac{d^2T}{dy^2}+\frac{\mu}{T}\left(\frac{V}{h}\right)^2=\frac{d}{dy}\left(\frac{k}{T}\frac{dT}{dy}\right)+\frac{k}{T^2}\left(\frac{dT}{dy}\right)^2+\frac{\mu}{T}\left(\frac{V}{h}\right)^2=0$$or, equivalently $$-\frac{d}{dy}\left(\frac{q}{T}\right)+\frac{q^2}{kT^2}+\frac{\mu}{T}\left(\frac{V}{h}\right)^2=0$$

This equation tells us that the net rate of entropy change per unit time (zero, for this steady state situation) is equal to the net local rate of entropy transport (per unit volume) from heat conduction (the first term) plus the net local rate of entropy generation per unit volume (the 2nd and 3rd terms). The latter is equal to the rate of entropy generation from heat conduction (the 2nd term) plus the rate of entropy generation from viscous dissipation.

If we integrate this equation between y = 0 and y = h, we obtain the overall entropy balance per unit area of the plates: $$-\left[\frac{q}{T}\right]_{y=0}=k\int_0^h{\frac{1}{T^2}\left(\frac{dT}{dy}\right)^2dy}-\mu\left(\frac{V}{h}\right)^2\int_0^h{\frac{dy}{T}}$$ The LHS of this equation represents the rate of entropy removal across the lower plate boundary per unit area of the plates, and the right hand side represents the average rate of entropy generation in the fluid per unit area of the plates. Since the net rate of entropy change in this steady state system is zero, all the entropy that is generated must be removed across the lower system boundary. Please see if you can integrate the heat balance equation in post #27, subject to the boundary conditions at the plates, to obtain the temperature profile of the fluid between the plates. Then, please see if you can integrate our overall entropy balance equation to obtain an analytic expression for the RHS of the above equation.

 

[Stephen Tashi]

Entropy is a material property of a substance, and is a unique function of its temperature and specific volume, independent of any process path.

As I understand the main topic of the thread ("Show that entropy is a state function"), the question is whether this can be proven from more fundamental assumptions and definitions - or should it be taken as an assumption (i.e. Assume Entropy is "an extensible property of matter").

The approach implied by "University Physics" is (apparently) to show changes in entropy are the same for any path between two states on a PV diagram and then argue that this somehow implies that a transition that cannot be represented by any path on the diagram produces the same change in entropy.

As I understand the approach of post #27, we assume that any process where the heat $q$ and temperature $T$ along a path can be defined, allows us to define the change in entropy between two situations. In that example, the "path" seems to be a path in space that connects different physical systems (different locations on the rod) rather than a path showing the transition of a single physical system between two states.

For a single physical system, we can consider paths are more general that the set of paths that can be represented on a PV diagram. For example, in what Wikipedia calls a Joule expansion of gas ( https://en.wikipedia.org/wiki/Joule_expansion ) I don't see how to define the volume and pressure for a gas in free expansion, but the gas has a definite total kinetic energy, which allows us to define temperature $T$ as the average kinetic energy where the average is taken over the number of molecules instead of over the (undefined) volume of the gas.

From the viewpoint of teaching thermodynamics, is there a way to prove that entropy change is the same for any transition of a physical system that follows a path on a $(q,T)$ diagram?

 

[Chestermiller]

As I understand the main topic of the thread ("Show that entropy is a state function"), the question is whether this can be proven from more fundamental assumptions and definitions - or should it be taken as an assumption (i.e. Assume Entropy is "an extensible property of matter").

The approach implied by "University Physics" is (apparently) to show changes in entropy are the same for any path between two states on a PV diagram and then argue that this somehow implies that a transition that cannot be represented by any path on the diagram produces the same change in entropy.

As I understand the approach of post #27, we assume that any process where the heat $q$ and temperature $T$ along a path can be defined, allows us to define the change in entropy between two situations. In that example, the "path" seems to be a path in space that connects different physical systems (different locations on the rod) rather than a path showing the transition of a single physical system between two states.

No, this is not what it addresses. (@Philip Koeck, you didn't think really this is what the analysis was addressing, did you?). It addresses the rate of change of entropy for the overall system with respect to time during an irreversible process operating at steady state (which, for this system, is zero). It shows that the net instantaneous rate at which entropy enters and leaves the rod at its two ends is equal to the rate of entropy generation within the rod.

@Stephen Tashi : Please do me a favor. Before commenting further, please see the reference I cited in post # 24: "Chapter 11, Transport Phenomena, Bird, Stewart, and Lightfoot, Problem 11D.1." This shows the details of how to determine the instantaneous rate at which entropy of a system is changing during an irreversible process (not just looking at the initial and final states or even referring to an alternative reversible path between these end states). It provides very powerful insight into the mechanisms responsible for entropy change in irreversible processes. Plus, it also clearly shows why entropy is a state function. Please get back to me with any questions you might have on the BSL development.

 

[Philip Koeck]

As I understand the main topic of the thread ("Show that entropy is a state function"), the question is whether this can be proven from more fundamental assumptions and definitions - or should it be taken as an assumption (i.e. Assume Entropy is "an extensible property of matter").

The approach implied by "University Physics" is (apparently) to show changes in entropy are the same for any path between two states on a PV diagram and then argue that this somehow implies that a transition that cannot be represented by any path on the diagram produces the same change in entropy.
…….…………….

The discussion in "University Physics" shows nicely that all reversible cyclic processes can be approximated by series of Carnot cycles.
They could just as well write that all sufficiently slow (quasistatic) cycles can be approximated by Carnot cycles, just as you say. The only thing that's important for the argument is that the processes can be plotted on a pV diagram.
This obviously means that entropy change must be the same for any quasistatic path between two states, both reversible and irreversible.
The difficult step, as you point out, is to say that this must be due to the fact that this "change of entropy" is actually the change of a state function called entropy. One could still insist that entropy change is something like work or heat, but for some reason it's path-independent as long as the path is quasistatic.

On the other hand, as soon as I know that entropy is a state function I can even calculate its change for fast, irreversible processes like the free expansion, simply by finding a suitable alternative path.

I would say that this interpretation, that entropy is a state function, is very economical and maybe not so far fetched.
I don't know, however, whether it's more like a powerful generalization (or assumption) or whether one is logically or mathematically forced to take this step.
Could one be very annoying and claim that entropy change is the same for all quasistatic paths between two states but different for a fast process between the same states?

 

[Philip Koeck]

with respect, this makes no sense to me. Bird, et al have identified the only two mechanisms for entropy change in a system as

1. entropy transfer by heat flow across the boundaries of the system

2. entropy generation from irreversible transport processes occurring within the system

In a reversible process, only the first mechanism is significant.

Entropy is a material property of a substance, and is a unique function of its temperature and specific volume, independent of any process path. So, of course, if specific volume changes, it follows that entropy may change. But this has nothing to do with the fundamental mechanism for the change.

Saying that volume change is a cause of entropy change is analogous to saying that change in elevation is a cause of gravitational potential energy change. It is a force acting through a distance that is the fundamental cause (mechanism) of potential energy change.

I realize that basic textbooks contain many simplifications and these discussions on PF are really helping me to get past these simplified ideas.
The reason why I thought volume change is a cause of entropy change comes from an example in "University Physics" where they try to show that Carnot's entropy and statistical entropy, given by Boltzmann's formula, are the same thing.
They look at the free expansion and say that if the volume increases by a factor 2 each molecule has twice as many places to be. Then by taking into account the number of molecules you can find the ratio of the number of microstates before and after the expansion and then use Boltzmann's formula to calculate entropy change, which turns out the same as for the non-statistical calculation.

You mention irreversible transport processes. In a free expansion, could one regard the movement of molecules into previously empty space as such a transport process?

 

[Chestermiller]

I realize that basic textbooks contain many simplifications and these discussions on PF are really helping me to get past these simplified ideas.
The reason why I thought volume change is a cause of entropy change comes from an example in "University Physics" where they try to show that Carnot's entropy and statistical entropy, given by Boltzmann's formula, are the same thing.
They look at the free expansion and say that if the volume increases by a factor 2 each molecule has twice as many places to be. Then by taking into account the number of molecules you can find the ratio of the number of microstates before and after the expansion and then use Boltzmann's formula to calculate entropy change, which turns out the same as for the non-statistical calculation.

You mention irreversible transport processes. In a free expansion, could one regard the movement of molecules into previously empty space as such a transport process?

It is more basic than that. The irreversible transport processes occur locally during an irreversible change, and there are only three of them: heat conduction, viscous momentum transport, and molecular diffusion. All of these are related to the movement of molecules.

 

[Chestermiller]

Could one be very annoying and claim that entropy change is the same for all quasistatic paths between two states but different for a fast process between the same states?

If the two end states are thermodynamic equilibrium states, each one must be characterized by a unique distribution of quantum mechanical states (i.e., a unique entropy). So how could the entropy change be different for the fast process between the same two states?