How Do You Calculate Entropy and Reversibility in Irreversible Processes?

[Gavroy]

okay, then there is no problem, was not sure you meant that.

nevermind, i think my underlying question is the following: the whole carnot cycle is reversible as the change in entropy is 0. but am i right by saying that e.g. the isothermic expansion process itself is irreversible as the change in entropy is >0 ?

could you answer this?

 

[Studiot]

I haven't been talking about 'cycles' - they are more difficult and I have already suggested that thermodynamics is tricky.

I think a good part of your difficulty is in identifying/defining the system appropriately - this is very common. Hence my earlier question.

To fully define a system you need to specify

1) The system components
2) the system boundary (between the system and the surroundings)
3) The system process

The surroundings are everything that is not part of the system.

Work or heat or other forms of energy may be exchanged between the system and the surroundings by the system process. They must cross the system boundary to be included in the first law.
This gives us two ways to calculate the energy transfer - the work etc done on the surroundings or the work etc done by the system. These must be equal.

A system may be open in which case both energy and mass may be exchanged across the system boundary

If is system may closed if energy but not mass is exchanged across the boundary.

If neither mass nor energy may be exchanged across the system boundary the system is isolated.

Since no energy may be input to an isolated system its total energy is constant.
Furthermore no heat may be exchanged, so for an isolated system and a reversible process

ΔS = Q/T= 0/T = 0 : S = a constant.

This is probably what you were thinking about and probably enough to be going on with.

Good night for now, we will get to Darwin's entropy creation tomorrow.

 

[Darwin123]

where does your colder environment come from?(without increasing entropy somewhere?)
thank you to both of you for your efforts, but i think my underlying problem with this is the following: if i have a closed system and one hot body and one cold body(e.g. ice) in it and the cold body melts(caused by the heat of the other body), then this process is irreversible in this closed system. so entropy increases and therefore this melting is irreversible. where am i wrong? why is increasing entropy not directly linked to irreversibility? i don't get this, in my opinion, this is exactly the second law of thermodynamics.

"where does your colder environment come from?(without increasing entropy somewhere?)"
Your colder environment has a low density of entropy. How is got that way doesn't matter.
However, I hypothesized that you live in a cold climate. If that is so, then it probably got that way by radiating into out space which is colder yet. Up North, the sun hits at a low angle. Black body radiation sucks entropy from the ground making it cold.
Your colder environment may have also been caused by a cold front. Weather conditions cause advection of cold air from the north into a southern region that formerly had only warm air. The warm air may have been forced up and north. This is an example of work forcing entropy to go in the "nonspontaneous" direction.
"If I have a closed system and one hot body and one cold body(e.g. ice) in it and the cold body melts(caused by the heat of the other body), then this process is irreversible in this closed system. so entropy increases and therefore this melting is irreversible."
I gave two scenarios where this is not true.
1) Maybe you live in a subzero climate. Just take the ice cream outside!
-Heat conduction without work is reversible as long as the difference in temperature is infinitesimal.
2) Replace the cold reservoir of a Carnot engine with the melted ice cream, and then do work on the handles.
-It is always possible to move entropy from a low temperature to high temperature by doing work. The second law states that entropy can't spontaneously move from low to high temperature. However, doing work is not spontaneous.

"why is increasing entropy not directly linked to irreversibility?"
Because there are two ways to increase entropy. First, one can produce entropy. Second, one can transfer entropy.
Your problem is your inability to believe that entropy can be moved. Entropy is an intensive property. Therefore, it can be localized. Every point has a different entropy density. Entropy in many ways is like a fluid.
If a fixed quantity of entropy can move one way, then it can move the other way. This is what the word "reversibility" means in this case. If a given amount of entropy spontaneously transfers from high temperature to low temperature, then work can be used to transfer the same amount of entropy from low temperature to high temperature.

"I don't get this, in my opinion, this is exactly the second law of thermodynamics."
I see a problem with this. I hope that you misunderstand the word "opinion."
In any case, scientists are always talking about the transfer of entropy. Consider refrigerator air conditioners.
Refrigerator air conditioners move the entropy from inside the room to the outside of the room. Work has to be done in the form of electric power to cool a room. Designers always include a vent where the heat (i.e., entropy) goes to the outside. Although entropy is created, this is an unintended. An ideal air conditioner would not create entropy.

Entropy can not be destroyed. However, entropy can always be moved.You seem to have problems with moving entropy.
This works for some people (i.e., me). Try to imagine entropy as an indestructible gas. Temperature is the pressure of the entropy gas.
A gas can move spontaneously from a region of high pressure to low pressure. Entropy can move spontaneously from a region of high temperature to low temperature. However, work is necessary to move gas from low pressure to high pressure. Work is also necessary to move entropy from low temperature to high temperature.
Historically, entropy was first described as a gas. Lavoisier listed entropy (i.e., caloric) on an early version of the periodic table. Lavoisier actually thought entropy was a chemical gas! Entropy was later discovered to be a type of disorder, not a chemical gas. However, this didn't change the fact that entropy sometimes acts like a gas.
Something in "your opinion" won't accept the fact that entropy sometimes behaves like a gas. I suggest that you ask another question about entropy, not about reversibility. Unless you understand entropy, you can't understand reversibility.

 

[Darwin123]

when you freeze something, this substance produces heat.(heat of fusion)

nevermind, i think my underlying question is the following: the whole carnot cycle is reversible as the change in entropy is 0. but am i right by saying that e.g. the isothermic expansion process itself is irreversible as the change in entropy is >0 ?

No. Every part of the Carnot cycle is irreversible. There is no part of the Carnot cycle where entropy is created, whether you run it backwards or forwards.
The Carnot cycle moves some entropy. If the entropy is allowed to flow "spontaneously", then the entropy moves from hot reservoir to cold reservoir. If work is done on the Carnot engine, then entropy flows from cold reservoir to hot reservoir.
The Carnot cycle is reversible at any time because entropy is never created. The change in entropy in your last sentence is brought into the cylinder by motion.
The entropy is not created in a Carnot cycle. It is only moved.
ΔS here represents the entropy that is moved, not the entropy that is created. However, I acknowledge that textbooks use the symbol ΔS ambiguously. This is a matter of context. Sometimes, ΔS represents the entropy moved and sometimes ΔS represents the entropy that is created.

 

[Gavroy]

so all isobaric, isochoric, isothermic and adiabatic processes are reversible(if we are talking about ideal processes using an ideal gas?)

 

[Darwin123]

okay, i think i did not understand this. in this case, we have a change in entropy, why do you call this process reversible? i thought that a proper definition of reversible is ΔS=0 and nothing else, where am i wrong?

You are wrong because the proper definition of reversible is not ΔS=0. The physical definition of ΔS is not clear in that equation. In order for the definition to be valid, one would have to state in words what ΔS means.
Here is the proper definition of reversible. A reversible process is defined as a process where entropy is not created. If ΔS is defined as the entropy that is created in a reversible process, then ΔS=0. However, if ΔS is defined as the entropy that is transferred to a body in a reversible process, then ΔS can take on any real value at all.
The symbol, ΔS, can be used to designate two entirely different types of process. First, ΔS can be the change of entropy due to the creation of entropy. Second, ΔS can be the change of entropy due to entropy transport. If you think the symbol ΔS always means the creation of entropy, then you are wrong.
A process is reversible when entropy isn't created inside a body. However, it is still possible for the entropy in a body to change because entropy is transported into (ΔS>0) or out of it (ΔS<0). When describing a reversible process, it is customary to let ΔS mean the change due to entropy transport.
A process is adiabatic when entropy isn't transported into or out of a body. However, it is still possible for entropy to change due to the creation of entropy. When describing an adiabatic process, it is customary to let ΔS be the change due to entropy creation.
A process can be both irreversible and adiabatic. In which case, the change in entropy of a body has two components. First, there is the change in entropy due to entropy transport. That can be positive or negative. Second, there is the change in entropy due to entropy creation. That is always positive. It is helpful in such a case to separately identify each component.
In a typical textbook problem, one is asked to calculate ΔS for a Carnot engine that is doing work. The Carnot cycle is by definition reversible. Therefore, there is no creation of entropy. However, entropy is transferred from the hot reservoir to the cold reservoir. By custom, ΔS is the change in entropy of the cold reservoir due to the transport.
The Carnot engine works by the spontaneous transfer of entropy from the hot reservoir to the cold reservoir. If ΔS is the change in entropy of the cold reservoir due to entropy transfer, then the Carnot engine causes ΔS>0. If ΔS is the creation of entropy by the Carnot engine, then ΔS=0. If ΔS is the change in entropy of the hot reservoir due to entropy transfer, then the Carnot engine causes ΔS<0. Note there are three "correct" answers that vary with the definition of ΔS.
The Carnot refrigerator works by the forced transfer of entropy from the cold reservoir to the hot reservoir. If ΔS is the change in entropy of the cold reservoir due to entropy transfer, then the Carnot engine causes ΔS<0.
You are wrong if you think that the symbol ΔS means the same thing to every reader every time it is written down. Sometimes ΔS stands for the creation of entropy, sometimes ΔS stands for the transfer of entropy, and sometimes ΔS stands for some combination of creation and transfer. ΔS has no meaning unless the user states in words what ΔS means. No proper definition can use the symbol ΔS unless it is defined elsewhere in words.
Do you have any questions about this answer?

 

[Darwin123]

so all isobaric, isochoric, isothermic and adiabatic processes are reversible(if we are talking about ideal processes using an ideal gas?)

Yes, if by ideal you mean this.
An "ideal process" is one where no entropy is created.
If this is what you meant by ideal, then all the processes that you described above are reversible. If fact, they would still be reversible even if one didn't use an ideal gas.
I recommend that you use the word ideal less often. This is another word that changes its meaning. When talking about an engine or refrigorator, ideal usually means without changing entropy.
Allow me to rewrite the idea behind your sentence.
All isobaric, isochoric, isothermic and adiabatic processes are reversible if the processes do not create entropy.
The previous sentence is correct and unambiguous.

 

[Darwin123]

well, my question is: why is dS=dQ_rev/T, what is this reversible for? why would this definition be wrong if i calculate the change in entropy of a nonreversible process? or why do people stress this "reversible" in this definition?

If the measurement is done on an irreversible process, the value for the change of entropy may not be reproducible. Frictional forces will cause the answer to depend explicitly on the speed of the process and other things. Let me give a concrete example.
Suppose you do an experiment where you want to measure the amount of heat given off by an ice cube that is floating on the surface of water in a container. You want to measure the difference in temperature in the water that occurs when the ice cube is completely melted.
1) Suppose the liquid water in the container is initially just slightly more than the melting temperature of ice. Assume that the container is very large compared to the ice cube. Thus, there is far more liquid water in the tank than ice.
This type of measurement is approximately reversible because entropy is not created. When energy enters the ice by conduction, there is little creation of entropy. You can calculate the creation in entropy by calculating dS where the energy is transferred at nearly equal temperatures. The creation of entropy is close to zero.
As the ice melts, the liquid water it gives off is exactly at the melting point of ice. One may have to wait a long time for the ice to melt. However, the ice cube will remain still. The water that comes off the melting ice stays on top of the surface.
2) Suppose the liquid water in the container is initially much hotter than the ice. Assume that the water is a few degrees short of boiling. One places the ice cube in the water.
This type of measurement is completely irreversible because entropy will be created. When the energy enters the ice by heat conduction, entropy will be created.
As the ice melts, the liquid water it gives off is at the melting point of ice. However, this water is surrounded by water at a higher temperature. The high temperature water has a lower density than the ice water. Therefore, the ice water sinks. The convection current will generate turbulence, so the ice block moves around during the measurement. The motion of the ice block will be effectively random. The amount of entropy created will vary with the motion of the ice block. So the entropy generated is both large and variable. The measurement of entropy won't be reproducible.
Note that I am including the expansion of ice. At temperatures near boiling, the hot water will be less dense than ice.
I think that is one way to think of it. Irreversible processes are usually associated with some random or chaotic process. Therefore, the measurement of thermodynamic properties using irreversible processes usually becomes irreproducible.
Another way to think of it is in terms of frictional forces. Frictional forces and their ilk often result in chaotic behavior. Frictional forces also create entropy.To avoid the chaotic behavior, avoid processes that create entropy. In other words, for reproducibility stick with reversible processes.

 

[Studiot]

Darwin, I asked you for a quantitative explanation of your entropy transferred and entropy created proposition, a question you skillfully avoided.

so all isobaric, isochoric, isothermic and adiabatic processes are reversible(if we are talking about ideal processes using an ideal gas?)

So in response to my question how about working through this example in the light of your proposition and explaining how Gavroy is mistaken.

10 litres of a monatomic ideal gas is expanded from an intial pressure of 10 atmospheres to 1 atmosphere.

Calculate and compare the entropy and energy changes for

(1) Adiabatic reversible expansion

(2) Adiabatic irreversible expansion


These can all be calculated by using 'ideal' process equations and clearly show that they are different

For example I calculate the work done to be

(1) 9141 joules - reversible

(2) 5474 joules - irreversible

Can you explain this?

 

[Darwin123]

Darwin, I asked you for a quantitative explanation of your entropy transferred and entropy created proposition, a question you skillfully avoided.



So in response to my question how about working through this example in the light of your proposition and explaining how Gavroy is mistaken.
10 litres of a monatomic ideal gas is expanded from an intial pressure of 10 atmospheres to 1 atmosphere.

You also did not specify the initial temperature, nor the initial molar amount of gas. I don't think it matters in the problem. I was able to work the problem as presented for a monotonic gas.

The power law exponent of volume for a monotonic gas is 1.5. In other words, for a monotonic gas undergoing a adiabatic expansion or contraction:
P_1 V_1^3/2=P_2 V_2^3/2
where P is the pressure, V is the volume, the initial condition is _1 and the final condition is _2. Sorry I don't have a LaTex editor to write the equations in a more formal way.
Given your initial pressure, your final pressure, and the initial volume of 10 Liters.
Final volume = 46 Liters
Note that 1 atmosphere = 1.01×10^5 Newtons/meter^2.
Using the adiabatic expansion formula, the initial volume and the final volume, I calculated the work done by the adiabatic gas. The work done by the gas is the volume integral of pressure. Again, I am sorry that I don't have a LaTex editor. I did the integral with the adiabatic expansion formula for pressure.
The work done by the gas is during the expansion is 1.07×10^5 Joules.

Calculate and compare the entropy and energy changes for

(1) Adiabatic reversible expansion

In a reversible adiabatic expansion, no entropy is transported. Therefore, there is no heat transfer. There are no frictional forces. There is only work exerted by the elastic force of the gas.
The change in energy of the gas is equal to the negative of the work done by the gas. The only force exerted by the gas Therefore, the change in energy of the gas during the adiabatic transfer is -1.07×10^5 Joules.
In a reversible process, no entropy is created. Since no entropy is created, and no entropy transferred, the change of entropy of this gas during the expansion is 0 (i.e., zero).
ΔE=-1.07×10^5 Joules
ΔS=0.

(2) Adiabatic irreversible expansion

Here is where the problem comes in. You did not describe how this expansion becomes irreversible. I hypothesize that you were thinking of a frictional force on the movable piston. However, you did not give me any idea of the size of the force. You did not give me a friction coefficient, a normal force, or anything that can be used to determine the size of the frictional force. The frictional force can be anything.

These can all be calculated by using 'ideal' process equations and clearly show that they are different

I have never heard of an "ideal frictional force". Therefore, I have no "ideal" formula to work out the problem. Is kinetic friction and "ideal" frictional force? Is static friction an "ideal" frictional force? How large is this "ideal" frictional force?

For example I calculate the work done to be

And how did you calculate these numbers for work and entropy?
The reversible value is different from the value that I calculated. However, I suspect that one of us made an arithmetic mistake. I would certainly like to compare your method for calculating the change in entropy for a reversible expansion to my method. My value, once again, was -1.07×10^5 Joules.

(1) 9141 joules - reversible

Now here is something I can't grasp. Without knowing the frictional forces, or whatever process makes it irreversible, how can you calculate the loss in energy?


(2) 5474 joules - irreversible

Can you explain this?

No. I would be interested in your explanation, though.

 

[Studiot]

Sorry you are right I omitted the initial temperature of zero ° C, but specified the volume.

adiabatic reversible expansion


\begin{array}{l}<br /> \gamma = \frac{{{C_p}}}{{{C_v}}} = \frac{{\frac{3}{2}R + R}}{{\frac{3}{2}R}} = \frac{5}{3} \\ <br /> {V_2} = {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{{\gamma ^{ - 1}}}}{V_1} = 39.81litres \\ <br /> {T_2} = \frac{{nR}}{{{P_2}{V_2}}} = 108.8^\circ K \\ <br /> \end{array}

First law, q = 0 , ΔU = w


w = \Delta U = n{C_v}\Delta T = n\frac{{3R}}{2}\Delta T = 9141J

 

[Darwin123]

Sorry you are right I omitted the initial temperature of zero ° C, but specified the volume.

adiabatic reversible expansion


\begin{array}{l}<br /> \gamma = \frac{{{C_p}}}{{{C_v}}} = \frac{{\frac{3}{2}R + R}}{{\frac{3}{2}R}} = \frac{5}{3} \\ <br /> {V_2} = {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^{{\gamma ^{ - 1}}}}{V_1} = 39.81litres \\ <br /> {T_2} = \frac{{nR}}{{{P_2}{V_2}}} = 108.8^\circ K \\ <br /> \end{array}

First law, q = 0 , ΔU = w


w = \Delta U = n{C_v}\Delta T = n\frac{{3R}}{2}\Delta T = 9141J

Okay, it was my error. I forgot the exact formula for an adiabatic expansion.
γ=5/3, not 3/2.
Okay, you calculated the change in energy correctly for the reversible reaction. Since you wrote q=0, I take it you also calculated the change in entropy for a reversible process correctly (ΔS=0).
I still don't get how you were able to get a precise value for work in the case of an irreversible process. Irreversibility is graded. There can't be one value of anything for a irreversible reaction.

 

[Studiot]

irreversible adiabatic

As before q=0, hence ΔU=w

and also for an expansion at constant pressure w = -pΔV


+ w = \Delta U = n{C_v}\Delta T

and


- w = {P_2}({V_2} - {V_1}) = {P_2}\left( {\frac{{nR{T_2}}}{{{P_2}}} - \frac{{nR{T_1}}}{{{P_1}}}} \right)

equating work expressions and solving for T₂

T₂ = 174.8°K

Substitute back into gives the work as

5474J

I will have to sign off now for a couple of hours.