# Entropy and reversibility

1. Jul 20, 2012

### Gavroy

hi

i am a little bit confused about the definition of entropy that says: dS=dQ_rev/T

what does this dQ_rev mean? is this definition wrong, if we are talking about irreversible processes?

My idea was that when you have an irreversible process like the isobaric expansion of a gas, then you cannot just say that
delta S=(delta U+p delta V)/T, instead you have to integrate, as the entropy is not a state function anymore, but the definition should stay more or less the same.

2. Jul 20, 2012

### Studiot

Entropy is always a state function. (as are all functions that are state functions. Functions are either always state functions or always not state functions)

Heat exchanged, however, is not a state function.

A system state is fully described by listing the values of all state functions (P,V,T, S, U, H, G etc)

Every state so described is unique.

For two specific states the difference in value of a particular state function is independent of the path by which the system passes from one state to the other.

It is not necessary to list them all since equations are available containing only state functions so if some are known others can be calculated.

Further equations such as ΔU = Q+W are also available but because these contain non state functions they are path dependent.

On the left hand side U is a state function so these passage from state U1 to state U2 will involve the same input or extraction of energy, no matter how we do it.
The right hand side tells us that we can achieve this passage in an infinite variery of ways by adding heat or expending work in different amount.
The equation as a whole tells us that the combination or sum of this heat and work exchange is constant.

Another way to look at this is to note that state functions are purely set by the system.

Non state functions, on the other hand, are can be forced by the environment.

Forced changes result in other effects so if we forcibly add more heat to a system than is necessary for it to pass from state1 to state2 then the extra heat will appear as work on the surroundings.
This process is then irreversible since we cannot extract the amount of heat we put in as some of it has become work.
This is what is meant by saying that the reversible heat change is the minimum possible.

Does this help?

3. Jul 20, 2012

### Gavroy

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?

4. Jul 20, 2012

### Studiot

Before you can answer that you need to understand that entropy is a state function.

Is that now clear?

5. Jul 20, 2012

### Gavroy

yes, it is.

6. Jul 20, 2012

### Studiot

OK so understand there is only one value for ΔS in a given situation?

Yet we can input as much heat as we like, above a minimum value = TΔS?

7. Jul 20, 2012

### Gavroy

that is more or less the definition of a state function.

8. Jul 20, 2012

### Studiot

Well the full definition of entropy is

ΔS ≥ Q/T, with equality holding for reversible heat transfer.

This means that for an irreversible process you have added some energy not accounted for by Q

9. Jul 20, 2012

### Gavroy

and how do i calculate entropy for irreversible changes in general, i guess there has to be a more general definition than an inequality

10. Jul 22, 2012

### Studiot

In order to calculate entropy changes we have three strategies available.

Firstly we can try to evaluate Qreversible directly and put it into the equation.

So for instance the entropy of fusion, which takes place at constant temperature and is an equilibrium or reversible process, is given by

ΔS = Latent heat/Tmelting for unit quantity of substance.

Secondly we can use other formulae which contain only state functions and constants to avoid including Q or W directly. These then are valid for both reversible and irreversible processes.

For instance for a perfect gas

$$\begin{array}{l} \Delta S = {c_v}\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) + R\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right) \\ \Delta S = {c_v}\ln \left( {\frac{{{P_2}}}{{{P_1}}}} \right) + {c_p}\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right) \\ \Delta S = {c_p}\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) - R\ln \left( {\frac{{{P_2}}}{{{P_1}}}} \right) \\ \end{array}$$

Thirdly we can call in the dirty tricks brigade and get really sneaky. Knowing that entropy is a state function if faced with an irreversible path we can seek an alternative reversible path from stateI to state II, for which ΔS will therefore be the same.

For instance the simple transfer of heat by temperature difference is irreversible if we allow two bodies to come into direct contact.

The simple equation, similar to the latent heat above, will not work

Both values of heat calculated by

Q = CvΔT or Q = CpΔT

are irreversible.

However if we introduce a transfer fluid and place this in contact with the hotter body at T2 and allow the fluid to accept heat Q by reversible isothermal expansion.
The fluid is then removed from the first body and allowed to allowed to cool to T1 - the temperature of the second body, by adiabatic reversible expansion.
The gas is then brought into contact with the second body at T1 and compressed isothermally( and reversibly), until it has transferred heat Q to the second body.

The net entropy change is therefore

$$\Delta S = \frac{Q}{{{T_1}}} - \frac{Q}{{{T_2}}}$$

Last edited: Jul 22, 2012
11. Jul 22, 2012

### Gavroy

thank you, that really helped...

12. Jul 22, 2012

### Studiot

13. Jul 22, 2012

### Darwin123

In this example, "dS" is the entropy that is moved into the system, not the entropy that is created. Therefore, "dQ_rev" is the energy that entered the system by heat conduction.
Entropy can be either moved or created. The main issue in many thermodynamics problems is what part of the change in entropy is moved, and what part of the change in entropy is created.
If entropy is created, the process is irreversible. The reason is the Second Law. Entropy can never be destroyed. So whatever is created sticks around.
If entropy is only moved, then the process is reversible. Entropy can move spontaneously from high temperature to low temperature. "Spontaneous" here means without doing work on the system. However, entropy from low temperature to high temperature when work is done by the system.
I suggest that the equation should have been written for clarity as,
dS_Moved=dQ_Rev/T
to highlight the fact that there is more than one way entropy can change in a system. The total change of entropy in the system would be "dS", where,
dS=dS_Rev+dS_Irrev,
where dS_Rev is the change in entropy due to the motion of entropy and dS_Irrev is the change in entropy due to the creation of entropy.

Just for completeness, I will mention two types of processes that created entropy. First, frictional forces of any kind can create entropy. Second, increases in specific heat can also create entropy. Chemical reactions can "create" entropy because the products of the chemical reactions have larger specific heats than the reactants.
In the case of the Carnot cycle, there are no processes that create entropy by definition. Therefore, "dS=dS_Rev".
I started to understand the notion of moving entropy when I read the following article.
"Entropy in the teaching of thermodynamics" by Hans Fuchs. American Journal of Physics 55(3), 215-329 (1987).
I understood it a bit better when I read an English translation of:
"Reflections on the Motive Power of Fire" by Sadi Carnot (1824).

There are several places to find the second reference. However, you have to read the word "caloric" as "entropy". Once you do that, you may find the essay makes a lot more sense.

14. Jul 22, 2012

### Studiot

That is an interesting point of view, however it is not complete.

Can you describe and quantify how you create entropy?

15. Jul 22, 2012

### Gavroy

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?

16. Jul 22, 2012

### Studiot

I hope I didn't say anything to lead you to think that.

Look again at post#8.

The latent heat of fusion is the heat transferred to melt something. During this process the temperature does not change.
I am ignoring the small volume difference between the solid and liquid, both at the melting point.
Strictly speaking, the something must be a pure substance since only pure substances have a single well defined melting point.
You can recover that heat by solidifying the liquid so the process is reversible.

And by the equality in post#8 the entropy of the change is Q/T (not zero)

For example the entropy of fusion for water / ice is about 22 Joules degree-1 mole-1

17. Jul 22, 2012

### Gavroy

okay alright, but in my opinion melting something is not reversible, e.g. when ice-cream melts, there is no possibility to get your melted ice back in the same condition, without creating entropy.

18. Jul 22, 2012

### Studiot

Is ice cream a pure substance?

19. Jul 22, 2012

### Gavroy

okay, if i think of ice as frozen water, then melting it at room temperature will be irreversible too.

20. Jul 22, 2012

### Studiot

What do you think irreversibility means?

In you physics/science classes at high school did you ever plot a temperature - time curve for ice/water/steam?
This is usually done as a cooling curve with a thermometer.

And while we are on the subject, do you understand about open, closed and isolated systems?

21. Jul 22, 2012

### Darwin123

The equation under discussion is:
ΔS = Latent heat/Tmelting for unit quantity of substance.

The equation isn't a process. Therefore, the equation can't be considered either reversible or irreversible. The process by which the "latent heat" is introduced into the water can be reversible or irreversible.
If all the entropy is moved or removed from the water by the process, then the process is reversible. If one uses a frictional force to make the entropy, then the process is irreversible.
Consider the following reversible experiment for determining ΔS.
Start with a Carnot engine between a high temperature reservoir and a cold temperature reservoir. Attach a meter to the piston so that one can measure the work done in Joules. Stick a thermometer into each reservoir.
Now, replace each reservoir a container of water with finite volume. This isn't an ideal Carnot cycle anymore. The operation of the engine will change the temperatures of the reservoirs by a tiny amount that can be measured with the thermometers. We will call this the Carnot calorimeter.
Let the high temperature reservoir be an insulating tank of water of finite volume with a temperature of 25 C°. Let the low temperature reservoir be a small block of ice in an insulating container at a temperature of 0 C°. The weight of the water in both containers is measured previous to the experiment.
Let the entropy spontaneously move from the hot reservoir to the cold reservoir. Keep going until all the ice is melted in the cold reservoir. Then stop and measure the temperatures of each reservoir. Read the meter to see how much work the Carnot engine did.
By using the formulas that you know for a Carnot cycle, you can determine the Latent heat of the ice. You measured the temperature of each container before and after the ice melted. You know how much work the Carnot engine did. The temperature of the cold reservoir didn't change. However, the temperature of the hot reservoir went down by a few degrees. You can determine the latent heat using the weight of the water in the hot reservoir, the change in temperature of the hot reservoir, and the amount of work done by the engine. Using the temperature of the cold reservoir, one can determine the entropy that has moved from the hot reservoir to the cold reservoir (ΔS).
This process is reversible, now that the ice is melted.
Start doing work on the Carnot engine, measuring the amount of work done with the meter. Entropy will move from the cold reservoir to the hot reservoir using the Carnot engine. Entropy is now being forced to move from the cold reservoir to the hot reservoir. This is not a spontaneous process because you are doing work on the Carnot reservoir. The water in the cold reservoir will freeze. Keep going until all the water in the cold reservoir is melted.
Using the work done in freezing the water, the weight of water in both reservoirs, and the change in temperature of the hot reservoir, one can calculate the latent heat of the water. Again, one calculates the amount of entropy moved from cold reservoir to hot reservoir (ΔS).
The process described for measuring ΔS is reversible. Entropy can spontaneously move from the hot reservoir to the cold. Entropy can be forced to move from the cold reservoir to the hot.
There are probably other reversible processes that can be used to determine ΔS. If the process is reversible, then ΔS will be a function of the material (in this case, water). If you don't use a reversible process, then some correction has to be made for the entropy created.

22. Jul 22, 2012

### Darwin123

Sure.
You can get it back by moving the entropy to an environment slightly lower in temperature than the original ice cream. If you do that, entropy will be moved out of the melted ice cream.
The operative phrase here is "slightly lower". If you put it in an environment that was much colder, entropy would be created by the temperature gradient.

23. Jul 22, 2012

### Studiot

This post is closer to the mark, but I am still waiting for your description and quantifiction of the creation of entropy.

24. Jul 22, 2012

### Darwin123

Processes that create entropy include:
1) Mechanical processes that involve frictional forces of any kind.
-Static friction, kinetic friction, electrical resistivity, etc.
2) Chemical reactions or phase changes that cause an increase in the total "specific heat" at a fixed temperature.
-An increase in specific heat can cause the internal energy of a substance to be "trapped".
3) Movement of entropy without doing work.
-Entropy that moves by heat conduction alone over a large temperature gradient, from high temperature to low temperature, can cause the creation of more entropy.

These types of processes are not mutually exclusive. Types 1 and 2 can lead to type 3.

25. Jul 22, 2012

### Gavroy

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.