# Regarding the second law of thermodynamics.

1. Aug 20, 2012

### A Dhingra

Hello everyone,
These days i am studying thermal physics in my classes so had a few queries to get clear about.

The Kelvin statement says:
“ No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. ”

Yet the isothermal expansion is possible, i.e., all the heat supplied to the system by its surroundings get completely converted to work.

And Wikipedia explains this as : Note that it is possible to convert heat completely into work, such as the isothermal expansion of ideal gas. However, such a process has an additional result. In the case of the isothermal expansion, the volume of the gas increases and never goes back without outside interference.

So can someone please explain how is this isothermal possible still following the second law?
Also please tell me if there is any restriction to the isothermal compression of a gas by the second law?

Waiting for your reply..

2. Aug 20, 2012

### Studiot

You are correct in that for an isothermal irreversible expansion the first law requires that all the heat is converted to work.

However, I don't know where you obtained your 'Kelvin Statement' but Lord K actually said something somewhat different.

Importantly most discussions about the second law refer to cyclic processes and Lord K's statement was no exception.

So you would need to consider a cyclic process.

Consider also the statement that the sole result is the conversion of heat into work.

This is not so even in your isothermal example as entropy is subtracted from the surroundings and added to the gas.

The entropy change per mole is

$$\Delta S = R\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right) = R\ln \left( {\frac{{{P_1}}}{{{P_1}}}} \right)$$

It is this entropy change that prevents what is known as perpetual motion machines of the second kind being based on an isothermal expansion contraction cycle.

Last edited: Aug 20, 2012
3. Aug 20, 2012

### Darwin123

My favorite physics bugaboo. The "heat trap".
The word "heat" is ambiguous. It is used ambiguously even by science teachers. Even though they probably understand it completely from a scientific viewpoint, teachers continue to teach several informal definitions of the word.
The word heat is context dependent. This is yet another horrible example.
Lord Kelvin was using the word "heat" to mean "energy carried by moving entropy." The Wikipedia article is using the word "heat" to mean the "internal energy" of the system. These are completely different definitions of the word "heat".
An isothermal process can turn all the internal energy of the system into "work energy". This is allowed by the Second Law. However, an isothermal process can't turn all the energy carried by moving entropy into work. That is forbidden by the Second Law.
Note that Wikipedia makes a good definition for isothermal process that completely contradicts its own statement that you quoted so well.
http://en.wikipedia.org/wiki/Isothermal_process
“An isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0. This typically occurs when a system is in contact with an outside thermal reservoir (heat bath), and the change occurs slowly enough to allow the system to continually adjust to the temperature of the reservoir through heat exchange.”
Here, "heat exchange" means the "energy carried by moving entropy". It does not mean "internal energy."
Here is a case where Wikipedia has used the word "heat" in at least two entirely contradictory definitions. I wish I could blame it all on Wikipedia. Most introductory textbooks use the word "heat" in an informal manner.
Just to clarify a bit further, "energy carried by moving entropy" usually means the same as "heat conduction". However, "heat conduction" can be terribly ambiguous in certain contexts. So I prefer the phrase "moving entropy" to "heat conduction" or "heat transfer". When I talk about "internal energy", I use the phrase "internal energy". I have come to despise the word "heat".
I have learned to avoid the word "heat" as much as possible in my own responses. Although my answers tend to sound technical and pedantic, I am trying to avoid ambiguity. However, I can't avoid other scientists who use the word "heat". Therefore, whenever I hear or read the word heat I make a mental note as to what the word specifically means in that context.
I am sorry that I don't have any links to articles showing how words in physics are commonly misused by teachers. The articles and books that I have are out of print. I will list them anyway, in the hopes that there are used copies somewhere.
Book:
1) "The Teaching of Physics" by J. W. Warren (Butterworths, 1965).
2) "Entropy in the teaching of introductory physics" by Hans U. Fuchs in American Journal of Physics, 55(3), 215-219 (March 1987).

4. Aug 20, 2012

### Studiot

Good evening Darwin,

How exactly does your response help the OP in his classes?

You are not even correct in this as heat is a quantity clearly defined by the first law.

Please also note this quote from Wiki in relation to your comment about internal energy.

Last edited: Aug 20, 2012
5. Aug 20, 2012

### Darwin123

The OP should understand that the Second Law of Thermodynamics is specifically refers to heat conduction, regardless of how it is expressed. The word "heat" does not always refer to the heat conduction. For instance, the word heat very often refers to "the internal energy". There are some other definitions of the word heat.
In the older texts often quoted, the word "heat" means entropy rather than energy. For instance, Carnot usually means entropy rather than energy in his monograph where he describes the Carnot cycle. Clausius often meant entropy when he uses the word heat. The problem is that the second law of thermodynamics is often literally quoted from these older references without explanation. Then, they are given a word problem where the word "heat" is used in a more modern context.
The OP can not solve a quantitative problem using the word "heat" until he figures out what the word means in that context. If he doesn't learn to read the word in the context of the problem, he is very likely not to do the problem right.
My conjecture is that the OP read the word heat as meaning "heat conduction" when it really meant "internal energy". Look at what the OP wrote. Tell me if you think that I am wrong, and why. Come up with a better answer than I did.
Okay, maybe my answer should have been shorter.
"Heat transfer" ≠ "Internal energy"

6. Aug 20, 2012

### Studiot

Please do not tell me and in particular a student grappling with the complexities of thermo, that you consider the second law inapplicable to adiabatic processes?

You can boil a jug of water just as effectively in a microwave oven (by radiation) as on the gas (by conduction and convection). The first and second laws apply just as well to both methods.

Last edited: Aug 20, 2012
7. Aug 20, 2012

### Darwin123

No, I did not mean that. I don't even understand how my post could be interpreted that way.
An adiabatic process is one where no entropy is transferred into or out of the system. However, energy can be transferred into the system as work. However, entropy can be created by an adiabatic process.
If you tell a new student that heat can't be transferred in an adiabatic process, then he may think you are talking about energy. So a logical conclusion would be that one can't do work on an adiabatic system.
The microwave oven is an excellent example of what I mean. The energy in the microwaves that bombard the water can be divided partly into heat energy and partly into work energy. Yet, no one really cares how the energy is divided between these two modes. The reason no one cares is that the temperature of the water is determined by the internal energy of the water. It doesn't matter how much is heat and how much is work.
Although it doesn't matter for determining the temperature, one can make a good approximation by assuming that all the energy in the microwaves comes in the form of work. Unless the spectrum of the microwaves is a black body spectrum, one can't use temperature to characterize the microwave energy. If the microwaves did precisely follow a black body spectrum, then we could have a violation of the Second Law.
Suppose that the microwaves could be treated as "heat energy." The peak of the microwave intensity occurs at radio frequencies. Using the Wein law, or any blackbody law, we estimate the temperature of the radio oscillator in the oven. Because the peak wavelength is so long, the temperature of the oscillator is very cold.
However, the temperature of the water can go all the way to boiling with no difficulty. The peak wavelength of the boiling water is in the infrared region, far shorter than the peak in the microwave spectrum.
Thus, you have an example of "heat" spontaneously transferred from a cold reservoir (the oscillator) to a hot reservoir (the boiling water). The Second Law of Thermodynamics has been violated. - Not-
The microwaves do work on the water. The work doesn't count as heat. The internal energy of the water increases. The liquid water turns to water vapor because of the work done by the microwaves. "Heating the water" really means "working the water."
Referring to the microwaves as "heat" is confusing. The important thing is that the internal energy of the water is increasing.
Now, a teacher may ask how much "heat" the water has absorbed from the microwave. What is being asked for is the change in "internal energy", not the fraction of radiation that can be characterized by a black body spectrum.
You tell me. Are the microwaves bombarding the water "heat" or "work"? How can you demonstrate that the Second Law is not violated by a microwave, when you can't even tell me what fraction of the microwave energy is heat or work?

8. Aug 22, 2012

### A Dhingra

Is this the final answer?
I mean, a process in itself can occur which converts all the heat to work, but if the process becomes cyclic all heat can't be converted to work. Is it fine? And is the converse possible for a cyclic process, i.e., all the work done on the system is converted completely into heat released by the system?

9. Aug 22, 2012

### Studiot

Think about it.

You wish to convert all the input heat energy to work, how do you do this?

Wiki (which is not a reliable source) says glibly

'Use the isothermal expansion of a perfect gas'

OK but does it describe apparatus to do this? You can pretend friction does not exist so how do you do this?

10. Aug 22, 2012

### A Dhingra

i gave a thought to it...regarding converting all the heat to work in say isothermal expansion.

If we supply heat to a closed system containing a gas, and allow the heat supplied to raise the temperature by say a very small amount, then we remove the source of heat till the time the temperature comes down to its initial value, so during this time the gas would have expanded using the energy stored as its internal energy doing some work on the surroundings. As soon as the temperature goes down to its initial value heat is supplied again and the same sequence is repeated for very infinitesimally small change of temperature, so this way all the heat supplied to the system gets converted into work done by the system.

Or i can simply say if the isothermal process is carried out quasi-statically, then according to the first law of thermodynamics all the heat gets converted to the work done by the system.
So do you think this will do or not?

11. Aug 22, 2012

### Studiot

Yes of course you can do this but:

The quasi static process requires that the heat added is added at an infinitesimal temperature difference between the source and the working fluid. Otherwise the process of heat addition will not be reversible.

How long will it take to raise the temperature 1 degree at an infinitesimal amount δT per go?

12. Aug 22, 2012

### Darwin123

You said, regarding the Wiki article on isothermal processes,
"...so during this time the gas would have expanded using the energy stored as its internal energy doing some work on the surroundings."
I take it you agree with me. That Wiki article is using the word "heat" to mean "internal energy." Internal energy is determined by the absolute temperature of the system.
However, Lord Kelvin's statement is not using the word "heat" the same way. Lord Kelvin is using the word heat to mean energy carried by heat transfer (i.e., entropy transfer). Entropy transfer is determined by the difference between the temperatures of system and reservoir.
The definitions of "heat" are not the same. Knowing that, there is no wonder that the statement by Wikipedia seemingly contradicts the statement by Lord Kelvin. The "contradiction" was semantic, not logical.
The second law is expressions in several different ways. I prefer the expressions that don't use the word "heat".
Of course, we can also use a mathematical expression rather than an English (French, German, Japanese, ...) expression to pin the statement down. The second law can also be expressed as,
dS_Isolated/dt≥0,
where "S_Isolated" is the total entropy of an isolated system, and "t" is the time.
In any case, there is no mention of a "cyclic process".

13. Aug 22, 2012

### Darwin123

I didn't catch this on my first reading but you wrote,
"then we remove the source of heat till the time the temperature comes down to its initial value".
Therefore, the entire process that you just described is not completely isothermal. Wikipedia said that all the heat can be converted work in an isothermal process. If there is a part of the process where the temperature changes value, then it is not isothermal.
One can conceive of an isothermal process that is also cyclic. However, not all the heat absorbed in this process can be turned into work. The reason is that the system has to lose its energy in order to be cyclic. In order to be cyclic, the system has to be returned to the same internal energy and the same absolute entropy as it was before the heat was absorbed.
Let us consider the specific case of a Carnot cycle. This is a reversible reaction that ordinarily has four parts: two of the parts are isothermal and two are adiabatic. However, let us take the Carnot cycle to the limit where there is no adiabatic part of the cycle. Let us consider a Carnot cycle in which the adiabatic part of the cycle is suppressed.
When you draw the pressure versus volume (P vs. V) diagram of an ordinary Carnot cycle, the result is a closed curved. The area of this curve is equal to the work done by the engine during the cycle. However, reduce the adiabatic part of the Carnot cycle. The two segments of the plot belonging to the isothermal parts come closer together. At the point where the cycle is completely isothermal, the two adiabatic components on the curve coincide. However, this mean there is no work done by the Carnot cycle.
In a Carnot cycle, the engine is placed between two heat reservoirs. If the two heat reservoirs are of different temperature, the Carnot engine can do work. However, an isothermal Carnot cycle is one where both reservoirs have the same temperature. Obviously, the Carnot engine can't do positive work during the cycle. Whatever work is done by the engine has to done on the engine in order to return the engine to its original state.
Therefore, one can't solve the conundrum that you are describing simply by restricting the discussion to cyclic processes. There is no way a cyclic process that also isothermal can do any work.
You had to add a step in the process which was not isothermal in order to have the cycle do work. Now maybe that is the unstated hypothesis of Wikipedia. However, this is not what Kelvin was talking about. Wikipedia should have made clear what they meant by "heat" in their statement.
Unstated hypotheses are part of what I meant by "context sensitive." Sometimes, there is a hypothesis that the communicator does not explicitly state. It may be clear from the previous words that there is a cyclic process or that "heat" is now internal energy. If you read carefully through the entire Wikipedia article, there may be some statement inserted that changed the meaning of the word "heat". However, maybe there wasn't.
If there is fault in your not understanding, then I would give it to Wikipedia. However, it would be helpful in your endeavors if you take some responsibility to learn how to interpret the context of an article.
Wikipedia made a mistake. I don't think what I said is revolutionary. Rather, Wikipedia was not clear. I was just warning you that most textbooks and teachers make similar "mistakes" when they discuss thermodynamics.
The word "electricity" has similar problems to the word "heat". If you ask someone how much "electricity" is passing through a wire, then there is more than one answer. For instance, the word "electricity" can current or it can be power. The desired answer may be current, measured in Amperes. The desired answer may be in power, measured in Watts. Some people may even think the word "electricity" means the electric potential, which is measured in Volts.
Context is important. The word "electricity" does not have a fixed scientific meaning. In the context of your electric bill, electricity always means power. One doesn't pay for Amperes or Volts. One pays for electric power. If you want to electrocute someone, the answer is never in Watts. The lethality of electricity is determined by the Amperes.

14. Aug 22, 2012

### Studiot

@ Darwin

I'm glad you are finally coming to what I have been saying.

The only issue I have is your somewhat unconventional viewpoint & terminology which will not gain any marks for students taking conventional exams. You have already taken yours please consider those who have still to pass theirs.

The definition of heat is well defined and tabulated in thousands of pages of engineering data throughout the world. On account of a couple of centuries worth of data collection this is not going to change any time soon.

Your viewpoint is certainly interesting and thought provoking and I thank you for this.

@ A Dinghra

You need to realise that there are two types of quantities in thermodynamics.

Those which are properties of the system (normally state variables).

Those which are properties of the surroundings.

Properties of the surroundings include both the heat passed and the work done across the system boundary. The surroundings can force either on the system.

An engineers definition is

Heat passed is the energy passed across the boundary as a result of temperature difference between the system and its surroundings.

Work is the work done as a result of change (movement) boundary between the system and its surroundings.

The first law tells us that both of these contribute to the system state variable called internal energy.

Last edited: Aug 22, 2012
15. Aug 22, 2012

### Darwin123

Okay, maybe I was wrong.
I finally understand what Wikipedia could have meant.
Any one way reaction where,
PdV=TdS
can be said to turn heat completely into work.
Under conditions that are both isothermal and isobaric, an ideal gas can turn heat into work. Okay, suppose we have an ideal gas in a cylinder. The gas and the heat reservoir are the same temperature. There is no friction between the cylinder and piston, so the pressure inside the cylinder equals the pressure outside. The piston is allowed to expand slowly, so the is no sound or other coherent vibration in the cylinder. Under those conditions,
PdV=TdS.
This process is reversible. It could be irreversible only if the pressure inside the piston is different from the pressure outside the piston. That can occur because of static friction between the cylinder and the piston. In that case,
PdV<TdS.
So I was wrong. Wikipedia was right.
The shame!
They may very well have used the same definition of heat as Kelvin.
Kelvin specified a cyclic process. Wikipedia did not say whether or not the process was cyclic. Maybe that is where the OP made a mistake. Heat can not be turned completely into work in any cyclic process. By not including the words cyclic, the OP made it look like Wikipedia contradicted Lord Kelvin.
I am so sorry.

16. Aug 22, 2012

### Studiot

Don't forget that with your 'one shot' expansion you cannot create a useful machine.

What happens at the end of the expansion?

Kelvin et al hailed from the days of steam engines where they wanted that piston in a cylinder to go back and fore to drive a prime mover so they naturally thought in cyclic terms.

Engineering thermodynamics, in contrast to physicists thermodynamics, was developed in terms of a working fluid that is cyclically compressed and expanded, but never used up.

Hence the Carnot and other machine cycles.

As the working fluid is taken round its machine cycle it accepts heat from a hot reservoir and uses some of it to perform useful work. The remainder is rejected to a cold reservoir.

The second law is all about how much useful work can be extracted from such a system.

Last edited: Aug 22, 2012
17. Aug 22, 2012

### A Dhingra

I am sorry but i couldn't make out your question..
If i am talking of an isothermal expansion, then minor changes in the temperature will occur, but we will never allow the temperature rise to an appreciable value. And if this is what you meant then I think it will take infinite time to raise the temperature by 1 degree..
And is this the reason that we say that isothermal processes take infinite amount of time
?

18. Aug 22, 2012

### Darwin123

A gun is a useful machine. Sometimes, 'one shot' is all you need!

19. Aug 23, 2012

### Studiot

Yup you got it.

The isothermal expansion line on a PV diagram marks what is known as the maximum work possible.

Edit: @Darwin

Many treatments of the second law define and discuss "heat engines".
Is a gun a heat engine?

It is true that my treatment was a gross simplicification to assist understanding and that even in the early steam engines the working fluid was discarded not recovered so had to be replaced.
It is also true that the distinction between flow processses and non flow processes, such as we are discussing here, is very important in engineering thermodynamics.

Last edited: Aug 23, 2012
20. Aug 25, 2012

### Radhakrishnam

Thermodynamics does not answer questions concerning time. In other words, you can not find an answer to your question, in thermodynamics.

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