And here we go with the FAQ:

No!! Please take me back!!

Question:
I read in Feynman's "Lectures on Physics" that entropy, or more accurately, a differential change in entropy, is given by dQ/T, where dQ is a differential transfer of heat energy, and T is absolute temperature. Could the concept be extended to other kinds of energy?

Answer:
Heat and work are methods to transfer energy to a system. The energy is transferred as work, if the entropy of the system does not change. It is transferred as heat if it does change. Hence the answer to your question is "no". If you are to transfer energy without changing the entropy of the system then by definition you are doing work on the system

A perfectly-insulating horizontal cylinder has a sliding, perfectly heat insulating piston with no friction (impossible in practice) and no leakage (also impossible). The piston is initially centered in the cylinder, and the two halves are filled with gas to equal pressure and temperature. Now, the piston is slowly forced toward one end of the cylinder, e.g. by an electromagnet on the outside of the cylinder acting on a piece of steel in the piston. If one assumes no heat enters or leaves the system, the compressed gas gets hotter and the rarefied gas gets cooler. Of course, work is done to compress the gas in the one end.

Now, the electromagnet is suddenly turned off, thus releasing the piston. The difference in pressure forces the piston toward the center, but, because of its mass, it continues toward the opposite end, compressing and heating the gas at that end and rarefying and cooling the gas at the end where it started. With friction, leakage, imperfect heat insulation, etc. neglected, would the piston oscillate back and forth indefinitely?

Answer:
The keyword is "suddenly". With this, according to the Second Law (wich I absolutely believe in <g>) the answer is "no", the motion of the piston will be damped because of internal friction in the two gas portions, which increase the entropy of the system.

Question:
Do I conclude from this that collisions between molecules and the walls of the chambers, or each other, are not elastic?

Answer:
If you release the piston suddenly, the process is conducted irreversibly and this results in an increase of entropy. How this entropy is generated, is another question.

The conclusion that entropy increases follows from the Second Law. Hence, it must be true for ideal gasses, too. Therefore, it must hold also for elastic collisions between molecules.

In the process you suggested by your Gedankenexperiment, work is done upon the system. If you release that piston suddenly, this work must be converted into heat (at least partially), even if you neglect friction, and entropy is generated. How can this occur? Well, by releasing the piston quickly, some gas molecules are violently kicked to high speeds by the piston. In subsequent collisions between the molecules their energy is equilibrated out between all gas molecules, thus rising the temperature of the gas. This is how entropy is generated

Conversely, while moving the piston slowly (quasistatic) in one direction in the initial phase, when you were doing work on the system, no entropy was generated. Why? Let's have a closer look at the process:

Consider a gas molecule on its way in the direction of the resting piston. Eventually, it collides with the piston and, if the collision is perfectly elastic, it will leave the piston with its unchanged speed. If the piston is moved quasistatic, everything stays the same.

However, if the piston would be moved fast, the molecule would accumulate kinetic energy after the collision with the piston, that is, the temperature of the gas would be rised (after equilibrium is reached again). So you could generate entropy in the compression phase, too, if you would conduct that phase fast. Thus, no friction is needed to produce entropy. Hence, in your Gedankenexperiment, entropy is increased without addition of heat.

Answer:
I cannot help you with information theory, but here is why it violates the Second Law.

Think of a isolated box with an opening covered by your one way mirror. Place that box in a universe filled with radiation of 3K (any resemblance to a real universe is purely coincidental ). The device accumulates photons of the 3K radiation, hence you have more of them in the box than in a system at equilibrium. The number of high energy photons in the box will initally be too low compared to black body radiation.

Eventually, the radiation in the box achieves equilibrium (after isolating the opening), where the energies of the photons are distributed following the Planck formula of the black body radiator. This can only be achieved by increasing the number of high energy photons at the expense of low energy photons.

The result is an increase in temperature within the box. Thus, the net result is a spontaneous flow of energy against the temperature gradient, from the universe into the box, which contradicts the second law.

Answer:
I don't know about the universe (too big to wrap my brains around it), but cooling does not necessarily correspond to increased order. See for instance adiabatic cooling for which dS >= 0. While it is true that an adiabatically expanding ideal gas lowers its entropy by cooling, this lowering is fully compensated by the increase due to expansion, if the process is perfectly quasistatic. Then, dS = 0; otherwise, entropy increase due to expansion exceeds entropy decrease due to cooling.

Answer:
Yes. There are two main reasons for this. The more important is that the gas has to do work on its surroundings if it expands to normal pressure. It takes the energy needed from its internal energy, thus cooling down.

As an example, consider a monoatomic gas at pressure p_A = 1000 kPa, expanding to p_D = 100 kPa. Initial temperature be T_A = 300 K. What will the final temperature be after expansion? If I got the numbers right it will cool down to 119.4 K, which is quite a bit.

Question:
Does this not contradict the Second Law? Since the gas is cooling, its entropy decreases, doesn't it?

Answer:
You are right, cooling usually means decreasing of entropy of the cooling system. Overall entropy is not decreased: cooling is only possible if heat can be released to something; thereby heating this something and increasing its entropy as well. But in this case, don't forget about expansion. In your example, we have an adiabatic process, cooling is caused by expansion.

Question:
Uhm... so the gas cools and its entropy goes down, but it expands also and its entropy goes up? Could the two processes compensate?

Answer:
Yes and no. They compensate exactly if processes are conducted reversibly. "Reversibly" means: minimum possible entropy change. Let us first see what we can intuitively say about entropy changes:

• expanding at constant temperature (from A to D' in the diagram below) increases entropy of the gas
• cooling at constant volume (from D' to D) decreases entropy of the gas
• of course, in cooling, entropy is not overall decreased, it is exported instead: the cooling gas exports entropy to its surroundings.

If the process is conducted reversibly, entropy is not changed because effect of cooling and effect of expanding cancel out, otherwise entropy increases.

Question:
Now I have another problem: you seem to equate the adiabatic process to a sequence of isothermal and isochoric processes. Why is one allowed to do this? And are there conditions when one may equate such processes?

Answer:
Yes, there are conditions. Look at the diagram! Obviously, we have the same pressure and volume at state D no matter if we get to D via the adiabatic path or if we get there by the isothermal path from A to D' and then the isochoric (at constant volume) path from D' to D. Or at any other path. Pressure and Volume are said to be state functions. They depend only of the state of a system, not on the path the system got to that state.

Less obvious, but not less true, this is also valid for other state functions like entropy, energy.

Question:
Is there an example of a function that is not a state function?

Answer:
Look at the work needed to get from Chicago to Detroit. It makes a big difference in fuel consumption whether you go straigth or if you go via Los Angeles. Work and heat usually are not state functions, but depending on the path.

Because entropy (as well as pressure, volume, and others) is a state function, depending only on the state of the system and not on how the system got to that state, instead of looking at an adiabatic expansion from (p_A, T_A, V_A) to (p_D, T_D, V_D) we can as well look first at an isothermal expansion from (p_A, V_A) to (p_D', V_D) and then at an isochoric (at constant volume) cooling from (p_D', T_A) to (p_D, T_D), as shown in the picture

In this diagram, the state (p_A, T_A, V_A) is represented as point A, while (p_D, T_D, V_D) be point D. The process we are looking at is from A to D.

For the calculation it is easier to go via D'. And, for the calculation it is easier to assume ideal monoatomic gas. For real systems the only change is that there are many factors, instead of only two, contributing to entropy change. So here we go with the ideal gas:

From A to D' pressure decreases and volume increases at constant temperature. Entropy change is

S(isotherm) = nR ln ( VD/VA) = 11.5 J/mol/K

From D' to D volume is constant and temperature decreases from T_A to T_D. Entropy change is:

S(isochor) = CV ln (TD / TA) = - 11.5 J/mol/K,

where

• R is the universal gas constant, 8.314 J/mol/K
• C_V is heat capacity, for an monoatomic ideal gas C_V = 1.5 nR
• n is the number of moles

We see that the two effects compensate, as long as the process is conducted reversibly. If not, entropy increases.

Question:
Is the ideal gas a good approximation for CO₂?

Answer:
Not that good. There are two additional factors contributing to entropy:

• upon expansion, a real gas does not only work against external pressure but also against attraction between molecules. I would expect CO₂ to cool more than an gas behaving more ideally.
• gases composed of polyatomic molecules can store energy not only in translation, but in rotation and in vibration, too. Thus they have larger heat capacity than monotaomic gases. They don't cool down as much as the latter.
• the two effects can compensate partially. At high pressure I'd guess the first effect to be larger. That means, CO₂ escaping from a pressurizet bottle would probably cool down more than Ar, escaping from a bottle at the same pressure and temperature.

Answer:
Well, it is called "Maxwells demon" but it isn't a Maxwells demon in the sense that it isn't a gear that violates the second law. You are probably speaking of vortex tubes and may want to have a look at
vortex tube theory

Answer:
I interpret your question twofold:

• why is it delta: Delta stands for a difference. In the Second Law, entropy is not absolutely defined, only as a difference. Only the Third Law defines absolute entropies.
• why did they choose the symbol S: entropy was introduced by Clausius in 1854. I don't know why he choose S and not Z, Y, R or O. Might be pure coincidence, guess that his girl friend was named Susan or Sandra or so (ah, no, they hadn't that time, had they?) :-)

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Created: 97-10-12
last modified may 8, 1999 GVa
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