Entropy of ideal gas: Volume it *has* or is *allowed* to have?

In summary, the conversation discusses the concept of entropy and its definition in thermodynamic equilibrium. It also touches on the idea of fluctuations and how they may affect entropy, but ultimately concludes that entropy is only well-defined in equilibrium states. The question of defining the volume of a gas at a single point in time is also raised, highlighting the concept of ensembles versus individual sets of particles in classical thermodynamics.
  • #1
greypilgrim
538
36
Hi.

If an ideal gas of ##N## particles is allowed to expand isothermically to double its initial volume, the entropy increase is
$$\Delta S=N\cdot k_B \cdot \log\left(\frac{V_f}{V_i}\right)=N\cdot k_B \cdot \log\left(\frac{2V}{V}\right)=N\cdot k_B \cdot \log\left(2\right)\enspace .$$
This can also be derived with Boltzmann's entropy formula ##S=k_B \cdot \log(W^N)##, assuming the particles are independent and all have the same number of microstates ##W## which doubles as ##V## does. So after the expansion, the number of microstates of the whole gas is ##(2W)^N## and
$$S=k_B \cdot \log\left(\left(2W\right)^N\right)=N\cdot k_B \cdot \log\left(2W\right)\enspace ,$$
also leading to the entropy difference ##\Delta S=N\cdot k_B \cdot \log\left(2\right)##.

Now there will be fluctuations where more than ##N/2## particles are in the initial half of the containment (Poincaré's recurrence theorem even states there will be moments where all particles are back there). I read that this implies a (temporary) reduction in entropy. This is somehow what
$$\Delta S=N\cdot k_B \cdot \log\left(\frac{V_f}{V_i}\right)$$
says if ##V_i## and ##V_f## are taken as the volumes the gas actually has at a moment in time – and not the volumes the gas is allowed to have.

This gets even more confusing if we use Boltzmann's formula: The number ##(2W)^N## contains all microstates in ##2V##, including the ones where there's an imbalance of particle numbers in the halves. Hence those microstates are part of the macrostate with entropy
$$S=N\cdot k_B \cdot \log\left(2W\right)\enspace ,$$
so it will not change even when all particles are in one half of the containment.

As mentioned before, I think this discrepancy results in using the volume the gas actually has at a point in time versus the volume it is allowed to have. But is the first even a well-defined property, especially in thermodynamic equilibrium that is needed for entropy to make sense?
 
Science news on Phys.org
  • #2
greypilgrim said:
Now there will be fluctuations where more than ##N/2## particles are in the initial half of the containment (Poincaré's recurrence theorem even states there will be moments where all particles are back there). I read that this implies a (temporary) reduction in entropy.

Entropy is only defined for equilibrium states, so, for such a fluctuation to occur, the majority of particles would have to be in the initial half on the containment and also be in equilibrium for the gas to have a defined entropy in such a state. Unless you slammed a door shut behind them after the majority of particles entered the initial half of the containment, they wouldn't be in equilibrium. (A Maxwell's Demon could close the door.)

As mentioned before, I think this discrepancy results in using the volume the gas actually has at a point in time versus the volume it is allowed to have. But is the first even a well-defined property, especially in thermodynamic equilibrium that is needed for entropy to make sense?

If your definition of "at a point in time" is that you know the exact position and velocity of each gas molecule at that time, it isn't clear how to define the volume of a gas in that situation To take an extreme example, suppose the gas consists of 3 molecules in a 1 cubic meter chamber. If they are at known locations then how is the volume of the gas defined? Is it 1 cubic meter regardless of whether the molecules are near the walls of the chamber? Or is it the volume of a flat triangular shaped box, with one molecule near each corner of the box?

One could speak of a gas "at a point in time" and keep the model that exact position and velocity of each gas molecule has some probability distribution instead of being precisely determined. Classical thermodynamics tries to juggle two contradictory concepts. On the one hand, a gas is conceptually a set of particles with definite positions and velocities, and on the other hand, the major theorems of thermodynamics do not apply to anyone particular set of such particles, but rather to "ensembles" of sets of particles. So when you speak of a "gas", you must decide whether you mean an ensemble or a particular set of particles.
 
  • Like
Likes greypilgrim
  • #3
Ok thanks, your answer agrees with my understanding of a reasonable notion of entropy. But wasn't it exactly non-equilibrium situations like the ones I mentioned that made Zermelo and many others question the second law of thermodynamics? I don't think Zermelo simply forgot that entropy is only defined in thermodynamic equilibrium.
 
  • #4
greypilgrim said:
But wasn't it exactly non-equilibrium situations like the ones I mentioned that made Zermelo and many others question the second law of thermodynamics?

I'm not familiar with Zermelo's questions. Are they technical questions or philosophical questions?
 

Related to Entropy of ideal gas: Volume it *has* or is *allowed* to have?

1. What is the relationship between the entropy of an ideal gas and its volume?

The entropy of an ideal gas is directly proportional to its volume. This means that as the volume of an ideal gas increases, its entropy also increases. This relationship is described by the second law of thermodynamics.

2. How does the allowed volume of an ideal gas affect its entropy?

The allowed volume of an ideal gas refers to the range of volumes in which the gas can exist. The larger the allowed volume, the higher the entropy of the gas. This is because a larger allowed volume means that the gas has more possible microstates, or arrangements of its particles, resulting in a higher degree of disorder and thus a higher entropy.

3. Can the volume of an ideal gas be negative?

No, the volume of an ideal gas cannot be negative. This is because volume is a physical quantity that represents the amount of space occupied by a gas, and it cannot have a negative value. In addition, negative volume would imply that the gas is occupying a space that does not physically exist.

4. How does the entropy of an ideal gas change with temperature?

The entropy of an ideal gas is directly proportional to the temperature. As the temperature of an ideal gas increases, its entropy also increases. This is because at higher temperatures, the particles of the gas have more kinetic energy and thus have a higher degree of disorder, resulting in a higher entropy.

5. Can the entropy of an ideal gas ever decrease?

According to the second law of thermodynamics, the entropy of an isolated system, such as an ideal gas, can never decrease. This means that the entropy of an ideal gas can only remain constant or increase over time. This is because natural processes tend to move towards a state of higher disorder, resulting in an increase in entropy.

Similar threads

  • Thermodynamics
Replies
3
Views
1K
Replies
19
Views
1K
Replies
22
Views
2K
Replies
2
Views
883
Replies
3
Views
1K
  • Thermodynamics
Replies
1
Views
779
Replies
6
Views
1K
Replies
9
Views
1K
Replies
22
Views
2K
Replies
23
Views
1K
Back
Top