Undergrad Why is Entropy defined as a fraction of heat over temperature?

[vanhees71]

The thermal energy of a classical ideal gas is $U=f k_{\text{B}} N T/2$, where $f$ is the number of "relevant degrees of freedom" (i.e., $f=3$ for a monatomic, $f=5$ for a twoatomic, and $f=6$ for multiatomic gases); $k$ is Boltzmann's constant and $N$ is the number of particles/molecules, and $T$ the (absolute) temperature.

Then you have the ideal-gas law $p V=N k_{\text{B}} T$, which is obviously different from $U$ by a factor.

 

[binis]

Clausious definition is differential: dS=dQ/T, it's not always easy to integrate this.Keeping T stable, if you extract an amount dQ from the system, you increase the order of the molecules. If you induct an amount dQ in the system, then you increase the disorder of the molecules (random motion).

 

[Chestermiller]

Clausious definition is differential: dS=dQ/T, it's not always easy to integrate this.Keeping T stable, if you extract an amount dQ from the system, you increase the order of the molecules. If you induct an amount dQ in the system, then you increase the disorder of the molecules (random motion).

Can you please provide an example for this contention?

 

[Philip Koeck]

Clausious definition is differential: dS=dQ/T, it's not always easy to integrate this.Keeping T stable, if you extract an amount dQ from the system, you increase the order of the molecules. If you induct an amount dQ in the system, then you increase the disorder of the molecules (random motion).

I don't want to give away too much here, but do please follow up on Chestermiller's suggestion.
How would your statement work out for an ideal gas, for example?

 

[binis]

Can you please provide an example for this contention?

Think of boiling water. Molecules transit from a more "ordered" state (liquid) to a less "ordered" state (gas).

 

[Chestermiller]

Think of boiling water. Molecules transit from a more "ordered" state (liquid) to a less "ordered" state (gas).

Are you portraying this as a situation where it is difficult to determine the entropy change using the classical equation?

 

[binis]

I don't want to give away too much here, but do please follow up on Chestermiller's suggestion.

Steam liquefaction is an example

How would your statement work out for an ideal gas, for example?

Gas compression. For stable temperature and pressure it is dS= nRlnVf/Vi
n=number of molecules, Vi=initial volume, Vf=final volume
If you extract an amount dQ then dQ<0 => dS<0 => Vi>Vf
Smaller volume means more "ordered" situation.

 

[Chestermiller]

Steam liquefaction is an example

Gas condensation. For stable temperature and pressure it is dS= nRlnVf/Vi
n=number of molecules, Vi=initial volume, Vf=final volume
If you extract an amount dQ then dQ<0 => dS<0 => Vi>Vf
Smaller volume means more "ordered" situation.

You mean "compression," not "condensation," right? Your equation is only for an ideal gas at constant temperature, not for a real gas.

 

[Philip Koeck]

Gas condensation. For stable temperature and pressure it is dS= nRlnVf/Vi
n=number of molecules, Vi=initial volume, Vf=final volume
If you extract an amount dQ then dQ<0 => dS<0 => Vi>Vf
Smaller volume means more "ordered" situation.

Well, an ideal gas can't really condensate. There are no forces between the gas molecules.

Even in the real world we have to be able to discuss other processes than phase changes.

What I actually reacted to was that you specified constant T, but at the same time said that the random motion of the molecules (the average kinetic energy or the inner energy) increases due to added heat.
For an ideal gas that would be a contradiction, since the inner energy depends only on T.

So, clearly, if you add heat to an ideal gas in an isothermal process entropy increase is not due to increased motion of the molecules. There's something else going on.

As Chestermiller has just pointed out you've stated the expression for entropy change of an ideal gas at constant T. You could actually derive this starting from the basic formula for entropy change.

 

[binis]

What I actually reacted to was that you specified constant T, but at the same time said that the random motion of the molecules (the average kinetic energy or the inner energy) increases due to added heat.

I mean the randomness increases, not the motion.

So, clearly, if you add heat to an ideal gas in an isothermal process entropy increase is not due to increased motion of the molecules. There's something else going on.

Yes, the motion becames more random. Mind that the entropy concept is aligned with math probability theory.

 

[Philip Koeck]

Yes, the motion becomes more random.

How does that look when the motion of gas molecules is more random?

 

[binis]

How does that look when the motion of gas molecules is more random?

It looks less oriented.

 

[Philip Koeck]

It looks less oriented.

The motion of gas molecules never has a preferred direction when the gas is in equilibrium.
How can it become less oriented?

 

[binis]

The motion of gas molecules never has a preferred direction when the gas is in equilibrium.
How can it become less oriented?

When motion is limited by a smaller volume we consider as it is less random. Anyway,this is my perception about Clausious entropy. You know that there are also Boltzmann,Shannon,Renyi,Tsallis & other definitions. It is a large debate of what entropy really is.

 

[Philip Koeck]

When motion is limited by a smaller volume we consider as it is less random. Anyway,this is my perception about Clausious entropy. You know that there are also Boltzmann,Shannon,Renyi,Tsallis & other definitions. It is a large debate of what entropy really is.

I think we've arrived at an important conclusion.
Analysis of the Carnot process leads to a state function, S, and we can calculate the change of S in a reversible process using dS = dQ / T.
If we apply this to a reversible, isothermal process for an ideal gas it turns out that ΔS is related to a change of volume only, not an increase in random motion, for example.

 

[vanhees71]

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