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

• I
Mentor
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
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.

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Mentor
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.

• binis and Philip Koeck
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.

• weirdoguy and Motore
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?

• Chestermiller
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.

• weirdoguy and Motore
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.

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• binis