Probabilistic Definition of Entropy

[@rinsim]

When we use the Stirling approximation, the argument of the ln has to be big. While the total number of particles is certainly big, here the n_i are arbitrary. They can also be 0. How can we justify it? This is my attempt: the sum of n_i is equal to N, so if the total number of possible states is much lower than N, some n_i have to be big. So I can basically say: n_i small => ln(n_i!) ~ 0 n_i big => use Stirling approximation But even if the factorial grows incredibly fast, the ln grows incredibly slow, so I need to be careful here to make sure I'm really ignoring small numbers. Stirling says: ln(n!) ~ n*ln(n) - n so for n big this is big. Now this approximation actually works quite well even for very small numbers like 100 and ln(100!)~364 which is a small number in the summation. Any mistake in my reasoning?

[@PhysicalChemistry]

Yes, that's right. For macroscopic systems, N is almost incomprehensibly large, and even the n_i for "rare" states are large enough that Stirling is a good approximation. And if n_i is small enough (for super-rare states) that Stirling is not a good approximation, then these states contribute so little to the entropy that it doesn't matter. But it's true that Stirling is only an approximation. Strictly speaking, thermodynamics has assumed the limit where N is truly infinite, so all of the n_i are either zero or infinite, as well. That's a bit of a dodge, but it's what is needed to keep the math correct. There are systems for which the n_i for important states are NOT large enough to keep this and other approximations valid. In those cases, standard thermodynamics doesn't work well. For more details, I would recommend the book titled "Thermodynamics of Small Systems" by T. L. Hill.

[@rinsim]

@@PhysicalChemistry Thanks for the clarification!

[@sroydetroy6404]

@@PhysicalChemistry Just a quick question- I tried looking in the internet what a ghost state means and I didn't find a good answer. Would you kindly clarify? thanks in advance.

[@PhysicalChemistry]

​@@sroydetroy6404that's not a term I'm familiar with, either

[@sroydetroy6404]

​@@PhysicalChemistry But you said it in 0:47. I would like to know what it means.

[@thiagosena8911]

Professor, for any spontaneous process, according to the second law, deltaS must increase or be equal to zero, but never less than zero. Thus, considering a situation for two particles, in which particle 1 is at a temperature, T1, and particle 2 at a temperature, T2, with T2 > T1. If particles 1 and 2 receive the same amount of heat, the number of new microstates available for particle 1 will be greater than for particles 2, right? Therefore, as there are more ways to disperse energy to particles 1, will the entropy change, ∆S1 be greater than ∆S2? My thought process was considering entropy as a measure of the dispersion of energy between particles in their microstates. Thank you very much! I find your work to be very detailed.

[@PhysicalChemistry]

The number of accessible microstates available for systems 1 and 2 at various energies* will depend on the specific pattern of energy states of those systems. They don't have to be the same. Their (probabilistic) entropy may increase by different amounts under the same amount of energy increase. The second law doesn't necessarily work for individual particles, like it does for large macroscopic systems. If you want to invoke the second law, you're talking about themodynamic, not microscopic properties. In that case, you're absolutely right that ΔS₁ > ΔS₂ if T₁ < T₂. This video gives a more complete answer to your question: kzbin.info/www/bejne/p3yql5iXh8qqiZI And you're definitely right that you can think about entropy as a measure of the dispersion of energy across different microstates. *It's a bit dangerous to talk about the temperature of an individual particle. At the very least, we would need to clarify what we mean by that. But this doesn't really affect your main question, so I'll try to ignore this particular quibble.

[@shlokshankar1819]

what does it mean for n(sub i) molecules to be in a given state .I am asking this because we know that state of a given molecule is defined by its position and momentum. Even if number of molecules can have different momentum (which they will) they can't acquire the same position in space .So, how is it that they are acquiring the same state? Can you please help me out with this.

[@PhysicalChemistry]

For these equations, the states are energy levels. And they are discrete, not continuous. (E.g. a hydrogen molecule is in the 1s or 2s or 2p ... state. Other molecules are in their ground state or first excited state or ...) In this sense, it is okay if the molecules are in the same (energy) state. But even if they can't share the same state, the equations are still fine. You can think of n_i as the *average* number of molecules that are in that state, when averaged over a long time. (It will be between 0 and 1, if the state can only hold one molecule.) That's how (quantum mechanical) molecules behave,. But under conditions where the molecules behave classically, it is okay to think of them the way you describe: with any value of position and any value of momentum being possible. In that case, the equations in this video would need to be rewritten as integrals, instead of sums. For example, S = −k ∫ p(Γ) ln p(Γ) dΓ, where the Γ coordinates are integrated over all values the positions and momenta.

[@MattGiMa]

Why we want to maximize the entropy?