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Author  Topic: The Four Laws of Thermodynamics  
humoshi Junior Member (Idle past 3323 days) Posts: 25 Joined: 
So, I'm reading "The Four Laws" by Peter Atkins and I'm having a little trouble understanding some parts of the first chapter. I figure I'll have more trouble later on so the purpose of this thread is to get help with some questions. First, he discusses Boltzmann's distribution regarding energy levels. It has the equation: (Population of state of energy E)/(Population of state of energy 0) = e^(BE) Where beta is a constant inversely proportional to temperature. The graph is a histogram of the form 1/x^2. It shows that the majority of molecules occupy the lowest energy state of 0. All higher energy levels have a smaller population of molecules than the ground state 0. So far, so good. But then he talks about using this equation to get the MaxwellBoltzmann distribution for the various speeds of molecules. Namely, this: The problem I'm having is that the graphs aren't isomorophic. Shouldn't they be similar given that velocity and energy are directly proportional? How do you get from the first equation of the form 1/x^2 to the BoltzmannMaxwell distribution?
 
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Chiroptera Member (Idle past 10 days) Posts: 6531 From: Oklahoma Joined: 
(number of states with energy = E) times (population of a single state with energy = E). The graph that you posted is actually this quantity. 
Actually, energy is proportional to the square of the velocity. If I had a million dollars, I'd buy you a monkey. Haven't you always wanted a monkey?  The Barenaked Ladies
 
fallacycop Member (Idle past 3594 days) Posts: 692 From: FortalezaCE Brazil Joined: 
first things first.
No, the graph has the form e^(BE), as you stated in the first paragraph. To understand why the graph for molecule velocity distribution first rises and the falls, you must realise that this distribution takes two factors into consideration. The number of molecules with a given velocity is the product of the probability of a molecule being in a given state with that velocity (that's the one that goes like e^(BE)) multiplyed by the number of states possible with that exact velocity. That second factor grows, and the product of the two first go up and then comes down. Edited by fallacycop, : Typos
 
Taz Member (Idle past 1365 days) Posts: 5069 From: Zerus Joined: 
KE = kinetic energy m = mass v = velocity KE = (1/2)*m*v^2
 
humoshi Junior Member (Idle past 3323 days) Posts: 25 Joined: 
First, I apologize for the sloppiness in the first post. I wasn't very clear and I made a lot of mistakes. quote: I think I understand what you are saying here, but I have a few more questions if you will indulge me. The graph of the Boltzmann distribution in the book has energy (E) on the y access and the number of molecules in that energy state on the x access. The energy increases in discrete units and the lowest energy level has the most molecules in it. It looks something like: [/URL] >< !UB >< !UE> So, if I were to naively use this to make a graph of the velocities of those molecules, I'd count the number of molecules in the bottom state, use the kinetic energy formula to figure out the velocity, and plot it on a seperate graph. Then i'd continue to go up in energy levels doing the same thing. But this wouldn't produce the "MaxwellBoltzmann Distribution." Now, you're saying that different energy states can have molecules with the same velocity. That's what I don't understand, considering I don't see that in the Boltzmann Distribution. I just see the discrete energy levels rising with the relative number of molecules in that enery state going down. Thanks alot for the help! I need it. Edited by humoshi, : No reason given. Edited by humoshi, : No reason given. Edited by humoshi, : No reason given.
 
humoshi Junior Member (Idle past 3323 days) Posts: 25 Joined: 
Thanks for the help!
 
Chiroptera Member (Idle past 10 days) Posts: 6531 From: Oklahoma Joined: 
Here's a very simplified example. Suppose that at one energy, E1, the Boltzmann distribution predicts that a state will have 100 molecules in it. In a state of greater energy, E2, the Boltzmann distribution predicts that a state will have 50 molecules in it. However, suppose only one state has energy E1, but there are 4 states with energy E2. Then there will be 100 molecules with energy E1, but there will be 4*50 = 200 molecules with energy E2. So, in this case, twice as many molecules will have the greater amount of energy. That is (if the energy was determined by the kinetic energy only), twice as many molecules will have the higher velocity. Hope this helps  let's see what further questions that you have. If I had a million dollars, I'd buy you a monkey. Haven't you always wanted a monkey?  The Barenaked Ladies
 
humoshi Junior Member (Idle past 3323 days) Posts: 25 Joined: 
quote: I understand the math alright, but I really can't get a pictorial analog in my head. You are saying a state is defined by more than just its energy (E) and that different states can have the same E. So when you factor that in the ground energy state is no longer the most populated. The best picture I can get in my head is regarding energy levels in atomic orbital theory. Is it anything like how the there are three 2p orbitals and only one 2s orbital even though the three 2p orbitals are at a higher energy level. thanks.
 
Chiroptera Member (Idle past 10 days) Posts: 6531 From: Oklahoma Joined: 
That is one example. The canonical example, usually worked out in undergraduate thermo courses, are molecules in a cubical box. The quantum mechanical development of this situation also predicts a degeneracy of energy states that increases as energy of the state increases, and this gives exactly the Maxwell distribution. If I had a million dollars, I'd buy you a monkey. Haven't you always wanted a monkey?  The Barenaked Ladies



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