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Author Topic:   The Four Laws of Thermodynamics
humoshi
Junior Member (Idle past 3328 days)
Posts: 25
Joined: 01-29-2008

 Message 1 of 10 (454746) 02-08-2008 3:13 PM

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 Maxwell-Boltzmann 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 Boltzmann-Maxwell distribution?

 Replies to this message: Message 3 by Chiroptera, posted 02-08-2008 3:51 PM humoshi has responded Message 4 by fallacycop, posted 02-08-2008 4:01 PM humoshi has responded Message 5 by Taz, posted 02-08-2008 5:13 PM humoshi has not yet responded

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 Message 2 of 10 (454752) 02-08-2008 3:43 PM

Thread moved here from the Proposed New Topics forum.
Chiroptera
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Posts: 6531
From: Oklahoma
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 Message 3 of 10 (454755) 02-08-2008 3:51 PM Reply to: Message 1 by humoshi02-08-2008 3:13 PM

Missing information
The Boltzmann distribution is the population in each individual energy state. But there may be many different states with the same energy. So to get the population at a particular energy, you actually need to calculate

(number of states with energy = E) times (population of a single state with energy = E).

The graph that you posted is actually this quantity.

-

 Shouldn't they be similar given that velocity and energy are directly proportional?

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
 This message is a reply to: Message 1 by humoshi, posted 02-08-2008 3:13 PM humoshi has responded

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fallacycop
Member (Idle past 3599 days)
Posts: 692
From: Fortaleza-CE Brazil
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 Message 4 of 10 (454757) 02-08-2008 4:01 PM Reply to: Message 1 by humoshi02-08-2008 3:13 PM

first things first.
 The graph is a histogram of the form 1/x^2

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

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Taz
Member (Idle past 1370 days)
Posts: 5069
From: Zerus
Joined: 07-18-2006

 Message 5 of 10 (454774) 02-08-2008 5:13 PM Reply to: Message 1 by humoshi02-08-2008 3:13 PM

 humoshi writes: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?

KE = kinetic energy
m = mass
v = velocity

KE = (1/2)*m*v^2

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humoshi
Junior Member (Idle past 3328 days)
Posts: 25
Joined: 01-29-2008

 Message 6 of 10 (455014) 02-09-2008 6:46 PM Reply to: Message 4 by fallacycop02-08-2008 4:01 PM

First, I apologize for the sloppiness in the first post. I wasn't very clear and I made a lot of mistakes.
quote:
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.

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 "Maxwell-Boltzmann 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.

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humoshi
Junior Member (Idle past 3328 days)
Posts: 25
Joined: 01-29-2008

 Message 7 of 10 (455015) 02-09-2008 6:47 PM Reply to: Message 3 by Chiroptera02-08-2008 3:51 PM

Re: Missing information
I kind of understand what you are saying. I tried to clarify my troubled thinking in my reply to fallacy cop. Maybe you can understand my problem a little better.

Thanks for the help!

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Chiroptera
Member (Idle past 15 days)
Posts: 6531
From: Oklahoma
Joined: 09-28-2003

 Message 8 of 10 (455029) 02-09-2008 8:40 PM Reply to: Message 6 by humoshi02-09-2008 6:46 PM

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
 This message is a reply to: Message 6 by humoshi, posted 02-09-2008 6:46 PM humoshi has responded

 Replies to this message: Message 9 by humoshi, posted 02-11-2008 10:15 AM Chiroptera has responded

humoshi
Junior Member (Idle past 3328 days)
Posts: 25
Joined: 01-29-2008

 Message 9 of 10 (455186) 02-11-2008 10:15 AM Reply to: Message 8 by Chiroptera02-09-2008 8:40 PM

quote:
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.

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.

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Chiroptera
Member (Idle past 15 days)
Posts: 6531
From: Oklahoma
Joined: 09-28-2003

 Message 10 of 10 (455218) 02-11-2008 1:48 PM Reply to: Message 9 by humoshi02-11-2008 10:15 AM

 The best picture I can get in my head is regarding energy levels in atomic orbital theory.

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
 This message is a reply to: Message 9 by humoshi, posted 02-11-2008 10:15 AM humoshi has not yet responded

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