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Topic: Stirling's approximation  (Read 4638 times)

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Offline Big-Daddy

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Stirling's approximation
« on: May 11, 2013, 01:47:53 PM »
I know this is more Maths than chemistry but can someone explain how it is actually done? I would like to apply to NA, that is why I ask in a chemistry forum ... (not the best reason but I'm sure you guys know)

Offline curiouscat

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Re: Stirling's approximation
« Reply #1 on: May 11, 2013, 02:36:48 PM »
Quote
I would like to apply to NA

That does't make much sense, sorry.

Offline Big-Daddy

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Re: Stirling's approximation
« Reply #2 on: May 11, 2013, 03:32:24 PM »
Quote
I would like to apply to NA

That does't make much sense, sorry.

Ah sorry, I meant NA!. I meant I would like to apply Stirling's approximation to the factorial of Avogadro's number.

But in general, I'd like how to see how n! goes down in Stirling's approximation, assuming n is very large.

Offline curiouscat

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Re: Stirling's approximation
« Reply #3 on: May 11, 2013, 03:43:09 PM »
ln(n!) = n ln(n) - n


Offline Big-Daddy

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Re: Stirling's approximation
« Reply #4 on: May 11, 2013, 04:41:32 PM »
Thanks.

The common use of this calculation as far as I know (in chemistry) is to do with entropy and the number of possible microstates, but for all normal values of entropy this results in a number between ω=101020 and ω=101026. Yet NA!=104.6957·1046. This isn't anywhere near the right ballpark ... if we had ω=104.6957·1046 microstates the entropy of our system would be roughly 5·1024 J·K-1·mol-1 which is obviously ludicrous. What am I doing wrong? Is it perhaps not NA!, but rather something else, that is meant to be here?

Offline Borek

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Re: Stirling's approximation
« Reply #5 on: May 11, 2013, 04:58:17 PM »
Google for "calculating number of microstates".
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Offline Corribus

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Re: Stirling's approximation
« Reply #6 on: May 11, 2013, 05:01:30 PM »
I have no idea what you're trying to do.  Maybe you could try explaining where you came up with this need.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Borek

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Re: Stirling's approximation
« Reply #7 on: May 11, 2013, 05:24:38 PM »
IMHO he tries to calculate the entropy of a mole of ideal gas using S=k×ln(W) and assuming W=NA!
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Offline Big-Daddy

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Re: Stirling's approximation
« Reply #8 on: May 11, 2013, 05:49:19 PM »
IMHO he tries to calculate the entropy of a mole of ideal gas using S=k×ln(W) and assuming W=NA!

Yes but W≠NA! according to my Internet search (and also basic logic) ... I will have to dig deeper to find out what it is exactly.

Offline Corribus

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Re: Stirling's approximation
« Reply #9 on: May 11, 2013, 11:02:25 PM »
W is a partition function and significantly more difficult to figure out than simply taking factorial of the number of particles in the system.  Stirling's approximation is involved in the determination, but as you see here:

http://www.nyu.edu/classes/tuckerman/stat.mech/lectures/lecture_6/node2.html

There is a lot more to it.  This is really the subject of graduate level statistical mechanics. 
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

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Re: Stirling's approximation
« Reply #10 on: May 12, 2013, 05:39:46 AM »
W is a partition function and significantly more difficult to figure out than simply taking factorial of the number of particles in the system.  Stirling's approximation is involved in the determination, but as you see here:

http://www.nyu.edu/classes/tuckerman/stat.mech/lectures/lecture_6/node2.html

There is a lot more to it.  This is really the subject of graduate level statistical mechanics.

Haha I looked at the third equation on the page and realized instantly where this was going  :P Anyway, thanks for clearing up Stirling's approximation. At least now I know that W≠NA! as I used to think.

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