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Topic: Statistical mechanics (thermodynamics), Gaseous Kr  (Read 2670 times)

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Offline Vrig

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Statistical mechanics (thermodynamics), Gaseous Kr
« on: April 15, 2014, 11:14:50 AM »
"Consider a gaseous krypton (molecular weight 84 g/mol) at 50 K and 1 atm.

Use statistical mechanics to test if the gas can be treated classically."

Not even sure where to begin, anybody who can hint where I should start? I thought (at first) that if it works classically then pV=nRT would work, but I have nothing to go on. I know p, T, R, not sure if I can assume anything about the amount of moles in the system as we have a pressure, increasing n would increase p (or V or p is kept constant).

EDIT:
Also I'm assuming we're in the classical limit where N << no. of states (Φ)
« Last Edit: April 15, 2014, 12:56:50 PM by Vrig »

Offline mjc123

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Re: Statistical mechanics (thermodynamics), Gaseous Kr
« Reply #1 on: April 15, 2014, 01:24:48 PM »
Krypton is not a gas at 50K and 1 atm. It melts at 115 K and boils at 120K.

Offline Vrig

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Re: Statistical mechanics (thermodynamics), Gaseous Kr
« Reply #2 on: April 15, 2014, 01:51:47 PM »
I think what they are trying to say is that the atomic interactions are non-existing and thus the system can be treated as a very dilute monoatomic mixture (and thus a "gas"). Not sure.

Offline Corribus

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Re: Statistical mechanics (thermodynamics), Gaseous Kr
« Reply #3 on: April 15, 2014, 02:12:15 PM »
It's not really clear what the question is after. It seems likely the partition coefficient will be involved somewhere in the answer but without a better indication of what the question wants you to do, it's hard to give direction. Is this the exact wording of the question?

Also, "atomic interactions are non-existing" is never the case.
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 Vrig

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Re: Statistical mechanics (thermodynamics), Gaseous Kr
« Reply #4 on: April 15, 2014, 02:55:56 PM »
Yes, those are the exact words. I'm assuming it has something to do with the partition function, derivating it with respect to V, or N, or whatever, to obtain E, p, etc is probably the way to go. Then from that going to either the molecular weight ("to confirm that it is OK to treat it clasically").

Yeah I know the interactions are never really zero but some assumptions can be made to make life a bit easier for us, non-interacting molecules (or atoms, because it's monoatomic) would be one of them.

I'm guessing that by confirming the molecular weight (as only M, T and p are given in the question) one shows that it can be treated clasically, but if one gets a different weight then you've shown that "it cannot be treated clasically".

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