[Statistical Physics] Microstates in a large system/Boltzmann entropy

AI Thread Summary
The discussion focuses on estimating the number of microstates for a 1m³ box of nitrogen at standard temperature and pressure (S.T.P.) using Boltzmann's entropy formula. The user calculates the number of molecules, N, as approximately 2.46x10²⁵ and uses the mass of a nitrogen molecule to find the entropy, S, as 6122. However, they encounter a "math error" when attempting to calculate the number of microstates, Ω, due to the large value produced. Clarifications are sought regarding the mass used in the equations and the units for entropy. The conversation highlights the challenges of handling large numbers in calculations related to statistical physics.
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Homework Statement



For a box containing 1m^{3} of nitrogen at S.T.P., estimate the number of microstates which make up the equilibrium macrostate.


Homework Equations



S = Nk_{b}(ln\frac{V}{N} + \frac{5}{2} + \frac{3}{2}ln\frac{2πmk_{b}T}{h^{2}})

where the entropy of a volume, V , of an ideal gas, containing N molecules of mass m at temperature T

S = k_{b}lnΩ


The Attempt at a Solution



First off I don't know which mass it is asking for in the equation. Is it the mass of each individual molecule? Or the mass of all the molecules? Or the molar mass? Either way I tried them all but still couldn't get an answer.

I first worked out what N was.

40.82 mols in 1m^{3} of an ideal gas
1 mol = 6.022x10^{23}
∴ N = 2.46x10^{25}

I let m = 4.652x10^{-26} kg (the mass of a nitrogen molecule)

Plugging those numbers into the first equation gives entropy, S = 6122

Now I know the number of microstates is going to be huge, but from the second equation:

lnΩ = S/k_{b} = 4.44x10^{26}

∴Ω = e^{4.44x10^{26}}

which brings about a "math error".

Am I going about this in the right way?

Cheers
 
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Maybe a mod could move this into the Advance forum?
 
That seems correct to me, except for the missing units. S=6122 (what units?).
You only get a math error if you try to plug that huge number into a typical calculator. Why wold you do that?
 
Ah that's good then. I just thought it was an unreasonably large number.

JK^{-1}
 
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