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

In summary, the problem asks for an estimate of the number of microstates in equilibrium for a box containing 1m^{3} of nitrogen at S.T.P. Using the equations for entropy, an estimate is found to be 4.44x10^{26}, which results in a "math error" when trying to calculate the exact number of microstates due to the large value.
  • #1
Flucky
95
1

Homework Statement



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


Homework Equations



S = Nk[itex]_{b}[/itex](ln[itex]\frac{V}{N}[/itex] + [itex]\frac{5}{2}[/itex] + [itex]\frac{3}{2}[/itex]ln[itex]\frac{2πmk_{b}T}{h^{2}}[/itex])

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

S = k[itex]_{b}[/itex]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[itex]^{3}[/itex] of an ideal gas
1 mol = 6.022x10[itex]^{23}[/itex]
∴ N = 2.46x10[itex]^{25}[/itex]

I let m = 4.652x10[itex]^{-26}[/itex] 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[itex]_{b}[/itex] = 4.44x10[itex]^{26}[/itex]

∴Ω = e[itex]^{4.44x10^{26}}[/itex]

which brings about a "math error".

Am I going about this in the right way?

Cheers
 
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  • #2
Shameless bump
 
  • #3
Maybe a mod could move this into the Advance forum?
 
  • #4
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?
 
  • #5
Ah that's good then. I just thought it was an unreasonably large number.

JK[itex]^{-1}[/itex]
 

1. What is a microstate in statistical physics?

A microstate in statistical physics refers to the specific configuration of particles in a system at a given time. It includes the positions, momenta, and spins of individual particles.

2. How does the concept of microstates relate to entropy in a large system?

In statistical physics, entropy is a measure of the number of possible microstates that a system can have. In a large system, there are a vast number of microstates that are indistinguishable to an observer, and this leads to an increase in entropy.

3. How does Boltzmann entropy differ from thermodynamic entropy?

Boltzmann entropy is a measure of the number of microstates in a system, while thermodynamic entropy is a macroscopic quantity that describes the disorder or randomness of a system. They are related through the Boltzmann constant and the logarithm of the number of microstates.

4. Can microstates be observed in a large system?

No, microstates cannot be directly observed in a large system due to the enormous number of particles and the limitations of measurement. However, their effects can be seen through macroscopic quantities such as temperature and pressure.

5. How does the concept of microstates contribute to our understanding of entropy in thermodynamics?

The concept of microstates helps to explain the statistical nature of thermodynamic entropy. It provides a theoretical framework for understanding how disorder and randomness arise from the microscopic behavior of particles in a system.

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