- #1

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- Homework Statement
- Derive the Sackur-Tetrode equation.

- Relevant Equations
- ...

Okay so I am learning Statistical mechanics from an Indian book "Thermal Physics,kinetic theory and statistical mechanics by Garg, Bansal and Ghosh".

I have derived the MB distribution function, and have evaluated the parameters α and β. With its help I derived the expression for entropy:

##S=k_B N ln(Z) + \frac{U}{T}##

Then we assumed that particles of an ideal monoatomic gas are distinguishable and derived the expression for its partition function:

##Z= V^N (\frac{mk_B T}{2 \pi \hbar^2})^{3N/2}##

So entropy becomes:

## S=Nk_B ( ln(V T^{3/2} )+ ln( \frac{mk_B e}{2\pi \hbar^2} ) )##

which tends to negative infinity as T tends to 0.

Also it gives rise to the Gibb's paradox.

To resolve it, Sackur suggested that we're over counting the microstates. So divide the thermodynamic probability W by N!.

Now, ##S=k_B ln(W)##

So, in the expression for entropy, there is another term: ##-k_B ln(N!)##

Now, ##S=k_B (N ln(Z) - ln(N!)) + \frac{U}{T}##

At this point the book says: "With this we can conclude that Boltzmann counting influences the partition function in the same way as it does thermodynamic probability. So ## Z^C = \frac{Z}{N!}## " ( Zc is Z corrected.)

How does it conclude that!?

Before: ##S=k_B N ln(Z) + \frac{U}{T}##

After: ##S=k_B (N ln(Z) - ln(N!)) + \frac{U}{T}##

For new Z to be Z/N! , there should have been an N multiplied with the ln(N!) and I could take it common.

I have derived the MB distribution function, and have evaluated the parameters α and β. With its help I derived the expression for entropy:

##S=k_B N ln(Z) + \frac{U}{T}##

Then we assumed that particles of an ideal monoatomic gas are distinguishable and derived the expression for its partition function:

##Z= V^N (\frac{mk_B T}{2 \pi \hbar^2})^{3N/2}##

So entropy becomes:

## S=Nk_B ( ln(V T^{3/2} )+ ln( \frac{mk_B e}{2\pi \hbar^2} ) )##

which tends to negative infinity as T tends to 0.

Also it gives rise to the Gibb's paradox.

To resolve it, Sackur suggested that we're over counting the microstates. So divide the thermodynamic probability W by N!.

Now, ##S=k_B ln(W)##

So, in the expression for entropy, there is another term: ##-k_B ln(N!)##

Now, ##S=k_B (N ln(Z) - ln(N!)) + \frac{U}{T}##

At this point the book says: "With this we can conclude that Boltzmann counting influences the partition function in the same way as it does thermodynamic probability. So ## Z^C = \frac{Z}{N!}## " ( Zc is Z corrected.)

How does it conclude that!?

Before: ##S=k_B N ln(Z) + \frac{U}{T}##

After: ##S=k_B (N ln(Z) - ln(N!)) + \frac{U}{T}##

For new Z to be Z/N! , there should have been an N multiplied with the ln(N!) and I could take it common.