Orders of Magnitude: Practially How big is infinite in Stat. Mech. Thermodynamic Limit? In this MIT OCW lecture notes doc
"The  canonical  partition  function  for  a  finite  collection  of  particles  is  always  an 
analytical  function.  Hence  phase  transitions,  and  their  associated  non–analyticities,  are 
only  obtained  for  infinitely  many  particles,  i.e.  in  the 
thermodynamic limit, $N \rightarrow \infty$
. The 
study  of  phase  transitions  is  thus  related  to  finding  the  origin  of  various  singularities  in 
the  free  energy  and  characterizing  them."
But there aren't infinite particles! 


*

*Is this another time where I should interpret infinity as very large compared to the system size? 

*How many particles should we practically consider infinite? A mole?

 A: The short answer is that it depends. It depends on the specifical problem you are considering and it depends on the temperature and the other parameters. 
In general, in equilibrium statistical mechanics, one is dealing with integrals of the type
$$
\langle O \rangle = \frac{1}{Z} \int dx \ O(x) \exp( - N F(x) ) \ ,
$$
where
$$
Z = \int dx \exp( - N F(x) ) \ .
$$
If $N \to \infty$, one can use Laplace's method (or the saddle point method) to get
$$
\lim_{N\to\infty}\langle O \rangle = O(x^\ast)
$$
where $F(x^\ast) = \min_{x} F(x)$.
The question can be rephrased in ``how important the neglected term are?''. Applying the Laplace's method (or saddle point in case of complex variables) to the first two equation we obtain
$$
Z = \int dx \exp( - N F(x) ) \approx \int dx \exp( - N F(x^\ast) - \frac{N}{2} F''(x^\ast)(x-x^\ast)^2 )  \ ,
$$
and therefore
$$
\langle O \rangle \approx \frac{ \int dx \ O(x) \exp( - N F(x^\ast) - \frac{N}{2} F''(x^\ast)(x-x^\ast)^2 ) }{
\int dx \ \exp( - N F(x^\ast) - \frac{N}{2} F''(x^\ast)(x-x^\ast)^2 ) 
} = \\ = \sqrt{\frac{N F''(x^\ast)}{2 \pi}} \int dx \ O(x) \exp( - \frac{N}{2} F''(x^\ast)(x-x^\ast)^2 )  \ .
$$
This last integral is the average over a Gaussian with mean $x^\ast$ and variance $1/(N F''(x^\ast))$. In the limit of $N \to \infty$, one gets $O(x^\ast)$. One the other hand, one can get the fluctuations around this limit. Using $y = x - x^\ast$ and expanding around $y = 0$
$$
\langle O \rangle \approx \sqrt{\frac{N F''(x^\ast)}{2 \pi}} \int dx \ \left( O(x^\ast) + O'(x^\ast) y + O''(x^\ast) \frac{y^2}{2} + ... \right) \exp( - \frac{N}{2} F''(x^\ast) y^2 ) = \\
= O(x^\ast) + \frac{1}{N} \frac{O''(x^\ast)}{F''(x^\ast)} + ... \ .
$$
Therefore the error is of order $1/N$. How important this term is depends on $O(x^\ast)$, $O''(x^\ast)$ and $F''(x^\ast)$.
