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Statistical Mechanics - Physics at Oregon State University

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3.7. IDEAL GAS IN FIRST APPROXIMATION. 61<br />

the more rigorous tre<strong>at</strong>ment which will follow will indeed justify it. Before this<br />

more rigorous theory had been developed, however, experiments had already<br />

shown th<strong>at</strong> one needed this factor! Using this formula we obtain the results for<br />

an ideal gas:<br />

Z(T, V, N) = 1<br />

N! (V nQ(T )) N<br />

U = kBT 2<br />

<br />

∂ log(Z)<br />

∂T<br />

V,N<br />

(3.95)<br />

= 3<br />

2 NkBT (3.96)<br />

F = −kBT log(Z) = kBT log(N!) − NkBT log(nQV ) (3.97)<br />

Using Stirling’s formula and only keeping terms in F which are <strong>at</strong> least proportional<br />

to N, we obtain<br />

<br />

<br />

n<br />

F = NkBT log( ) − 1<br />

(3.98)<br />

nQ(T )<br />

<br />

∂F<br />

p = −<br />

=<br />

∂V T,N<br />

NkBT<br />

V<br />

<br />

∂F<br />

S = −<br />

= NkB<br />

∂T V,N<br />

<br />

log( nQ(T<br />

<br />

) 5<br />

+<br />

n 2<br />

(3.99)<br />

(3.100)<br />

This last formula is called the Sackur-Tetrode rel<strong>at</strong>ion.<br />

If one ignores the N! in the definition of the partition function for N particles,<br />

several problems appear. First of all, the argument of the logarithm in all<br />

formulas would be nQ(T )V , and quantities like F and S would not be extensive,<br />

i.e. they would not be proportional to N (but to N log N)! Also, the second<br />

term in S would be 3<br />

2NkB. S can be measured precise enough to show th<strong>at</strong> we<br />

really need 5<br />

2<br />

Hence these experiments showed the need of a factor N! (or <strong>at</strong><br />

least for an extra N log N − N).<br />

The formula for the entropy is valid for an ideal gas, or for a gas in the<br />

classical limit. Nevertheless, it contains factors (through nQ(T )). Th<strong>at</strong> is<br />

really surprising. In many cases the classical limit can be obtained by assuming<br />

th<strong>at</strong> is small. Here th<strong>at</strong> does not work. So why does show up in a classical<br />

formula? Interesting indeed.<br />

The chemical potential follows from the Helmholtz free energy:<br />

n<br />

µ = kBT log( ) (3.101)<br />

nQ(T )<br />

A calcul<strong>at</strong>ion of the grand partition function is also straightforward. We<br />

find<br />

Z(T, µ, V ) = <br />

ˆN<br />

z ˆ N 1<br />

ˆN! (Z1) ˆ N = e zZ1 (3.102)

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