92 Microcanonical ensemble
3.5.1 The Gibbs Paradox
According to eqn. (3.5.21), the entropy of an ideal gas is
S(N,V,E) =kln Ω(N,V,E)=Nkln[
V
h^3(
4 πmE
3 N) 3 / 2 ]
+
3
2
Nk−klnN!, (3.5.29)or, using eqn. (3.5.24), we obtainSas a function ofN,V, andT:
S(N,V,T) =Nkln[
V
h^3
(2πmkT)^3 /^2]
+
3
2
Nk−klnN!, (3.5.30)Recall, however, that the 1/N! factor in eqn. (3.5.21) was addeda posteriorito correct
for overcounting the number of microstates due to the identical nature of the gas
particles. If this factor is not included, then the entropy, known as theclassical entropy,
becomes
S(cl)(N,V,T) =Nkln[
V
h^3(2πmkT)^3 /^2]
+
3
2
Nk. (3.5.31)Let us now work through a thought experiment that reveals the importance of the
1 /N! correction. Consider an ideal gas ofNindistinguishable particles in a container
with a volumeVand uniform temperatureT. An impermeable partition separates the
container into two sections with volumesV 1 andV 2 , respectively, such thatV 1 +V 2 =V.
There are areN 1 particles in the volumeV 1 , andN 2 particles in the volumeV 2 , with
N=N 1 +N 2 It is assumed that the number densityρ=N/V is the same throughout
the system so thatN 1 /V 1 =N 2 /V 2. If the partition is now removed, will the total
entropy increase or remain the same? Since the particles are identical, exchanges of
particles before and after the partition is removed will yield identicalmicrostates.
Therefore, the entropy should remain the same. We will now analyzethis thought
experiment more carefully using eqns. (3.5.29) and (3.5.31) above.
Using eqn. (3.5.31), the classical entropy expressions for each ofthe two sections
of the container are (apart from additive constants) are
S
(cl)
1 ∼N^1 klnV^1 +3
2
N 1 kS
(cl)
2 ∼N^2 klnV^2 +3
2
N 2 k, (3.5.32)and, since entropy is additive, the total entropy isS(cl)=S 1 (cl)+S(cl) 2. After the
partition is removed, the total classical entropy is
S(cl)∼(N 1 +N 2 )kln(V 1 +V 2 ) +3
2
(N 1 +N 2 )k. (3.5.33)Therefore, the difference ∆S(cl)is
∆S(cl)= (N 1 +N 2 )kln(V 1 +V 2 )−N 1 klnV 1 −N 2 klnV 2
=N 1 kln(V/V 1 ) +N 2 kln(V/V 2 )> 0 , (3.5.34)which contradicts the anticipated result that ∆S= 0. Without the 1/N! correction
factor, a paradoxical result is obtained, which is known as theGibbs paradox.