home sol 5
TRANSCRIPT
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8/8/2019 Home Sol 5
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Problem Set 5
Statistical MechanicsPhysics 341
Tim Pennycook
Problem 1.
To find the thermodynamic properties NA, NB , NAB, one can use the partition function, which iseasily constructed, since we are given the single particle partition functions for each type of particle:
Q =fNAA f
NBB f
NABAB
NA!NB!NAB !,
where the factorials come from the assumption that these particles are indistinguishable. Using theSterling approximation, the free energy,
F = kT ln Q
= kT(NA
ln fAN
Aln N
A+ N
A+ N
Bln f
BN
Bln N
B+ N
B+ N
ABln f
ABN
ABln N
AB+ N
AB)
= kT(NA ln fA/NA + NA + NB ln fB/NB + NB + NAB ln fAB/NAB + NAB)
Now to find the minimum of the free energy,
(ln Q) = (ln fA/NA 1)NA + (ln fB/NB 1)NB + (ln fAB/NAB 1)NAB .
Setting this equal to zero, and using the fact that NAB = NA NB ,
0 = (ln fA/NA 1)NA + (ln fB/NB 1)NB (ln fAB/NAB 1)(NA + NB)
= (ln fA/NA)NA + (ln fB/NB)NB (ln fAB/NAB)(NA + NB)
= lnfANAB
NAfAB
NA + lnfBNAB
NBfAB
NB .
Finally, using NA = NB
ln
fAfBNABNANBfAB
= 0 =
NABNANB
=fAB
fAfB,
or, multiplying by one, 1VnABnAnB
= fABfAfB , so
nABnAnB
= VfAB
fAfB.
Problem 2.
Q1 =1
h3
PS
epcd3pd3r
=4V
h3
0
epcp2dp
=8V
(ch)3,
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8/8/2019 Home Sol 5
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so for N indistinguishable particles,
QN(V, T) =1
N!
8V
kT
hc
3N,
so using the free energy F = kT ln Q,
P =
F
V
T,N
=kN T
V, U =
ln Q
=
3N
= 3NkT,
confirming P V =1
3U, and U/N = 3kT . The specific heats,
CV =
U
T
N,V
= 3kN, C P =
U + P V
T
N,P
= 4kN,
so we see that indeed = CP/CV = 4/3.
Problem 3.
The multiplicity ofj, gj , is obviously equivalent to the number of ways of putting j balls in s boxes,which we know to be
gj =(j + s 1)!
j!(s 1)!,
the desired expression. The one particle partition function,
Q1 = es/2
j=0
ej = e1
2sw1 ew
s,
so the system of N such oscillators will have the partition function
Q1 = e
1
2sNw
1 ew
sN
,
which, as expected, is the same as for a system of sN one-dimensional oscillators. The partitionfunction uniquely defines the thermodynamic properties, so those for the N s-dimensional oscillatorswill be identical with those of sN one-dimensional oscillators:
= sN12
+ kT ln(1 e) , S = sNk12
coth12
ln 2 sinh12
P = 0, U =
1
2sN coth
1
2
,
and
C = sNk
1
2
2cosech2
1
2
.
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