UNIVERSITETET I OSLO
Det matematisk-naturvitenskapelige fakultet
Final exam in
FYS3130/FYS4130 Statistial mehanis
Date of exam:
Tuesday, 14. June, 2005, Time: 9:00 - 12:00
The problem set ontains:
2 pages
Two sheets (size A4) of handwritten notes, approved
mathematial formulae (Rottmann) and approved alulator.
You are allowed to bring:
Please make sure that the problem set is omplete before you start solving the problems!
Problem 1. Virial expansion
The Equation of State of a real gas an be written in a form
Pv
B B B
= 1 + 2 + 23 + 34 + : : :
RT
v
v
v
(1)
alled the virial expansion, whih is valid for dilute gases with molar density 1=v =
n=V . Find rst virial oeients B2 , B3 , B4 for the van der Waals gas equation of
state
P+
Hint:
Assume that b=v 1.
a
(v
v2
b) = RT
(2)
Solution.
Van der Waals equation fo state an be written as
v
Pv
=
RT v b
a
RT v
Expand the term v=(v b) into Taylor series with respet to x = b=v around x0 = 0:
v
1
1
=
=
v b 1 (b=v ) 1 x
1
1
= f (0) + f 0 (0)x + f 00 (0)x2 + f 000 (0)x3 + =
2!
3!
1 2 1 3
= 1 + x + 2x + 6x + =
2
6
2
3
b b
b
= 1+ + 2 + 3 +
v v v
f (x) = f (b=v ) =
=
(3)
(4)
(5)
(6)
Collet together terms in r.h.s. of van der Waals equation of state with respet to
1=v :
Pv
a 1 b2 b3
=1+ b
+ + +
RT
RT v v 2 v 3
and ompare it with (1) one an obtain:
a
; B3 = b2 ; B4 = b3
B2 = b
RT
Problem 2. Flutuations in Grand Canonial Ensemble
a) Using the partition funtion for Grand Canonial Ensemble, prove that dispersion of number of partiles N an be expressed as
N
N) =
(N ) = (N
2
2
where = 1=T , = =T and N is the average number of partiles.
b) Let Ni be the average oupany of a single orbital of a boson system. Show
that
(Ni )2 = Ni (1 + Ni ):
) Show that for a single orbital of a fermion system
(Ni )2 = Ni (1 Ni ):
d) Show that for a Maxwell-Boltzmann gas (Ni )2 = Ni :
Solution.
(a)
as
For Grand anonial ensemble (T
Q=
X
i;N
e
V
) partition funtion Q an be written
(Ei N )
=
X
i;N
eN
Ei
where = 1=kT , = .
Then average number of partiles in the system is
1 X N
Ne
N=
Q i;N
1 X 2 N
N2 =
N e
Q i;N
Ei
1 Q
=
Q Ei
1 2Q
=
Q 2
Therefore,
(N )2 = (N
N )2 = N 2
1 Q
=
Q 2
2
2
N =
1 Q
Q2 2
=
Q
2Q
2 Q2
Q 2
N
=
q.e.d.
(b,,d) Let us onsider the system from the previous problem (1.a) to be the
single-partile state onerned and suppose that this state has energy . For boson/fermion/MB system with single orbital
Ni =
,
1
exp ( ) + where = 1 for bosons, = +1 for fermions and = 0 for MB lassial system.
Sine the system is mirosopi, Ni should be replaed by Ni . So, using the folmula
from (1a) we will get
N
exp ( )
=
(exp ( ) + )
(exp ( ) + )2
1
1
1 = N (1 N )
=
(exp ( ) + )
exp ( ) + =
1
2
( 1)e
=
Problem 3. Two-dimensional photon gas
Consider two-dimensional photon gas with degeneray g = 2 at temperature T.
Using the formula for the mean oupation number of the photon energy state ,
< n >=
1
exp (=kT )
1
where = p = h , p is momentum and is frequeny of the photon,
a) nd the number of photons dn per unit volume in the energy interval
(; + d) and derive two-dimensional Plank's formula for spetral radiation
energy density
(; T ) = A
2
eh=kT
1
as a funtion of frequeny, , if (; T )d = dn .
Dene the oeient A;
b) Consider two limiting ases: h kT and h kT , and get two-dimensional
analogues of Wien and Rayleigh-Jeans radiation laws, respetively.
) Derive the Stefan-Boltzmann law for spatial energy density u(T ) = U=V ,
where U is the total radiation energy.
Hint:
Z 1
0
x2 dx
= (3) (3); where
ex 1
(3) = 2 and (3) 1:202:
d) Find
heat apaity at onstant two-dimensional volume, Cv ,
entropy, S ,
the Helmholtz free energy, F,
pressure of the photon gas, P,
the Gibbs energy, .
e) Show that for two-dimensional gas
1
P V = U:
2
Solution.
Photon gas at temperature T with energy = h = p, with degeneray g = 2 in
the avity of volume V = V2
< n > =
d
2
=
=
dN2 =
dU2 =
(; T ) =
1
exp (=kT ) 1
2V2 gpdp 2V2 gd 2V2 gh2 d
=
=
=
h2
h2 2
h2 2
2V2 gd
2
1
2gd
V2 2
exp (h=kT ) 1
1
2gh 2 d
V2
2
exp (h=kT ) 1
1 dU2 2gh
2
= 2
V2 d
exp (h=kT ) 1
2gh
2
(; T ) = 2
exp (h=kT )
!
1
4h 2 e h=kT =2 ; if h kT ,
4kT =2 ;
if h kT ,
Z
U
2 d
4h 1
u= 2 = 2
=
V2
0 exp (h=kT ) 1
Z
4h kT 3 1 x2 dx
=
= 2
h
exp (x) 1
0
8k3 (3) 3 9:6k3 3
T ' 2 2 T = 1 T 3
=
2 h 2
h
The total energy emitted from the volume V2 = V is equal:
U = 1 V T3
U
CV =
= 31 V T 2
T V
Z
3
CV
dT + f (V ) = 1 V T 2 + f (V )
S = V
T
2
f (V ) = 0; beause S (T = 0) = 0
3 3
T3
F = U T S = 1 (1
)T V = 1 V
2
2
3
T
1U
F
= 1 =
P =
V T
2
2V
T3
T3
= U TS + PV = 1 V + 1 V = 0
2
2