USTH Master 1 AMSN

Thermodynamics for physics and chemistry

Exam of March 2023 (2 h - inspired from U.Paris Saclay )

1. (a) The density of states of a perfect 3D atomic gas in a container of volume V for N atoms
has been proven in the course to be

V 3√
dn = (2m) 2 E dE
4π2 ℏ3
where m is the mass of the atoms
The integrated density of states is then

2 V 3 3
n(E) = (2m) 2 E 2
3 4π2 ℏ3

V 3 3
n(E) = 2 3
(2m) 2 E 2
6π ℏ
so we have the correct power law from the given formula and correct homogeneity but
we lack some terms. We need to use

S = kB ln Zm
.
where the total density of states is the total accessible volume in the phase space for a
given energy divided semi-classically by the quantum of action h :

1
Z
Zm (E) = 3N d 3 x1 d 3 p1 . . . d 3 xN d 3 pN δ(E − H(⃗x1 ,⃗p1 , . . . ,⃗xN ,⃗pN ))
h R6N

the Hamiltonian for the perfect gas is

N
⃗pi2
H(⃗x1 ,⃗p1 , . . . ,⃗xN ,⃗pN ) = ∑
i=1 2m

thus
!
N
1 ⃗p 2
Z Z
Zm (E) = d 3 x1 . . . d 3 xN d 3 p1 . . . d 3 pN δ E − ∑ i
(2πℏ)3N | R3N {z } R3N i=1 2m
VN

For the impulsions we have an 3N dimensional sphere with radius

N
p = (∑i=1 ⃗pi2 )1/2

The surface element for impulsion is d p times p3N−1 times the volume element dΩ3N ,
hence

VN
Z Z ∞
Zm (E) = dΩ3N d p p3N−1 δ(E − p2 /2m)
(2πℏ)3N 0

1
 The integral over
dΩ3N
is the surface of a 3N dimensional sphere
3N 3N
2π 2 2π 2
S3N−1 = 3N = 3N
Γ( 2 ) ( 2 − 1)!
The Delta-Function gives after accounting
√ for the two possible signs in the square and
a change of variables since p = 2mE
m h √ √ i
δ(E − p2 /2m) = √ δ( 2mE − p) + δ( 2mE + p)
2mE

3N h √
VN 2π 2 m
Z ∞
3N−1
√ i
Zm (E) = √ d p p δ( 2mE − p) + δ( 2mE + p)
(2πℏ)3N ( 3N
2 − 1)! 2mE | 0 {z }
√ 3N−1
2mE

√
taking the value of the Dirac peak at 2mE the integral disappears and we get
3N
VN (2πmE) 2 3N
Zm (E) =
N!(2πℏ)3N
( 3N
2 )!
2E
with the Stirling approximation and dividing by N! in order to account (in an imperfect
quantum way) for the indiscernibility of particles
√
N! ≈ N N e−N 2πN

3N
VN (2πmE) 2 3N
Zm (E) = √ 3N 3N √
N N e−N ( 2πN2πℏ)3N ( 3N 2 − 2
2 ) e 3πN 2E
collecting terms
 N   3N2
V 4πmE 5N 3
Zm (E) = 2
e2 √
N 3N(2πℏ) 2 6πE
The entropy becomes

     
V 3N 4πmE 5N 3
S = kB ln Zm (E) = kB N ln + kB ln + kB + kB ln √
N 2 3N(2πℏ)2 2 2 6πE

Neglecting the last term we find

"  3/2 !#
5 V mE
S(E,V, N) = NkB + ln
2 N 3πℏ2 N

(b) Show that S can be written as S = 3NkB ln(a∆x∆p/h) , where ∆x is a distance and ∆p
a momentum. Comment.
√
A momentum ∆p is homogeneous to mE. So getting this in the equation we have

2
 "  3  3/2 !#
5 V (2π∆p) 1
S(E,V, N) = NkB + ln
2 N h 3πN

Writing V = ∆x3 we get

" 3  3/2 !#
∆x3

5 2π∆p 1
S(E,V, N) = NkB + ln
2 N h 3πN

and
" 1/2 !#
4π2
 
5 ∆x ∆p
S(E,V, N) = NkB + 3 ln
2 N h 3πN

giving the desired result with

 1/2
51 4π
a=
6N 3N
We find Boltzmann’s statement about cutting phase space in cells of size h.
(c) Calculate the micro-canonical temperature T and pressure P. Derive an expression of
the entropy as a function of the number density n = N/V and the de Broglie wavelength
h
Λ = √2πmk T
.
B
The temperature by definition of the entropy in the second principle of thermodynamics
is

1 ∂S
=
T ∂E
here

∂S 3N 3nR 1
= kB = =
∂E 2E 2E T

2E
T=
3nR
with R the perfect gas constant and nm the number of moles (N = nm NA with NA equal
to Avogadro’s number)
We had already met this result for perfect gases.
The pressure is the derivative of the entropy as

∂S
P=T
∂V
here
1
P = T kB N
V
and
PV = nm RT

with the number density

n = N/V

3
 and the expression of the energy as a function of temperature E = 32 NkB T we get from
"  3/2 !#
5 V mE
S(E,V, N) = NkB + ln
2 N 3πℏ2 N
  !3/2 
5 1 m 23 kB T
S(E,V, N) = nV kB  + ln  
2 n 3πℏ2

(2πℏ)2
with de Broglie’s wavelength kB T = 2πmΛ2
we get
 
(ℏ)2
!3/2 
5 1 Λ2
S(E,V, N) = nV kB  + ln  
2 n ℏ2

and simplifying
"  3/2 !#
5 1 1
S(E,V, N) = nV kB + ln
2 n Λ2
  
5 1
S(E,V, N) = nV kB + ln
2 nΛ3
2. (a) Formulate the extensivity property that must be satisfied by the micro-canonical en-
tropy S(E,V, N) of the gas.
When the size of the system doubles, the entropy should double.
(b) We consider two identical volumes of the same gas separated by a wall. Calculate the
difference between the entropy of this system and the entropy of the system with the
wall removed.

∆S = S(2E, 2V, 2N) − 2S(E,V, N)

Why do we say that this result is paradoxical ?

Let’s go back to our calculation for the entropy above but without the N! factor. We
had
3N
V N (2πmE) 2 3N
Zm (E) =
(2πℏ)3N ( 3N
2 )!
2E

S = kB ln Zm (E)

thus

∆S = S(2E, 2V, 2N) − 2S(E,V, N)

3N
neglegting 2E as above
3N
(2V )2N (4πmE)3N V N (2πmE) 2
= kB ln − 2k B ln
(2πℏ)6N (3N!) (2πℏ)3N ( 3N
2 )!

4
 3N 3N
∆S = kB (2N ln(2V )−6N ln 2πℏ+3N ln 4πmE −ln (3N)!−2(N lnV −3N ln 2πℏ+ ln(2πmE)−ln ( )!))
2 2

3N
∆S = kB (2N ln(2) − ln (3N)! − 2 ln ( )!))
2
taking Stirling’s approximation again
√
N! ≈ N N e−N 2πN

we get

3N 3N 3N 3N 3N
∆S = kB (2N ln(2) − 3N ln 3N − 3N + − 2(− ln − + ))
2 2 2 2 4

3N
∆S = kB (2N ln(2) − 3N ln 3N + 3N ln ))
2

∆S = kB N ln(2)

which is obviously not zero

the differences in factorials is indeed (taking only the two first terms of Stirling’s ap-
proximation)

ln(2N!) − 2 ln N! ≈ 2N ln 2N − 2N − 2N ln N + 2N = N ln 2

(c) Indistinguishability.– Compare S(E,V, N) to the Sackur-Tetrode formula Discuss the

extensivity in the two cases. Verify that if we calculate the entropy of mixing using the
indiscernibility factor N! we have ∆S = 0
With the Sackur-Tetrode formula we have
"  3/2 !#
5 V mE
S(E,V, N) = NkB + ln
2 N 3πℏ2 N

∆S = S(2E, 2V, 2N) − 2S(E,V, N)

and
 
5 2V 2mE 3/2 5 V mE 3/2
∆S = 2NkB ( + ln( ( 2
) )) − 2NkB ( + ln( ) )
2 2N 3πℏ 2N 2 N 3πℏ2 N

∆S = 0

so Sackur-Tetrode gives us extensivity thanks to the factorial.

3. Grand Canonical ensemble (Systems in contact with a particle reservoir)
We consider an ideal gas at thermodynamic equilibrium in a volume V . We fix the tempe-
rature T and chemical potential µ

5
(a) Show that the grand potential may be written as J(T, µ,V ) = V j(T, µ). Discuss the
physical interpretation of the "volumetric density of the grand potential" j.
The grand potential is defined by J(T, µ,V ) = −kB T ln Ξ where Ξ is the grand partition
function itself coming from the canonical partition function as
∞
Ξ= ∑ eβµN QN
N=0

with QN = zN and z = ∑i e−βεi

The grand partition function is extensive, therefore the volume can be factorized out,
giving a quantity with the same homogeneity than pressure.
(b) We consider a dilute gas of monoatomic classical particles, for which we may assume
that the Maxwell-Boltzmann approximation is justified. For this question no supple-
mentary hypothesis (the number of the degrees of freedom, their relativistic or non-
relativistic nature, their dynamics, etc.) will be needed. We introduce z, the single par-
ticle partition fonction. Justify that z α V . Show that the grand canonical partition
fonction is

Ξ = exp(eβµ z)
.
Deduce the average number of particles N̄ and the pressure p . Show that, under this
minimal hypothesis, it is possible to derive the equation of state of the ideal gas, pV =
NkB T
We have
∞
Ξ= ∑ eβµN QN
N=0

∞
Ξ= ∑ eβµN zN
N=0

∞
Ξ= ∑ (eβµ z)N
N=0

From the previous question and the course material we have

 
∂J
p=−
∂V T

then
!
∂ ∑∞ βµN zN
N=0 e
p=−
∂V
T

since the single particle partition function is extensive z = cV with c a constant

!
∞
βµN N N−1
p=− ∑e c NV
N=0 T

also from the course we have the average number of particles

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 ∂J ∂ ∑∞ eβµN (cV )N
N̄ = − = − N=0
∂µ ∂µ
∞
N̄ = − ∑ βNeβµN (cV )N
N=0

∞
N̄ = −β ∑ NeβµN (cV )N
N=0

N̄ = βV p

N̄kB T = V p

and we find the law of perfect gases.

(c) Let us call ε̄ the average energy per particle. Show that the average total energy is

Ē = N̄ ε̄

The average total energy is gotten from the grand partition function as

βµN zN
∑∞
N=0 Ee
Ē =
Ξ

1 ∂J
Ē = −
β ∂β
∞
∂
Ē = − ln ∑ eβµN zN
∂β N=0
N−1
∞ βµN zN + N ∂z
1 µN ln ∑N=0 e ∂β
Ē = − βµN zN
β ∑∞N=0 e

∂zN−1 ∂(∑∞ −βεi )N−1

i=0 e
=
∂β ∂β
∂zN−1 ∞
= −(N − 1)εi ( ∑ e−βεi )N−2
∂β i=0

Assuming that N is close to N − 1 or N − 2 we recognize the expression for the average

energy per particle
(d) Using
 
∂ µ ∂
Var(E) = − + Ē
∂β β ∂µ
show that

Var(E) = N̄ ε¯2

Compare with the variance of the energy in the canonical ensemble.

7
  
∂ µ ∂ 1 ∂J
Var(E) = − + (− )
∂β β ∂µ β ∂β

1 ∂2 J µ ∂2 J
Var(E) = − −
β ∂β2 β2 ∂µ∂β
(e) Show that the entropy is
 
∂ ln z
S = N̄kB 1 − − µβ
∂ ln β
The entropy is given by probabilities pi of each state with

S = −kB ∑ pN,i ln pN,i

N,i

eβ(Nµ−ei )
S = −kB ∑ pN,i ln
N,i Ξ

S = −kB ∑ pN,i (− ln Ξ + βNµ − βei )

N,i

S = −kB (∑ pN,i − ln Ξ + βNµ ∑ pN,i − ∑ pN,i βei )

N,i N,i N,i

Remind that ∑N,i pN,i = 1 and recognize the average energy to get

S = −kB (ln Ξ + βN̄µ − βĒ)

so we have one part of the desired result. on the other hand

∂ ln z ∂ ln z dβ
=
∂ ln β ∂β d ln β
but
dβ
=β
d ln β
so
∂ ln z ∂ ln z
=β
∂ ln β ∂β
and

∂ ln z ∂
= ln e−β−ei
∂β ∂β ∑ i

e−βei
= ∑ −βei
i z

= −βĒ

We will get the energy E from the free energy F and the entropy S by F = E − T S.
Those quantities read
F = J + µN

8
  
∂J
S=−
∂T V,µ

therefore
 
∂J
E = F + T S = J + µN − T
∂T V,µ

∂J dβ
E = J + µN − T
∂β dT

∂J 1
E = J + µN +
∂β kB T

∂J
E = J + µN + β
∂β
(f) Justify that z α V . With z(T,V ) = V /Λ3 express Λ (up to a dimensionlesss factor) for
non relativistic particles (ε = p2 /(2m)) and ultra-relativist particles (ε = pc). Recover
the expression of µ(T, N). Deduce that the entropy coincides with the Sackur-Tétrode
formula.
We have already seen in the course and above for the perfect classical gas that

V
z(T,V ) =
Λ3
with

h
Λ= √
2πmkB T
We should divide by the indiscernibility factor N!

zN βNµ
Ξ=∑ e
N N!

( ΛV3 )N βNµ
Ξ=∑ e
N N!
using the factorial we recognize the Taylor series of an exponential
 
V βµ
Ξ = exp e
Λ3
From this we get the grand potential

J = −kB T ln Ξ

the average particle number

∂J V βµ
N̂ = − = e
∂µ Λ3
from which we get the chemical potential

9
 Λ3
µ = kB T ln N̂
V
and the entropy
 
∂J
S=−
∂T V,µ
 
∂
S=− − kb T ln Ξ
∂T

∂ V βµ
S = kb T e
∂T Λ3
we have three terms depending on T
 
1 βµ 3 µ
S = kBVe 1+ −
Λ3 2 kB T 2

J 5
S= ( − βµ)
T 2
now since
Λ3
µ = kB T ln N̂
V
and

V βµ
N̂ = e = −βJ
Λ3
we recover the Sackur -Tetrode formula

5 Λ3
S = kB N̂( + ln N̂ )
2 V
4. We consider a container of volume V filled with a monatomic ideal gas of indistinguishable
atoms. This gas is in contact with a solid interface that may adsorb (trap) the gas atoms.
We model the interface as an ensemble of A adsorption sites. Each site can adsorb only one
atom, which then has an energy −εo
The system is in equilibrium at a temperature T and we model the adsorbed atoms, i.e.
the adsorbed phase, as a system with a fluctuating number of particles at fixed chemical
potential µ and temperature T . The gas acts as a reservoir.
(a) Derive the grand-canonical partition function ξ for a single adsorption site. Deduce the
grand-canonical partition function Ξ(T, A, µ) for all atoms adsorbed on the surface.
∞
ξ= ∑ eβµN zN
N=0

The canonical partition function is

∞
z = ∑ e−βεi
i=0

We have only two possibilities : zero atom or one atom so that

10
 z = 1 + e+βε0

and
∞
ξ= ∑ eβµN zN
N=0

ξ = 1 + eβµ z
since we have only two possibilities on each site : no atom or 1 atom.
for all atoms on the surface if we account for indiscernability

A A
zN 1 βµN
Ξ= ∑ eβµN = ∑ e (1 + eβε0 )N
N=0 N! N=0 N!
we recognize the expansion of the Taylor series of an exponential of argument r =
eβµ (1 + eβε0 ) so we have

Ξ = er

(b) We will now explore an alternative route. Derive the canonical partition function Z(T, A, N)
of a collection of N adsorbed atoms (Note : the number of adsorbed atoms N is much
smaller than the number of sites A). Recover the results for Ξ(T, A, µ) obtained in the
previous question.
by definition
∞
Z(T, A, N) = ∑ e−βε i

i=0

so since the energy for not absorbed atoms is zero giving a 1 in the exponential ; for N
atoms the energy is −Nε0 ; thus

Z(T, A, N) = (1 + eNβε0 )

A
1 βµN
Ξ= ∑ e Z
N=0 N!

A
1 βµN
Ξ= ∑ e (1 + eNβε0 ).
N=0 N!
We have once again the expansion of the Taylor series of an exponential of argument

r = eβµ (1 + eβε0 )

by making the approximation that N is much smaller than A and recognizing the ex-
pansion of a power series in (1 + x)N ≈ 1 + Nx

11
(c) Calculate the average number of adsorbed atoms N as a function of εo , µ, A, T . From
this, derive the occupation probability θ = N/A of an adsorption site.
by the course

∂ ∂
N̄ = kb T ln Ξ = kb T eβµ (1 + eβε0 )
∂µ ∂µ

N̄ = kB T βeβµ (1 + eβε0 )

N̄ = eβµ (1 + eβε0 )
(d) The chemical potential µ is fixed by the ideal gas. This may be used to deduce an
expression for the site occupation probability θ as a function of the gas pressure P and
temperature T (note that the number of atoms N is much smaller than the number of
gas atoms).
we had found for the perfect gas with NA gas atoms

Λ3
µ = kB T ln NA
V
as a function of pressure

PΛ3
µ = kB T ln NA
NA kB T
injecting this in the above equation for N̄

N̄ = eβµ (1 + eβε0 )
we get
3
ln NA NPΛk (1+eβε0 )
N̄ = e A BT

N̄ = PΛ3 kB T (1 + eβε0 )
(e) We define a parameter
  32  
2πmkB T ε0
P0 (T ) = kB T exp −
h2 kB T
Express θ as a function of P and P0 (T )
 
P ε0
N̄ = exp (1 + eβε0 )
P0 kB T
 
P ε0
N̄ = (exp + 1)
P0 kB T
P
θ = N̄/A = (1 + e−βε0 )
AP0
(f) How does the curve θ(P) behave for different temperatures ?
When the temperature increases β goes to zero so that the exponential goes to 1 ,but P0
goes to infinity so that θ goes to zero.
5 5
For low temperatures β goes to infinity, P0 behaves as T 2 , so that θ behaves as T − 2
and increases.

12