Download HW7 solutions - Itai Cohen

Cornell University, Physics Department
PHYS-3341 Statistical Physics
Fall 2014
Prof. Itai Cohen
Solutions to Problem Set 7
David C. Tsang, Woosong Choi
7.1
Dilute Gas in an Enclosure
(a) By symmetry we see
v̄z 0 .
By the maxwell-boltzmann distribution we have
R∞
r
R∞
2
2
|v |e−mvz /2kT dvz
2 0 vz e−mvz /2kT dvz
2kT /m
2kT
−∞ z
= R ∞ −mv2 /2kT
=p
|vz | = R ∞ −mv2 /2kT
=
z
z
πm
e
dvz
dvz
2 0 e
2πkT /m
−∞
R ∞ 2 −mv2 /2kT
2kT 3/2
1√
z
π
e
dv
2
v
kT
z
z
2
m
0
v 2 = R ∞ −mv2 /2kT
= √
=
1/2
z
m
2 0 e
dvz
π 2kT
m
(b) DefineR φ(vz ) as the # of molecules escaping per unit time per unit z-velocity. That is
∞
Φ = 0 φ(vz )dvz . Now φ(vz ) = Avz n(vz ) where n(vz ) is the number of molecules with
z-velocity vz per unit volume per unit z-velocity.
2
N
e−mvz /2kT
n(vz ) = R ∞ −mv2 /2kT dv
z
z
V −∞ e
r
R∞
2
AN 0 e−mvz /2kT dvz
AN
kT
AN
R∞
⇒Φ=
=
|v
|
=
z
2
2V 0 e−mvz /2kT dvz
2V
V
2πm
(c) Now assume we cover the hole with a plate and assume elastic collisions on that plate.
Then the froce on that plate per unit time is
F = Φ∆p = 2mΦvz,esc
where ∆p is the average change in momentum due to an elastic collision.
F
2m
⇒P =
=
Φvz,esc
A
A
(d)
vz,esc
AP
PV
=
=
2Φm
Nm
r
πm
kT
=
2kT
m
r
πm
=
2kT
r
πkT
π
= |vz |
2m
2
For an ideal gas with maxwell-boltzmann distribution we have vz,esc > |vz |. Another
way of seeing this is to recognize that the escaping molecules are “pushed” preferentially
on one side, giving them a greater average speed.
2
1
7.2
Two-Dimensional Gas
Reif §7.7: Monoatomic molecules adsorbed on a surface are free to move on this
surface and can be traeted as a classical ideal two-dimensional gas. At absolute
temperature T , what is the heat capacity per mole of molecules thus adsorbed
on a surface of fixed size?
In an ideal gas the particles are non-interacting , and there are no external potentials,
therefore V = 0. Hence we have the energy E = K the kinetic energy. The equipartition
theorem gives us
1
kT = N kT
Ē = K̄ = 2N
2
where 2 comes from the number of dimensions and N is the number of particles.
Thus we have
∂ Ē
C
= N k = νR ⇒
=R
C=
∂T
n
2
7.3
More Oscillators
Reif §7.10: A system consists of N very weakly interacting particles at a temperature T sufficiently high so that classical statistical mechanics is applicable.
Each particle has mass m and is free to perform one-dimensional oscillations
about its equilibrium position. Calculate the heat capacity of this system of
particles at this temperature in each of the following cases:
(a) The force effective in resotring each particle to its equilibrium position is
proportional to its displacement x from this position.
(b) The restoring force is proportional to x3 .
(a) We have an energy of the form
E=
N
X
p2
1
+ ko x2i
m 2
i
i
Thus there are two seperate quadratic degrees of freedom per particle, hence we have
by equipartition
∂E
Ē = N kT
⇒
C=
= Nk
∂T
2
(b) For this system we have the energy
E=
N
X
p2
1
+ ko x4i
m 4
i
i
The classical partition function is then
Z ∞
N Z
1 Y ∞ −βp2i /2m
4
e−βko xi /4 dxi
e
dpi
Z = N
h0 i −∞
−∞
N
N/2 Z ∞
2πm
−ko y 4 /4
−1/4
=
e
dy
(where y = β 1/4 x)
β
2
ho β
−∞
N/2 Z
N
2πm
−ko y 4 /4
=
e
dy β −3N/4
2
ho
Since the first two terms in the product do not depend on β we must have
Ē = −
∂ ln Z
3
= N kT
∂β
4
⇒
3
C = Nk
4
2
7.4
Specific heat of graphite
Reif §7.11: Assume the following highly simplified model for calculating the
specific heat of graphite, which has a highly anisotropic crystalline layer structure. Each carbon atom in this structure can be regarded as performing simple
harmonic oscillations in three dimensions. The restoring forces in directions parallel to a layer are very large; hence the natural frequencies of oscillations in the
x and y directions lying within the plane of a layer are both equal to a value
ωk which is so large that ~ωk 300k. On the other hand, the restoring force
perpendicular to a layer is quite small; hence the frequency of oscillation ω⊥ of
an atom in the z direction perpendicular to a layer is so small that ~ω⊥ 300k.
On the basis of this model, what is the molar specific heat (at constant volume)
of graphite at 300◦ K?
For harmonic oscillations we have energy E = ~ω(n + 12 ). Adding the terms for all directions
we get
1
E = ~ω⊥ (n1 + n2 + 1) + ~ωk (nk + )
2
But since the energy terms are indepedent in each direction, we can separate direction
dependence of the partition function
Z = Z⊥ Zk
3
where both Z⊥ and Zk are partition functions for harmonic oscillator(s). Now using the
partition function to find the specific heat is simple
cV
∂ 2 ln Z
∂ Ē
=−
∂T
∂T ∂β
2
∂ ln Z⊥ ∂ 2 ln Zk
= −
−
∂T ∂β
∂T ∂β
¯
∂ E⊥ ∂ Ēk
=
+
∂T
∂T
=
At 300◦ K, since ~ωk 300k and ~ω⊥ 300k following the argument at p.253 of Reif we
get
E¯⊥ = Ē1 + Ē2 ' ~ω⊥ (1 + 2e−β~ω⊥ )
Ēk ' kT
Taking the derivative of these with respect to T ,
dβ ∂ E¯⊥
1
∂ E¯⊥
=
= − 2 ~ω⊥ (−2~ω⊥ e−β~ω⊥ )
∂T
dT ∂β
kT
2 −β~ω⊥
= 2k(β~ω⊥ ) e
)
cV = k + ~ω⊥ (−2~ω⊥ )
∂ Ēk
= k
∂T
Combining this we get
cV = k 2(β~ω⊥ )2 e−β~ω⊥ + 1
Introducing the definition of Einstein temperature θE =
in a mole(since above result was only for one atom),
~ω⊥
k
and summing over all the atoms
" #
2
θE
−θE /T
cV = R 2
e
+1
T
2
4
7.5
Ferro Fluids
Reif §7.14: Consider an assembly of N0 weakly interacting magnetic atoms per
unit volume at a temperature T and describe the situation classically. Then
each magnetic moment µ can make any arbitrary angle θ with respect to a given
direction (call it the z direction). In the absence of a magnetic field, the probability that this angle lies between θ and θ + dθ is simply proportional to the solid
angle 2πsinθdθ enclosed in this range. In the presence of a magnetic field H in
the z direction, this probability must further be proportional to the Boltzmann
factor e−βE , where E is the magnetic energy of the moment µ making this angle
θ with the z axis. Use this result to calculate the classical expression for the
mean magnetic moment M̄z of these N0 atoms.
Following the argument given in the problem we have
P (θ)dθ ∝ 2πsinθdθeβµHcosθ
for one magnetic moment. Since they are only weakly interacting, we can consider the atoms
individually and sum up in the end. Since we have the proportionality of P (θ) we can find
the partition function as follows:
Z
π
dθ2πsinθdθe
Z=
βµHcosθ
Z
1
βµHcosθ
d(cosθ)e
= 2π
−1
0
eβµHcosθ
= 2π
βµH
1
=−
−1
4π
sinh(βµH)
βµH
To find the average magnetization of the magnetic moment in z-direction,
µ¯z =
∂ ln Z
1
= µcoth(βµH) −
∂(βH)
βH
thus for N0 atoms
M̄z = N0 µ coth(βµH) −
1
βµH
2
5
7.6
Mean Values
Reif §7.19: A gas of molecules, each of mass m, is in thermal equilibrium at
the aboslute temperature T . Denote the velocity of a molecule by ~v , its three
cartesian components by vx , vy and vz , and its speed by v. What are the following
mean values:
(a) vx
(b) vx2
(c) v 2 vx
(d) vx3 vy
(e) (vx + bvy )2 where b is a constant
(f ) vx2 vy2
(a) vx = 0 since vx is an odd function.
(b) vx2 =
kT
m
by the equipartition thereom.
(c) v 2 vx = 0 since v 2 vx is an odd function
(d) vx3 vy = 0 since vx3 vy is odd in both vx and vy .
(e) (vx + bvy )2 = vx2 + 2bvx vy + bvy2 =
(f ) vx2 vy2 = vx2 vy2 =
kT 2
m
kT
m
+0+
bkT
m
= (1 + b) kT
.
m
since the distributions of vx and vy can be considered independent.
2
7.7
Isotope Separation
Reif §7.26: A vessel is closed off by a porous partition through which gases can
pass by effusion and then be pumped off to some collecting chamber. The vessel
itself is filled with a dilute gas consisting of two types of molecules which differ because they contain two different atomic isotopes and have correspondingly
masses m1 and m2 . The concentrations of these molecules are c1 and c2 , respectively, and are matintained constant inside the vessel by constantly replenishing
the cupply of gas in it by a steady slow flow of fresh gas through the vessel.
(a) Let c01 and c02 denote the concentrations of the two types of molecules in the
collecting chamber. What is the ratio c02 /c01 ?
6
(b) By using the gas U F6 , one can attempt to separate U 235 from U 238 , the
first of these isotopes being the one useful in initiation of nuclear-fission
reactions. The molecules in the vessel are then U 238 F619 and U 235 F619 . (The
concentrations of these molecules, corresponding to the natural abundance
of the two uranium isotopes, are c238 = 99.3 percent and c235 = 0.7 percent.)
Calculate the corresponding ratio c0235 /c0238 of the molecules collected after
effusion in terms of their original concentration ratio c235 /c238 .
(a) The ratio of concentration of the molecules in the collecting chamber will be the ratio
of the molecules effused through the porous partition. From (7.11.13) average number
of particles that will hit, i.e. pass through, the holes per unit area is
nkT
p̄
=√
Φ0 = √
2πmkT
2πmkT
where p̄ is the mean pressure of the molecules, thus can be substituted to nkT using
the equation of state. T is the same for the two types of molecules thus Φ0 ∝ √nm is
the relation relevant to the ratio.
r
c1 m2
c01
=
c02
c2 m1
(b) For U F6 , using the result of (a) we get
c0235
c235
=
0
c238
c238
r
m238
c235
=
m235
c238
r
352
c235
= 1.004 ×
349
c238
Although the change in concentration ratio is small, by repeatedly using effusion we
can increase the concentration of U 235 in the mixture.
2
7