PHYS 5163 Homework Assignment 2 solved

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Problem 1:
This problem serves as a refresher of thermodynamics: it utilizes the concepts of thermal
and mechanical equilibrium as well as expressions for the temperature and the pressure in
terms of the entropy.

Some substance has the entropy function
S = λV 1/2
(NE)
1/4
, (1)
where N is in moles, λ is a constant with appropriate units, and E and V denote energy
and volume, respectively.

A cylinder is separated by a partition into two halves, each of
volume 1 m3
.

One mole of the substance with an energy of 200 J is placed in the left half,
while two moles of the substance with an energy of 400 J is placed in the right half.

(a) Assuming that the partition is fixed but conducts heat, what will be the distribution
of the energy between the left and right halves at equilibrium?

(b) Does your result from part (a) make sense? If so, provide an intuitive explanation of
your result. If not, explain why your result does not make sense.

(c) Assuming that the partition moves freely and also conducts heat, what will be the
volumes and energies of the samples in both sides at equilibrium?

(d) Does your result from part (c) make sense? If so, provide an intuitive explanation of
your result. If not, explain why your result does not make sense.

Problem 2:
Suppose that the Hamiltonian H contains a term that is linear in the parameter λ,
H(~p, ~q) = H0(~p, ~q) + λh(~p, ~q). (2)

There are many physical situations that can be described by this type of Hamiltonian. For
example, if we consider N particles in an external gravitational field, then the parameter
λ can be identified as the gravitational acceleration g and the function h would take the
form
h = m
X
N
i=1
zi
. (3)

For the purpose of this problem, we will leave λ and h unspecified.
Show that the microcanonical average of the observable h(~p, ~q) is given by
hhi = −T
dS
dλ . (4)

The following will be helpful:
d
dλ log f =
df

f
and d
dλθ(E − H0 − λh) = −hδ(E − H0 − λh). (5)

Problem 3:
Mathematically, there are quite a few similarity between the system considered here and
the ideal gas system considered in class.

The position of a two-dimensional diatomic molecule with fixed distance between the two
atoms can be described by the three coordinates (x, y, θ), where x and y are the Cartesian
coordinates of the center-of-mass of the molecule and θ gives the orientation of the molecular axis with respect to the x-axis. The conjugate momenta are denoted by (px, py, pθ).

Physically, px and py are the linear center-of-mass momenta and pθ is the angular momentum of the molecule about its center-of-mass. The energy  of the molecule is
 =
p
2
x + p
2
y
2m
+
p
2
θ
2I
, (6)
where I is the moment of inertia about the center-of-mass.

(a) For a system of N non-interacting two-dimensional diatomic molecules confined to a
two-dimensional area A, use the microcanonical ensemble to calculate the entropy S(N, E, A).

(b) Using the entropy, derive the equations of state of the system, i.e., the equations that
give the pressure and the energy per particle as functions of the temperature and density.

(c) Calculate the constant volume (actually, it is better to say constant area) specific heat
per particle, defined through
CV =
1
N

∂E(N, T, A)
∂T !
N,A
. (7)

Note for part (b): In two dimensions, “pressure” is “force per unit length that is required
to confine the particles to an area A”. It is given by a formula analogous to the threedimensional formula, namely
P = T

∂S
∂A
!
N,E
. (8)

Problem 4:
For a system of N one-dimensional massless particles in a “one-dimensional box” of length
L, calculate the entropy S(N, E, L) of the system.

Start by thinking about the energy of a single massless particle.

To evaluate the required N-dimensional integral, you might consider a recursive approach.
Alternatively, you might look explicitly at N = 1, 2, · · · to deduce a pattern.