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1. (a) (5 points) If a three dimensional system has 106 particles, how many dimensions
does its phase space have? You may neglect internal degrees of freedom
of the particles.
Answer: Number of dimensions of phase space = 6N = 6 × 106
(b) (10 points) A radioactive source emits alpha particles which are counted at an
average rate of one per second. What is the probability P of counting exactly
10 alpha particles in 5 seconds?
Solution: Use the Poisson distribution
P(N) = hNi
(c) (10 points) Consider a one dimensional line of noninteracting spins in a magnetic
field H = Hzˆ. Each spin has a magnetic moment ~µ. The energy of a
spin in a magnetic field is E = −~µ · H~ .
i. (5 points) What is/are the equilibrium spin configuration(s) at temperature
Answer: At T = 0 the system is in its ground state. This lowest energy
state is where the spins point up in the same direction as the magnetic
field.
H
ii. (5 points) What is/are the equilibrium configuration(s) at very high temperatures
Answer: At high temperatures, the spins point in random directions.
The spins want to minimize their free energy F = E − T S and maximize
their entropy S. Pointing in random directions corresponds to most of
the configurational states. S = kB ln Ω where Ω is the number of states
available to the system and kB is Boltzmann’s constant.
2. (25 points) A magnetic solid is subjected to an external magnetic field H~ = Hzˆ
in the z direction. In centimeter-gram-second (cgs) units, the Gibbs free energy is
given by
G(T, Hz) = E − T S −14πBH (1)
where T is the temperature, S is the entropy, E is the internal energy, and B is the
z-component of the magnetic induction. (We will ignore the pressure-volume term
for this solid.) Starting from
dG = −SdT −14π
BdH (2)
use a Legendre transformation to derive the Helmholtz free energy F(T, B) from
the Gibbs free energy G(T, H). In other words, derive the relation between G and
F. Also derive the associated Maxwell relation.
dG = −SdT −14π
BdH (3)
The Legendre transformation is
d(BH) = BdH + HdB
BdH = d(BH) − HdB
Now note that the second derivatives of F are independent of the order of differentiation:
3. (25 points) A mole of an ideal gas expands quasistatically at a constant temperature
T from an initial pressure Pi to a final pressure Pf (Pi > Pf ). Find the work W
done by the gas, the heat Q absorbed by the gas, and the entropy change ∆S of
the gas.
Answer: The work is given by
From the ideal gas law for one mole:
From the ideal gas law
To find the heat Q absorbed by the gas, we use the first law of thermodynamics:
dE = dQ − dW (12)
For an ideal gas, the energy E(T) = 3RT/2 is solely a function of the temperature
T. Since the temperature is constant, dE = 0. So
dE = 0 = dQ − dW
dQ = dW
Note that it is wrong to use dQ = CvdT because the volume is not constant.
Besides, this would give dQ = 0 because dT = 0 due to the constant temperature.
W = RT ln (pi/pf )
Q = RT ln (pi/pf )
∆S = R ln (pi/pf )
7
4. (25 points) A finite body with a temperature independent heat capacity C is initially
at a temperature T1 which is hotter than an infinite thermal reservoir whose
temperature is and remains fixed at T2. A reversible Carnot engine is connected to
run between these reservoirs. After the engine has run long enough, it has drained
heat Q from the finite hot reservoir and reduced its temperature to T2, whereupon
the engine has to stop. Find the maximum work Wmax that can be gotten from the
engine and the body.
Answer: Let dq1 be the infinitesimal heat extracted from the finite body, and let
dq2 be the infinitesimal heat put into the infinite hot reservoir at temperature T2.
Let dW be the infinitesimal work done by the Carnot engine. Energy conservation
dictates that
dq1 = dq2 + dW (13)
Let T be the temperature of the finite body. Note that T decreases as the engine
runs.
For a Carnot engine, the equality applies.
where the minus sign compensates for the fact that dT < 0 because the temperature
of the finite body decreases when it loses heat dq1 > 0.
where we used the fact that a Carnot engine is the most efficient engine and therefore,
produces the maximum amount of work Wmax.
Note from Eq. (15) that the heat extracted from the hotter reservoir is −Q which

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