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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

T = 0? You may make a sketch. Explain your answer.

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

(µH ≪ kBT)? You may make a sketch. Explain your answer.

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.

Answer: Start with

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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