MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
8.044 Statistical Physics I
Spring Term 2013
Solutions to Problem Set #7
Problem 1: Free Expansion of a Gas
a) No work is done, so ∆W = 0. No heat enters the gas so ∆Q = 0.
Thus ∆E = ∆W + ∆Q = 0. The internal energy is conserved.
b) E(T, V ) is a state function; compare it before and after expansion in equilibrium situations.
dE =
CV
∂S
∂V
=
=
T
dE =
E =
∂S
∂S
T dS − P dV = T
dT + T
− P dV
∂T V
∂V T
| {z }
CV
3
Nk
2
∂P
Nk
=
using a Maxwell relation
∂T V
V − bN
2
3
N
N k dT + a
dV
2
V
3
aN 2
N kT −
+ constant
2
V
When equating E before and after the constant cancels out.
3
aN 2
3
aN 2
N kTi −
=
N kTf −
2
(V0 /3)
2
V0
3
2aN 2
N k(Tf − Ti ) = −
2
V0
Tf
4a
= Ti −
3k
The gas cools!
1
N
V0
Problem 2: Use of a Carnot Cycle
a) Carnot Cycle ⇒ dSH = −dSC ⇒ d
/QH /TH = −d
/QC /TC . Use dQ = C0 dT.
∆Sbody 1 = −∆Sbody 2
Z
TF
TH
C0
dT = −
T
ln
Z
TF
TC
TF
TH
= − ln
TF
TH
=
C0
dT
T
TF
TC
= ln
TC
TF
TC
TF
p
=
TH TC
TF
b)
∆Wout = |∆QH | − |∆QC |
Z
TH
Z
TF
C0 dT −
Wout =
TF
C0 dT = C0 [(TH − TF ) − (TF − TC )]
TC
p
= C0 (TH − 2TF + TC ) = C0 (TH − 2 TH TC + TC )
p
p
= C0 ( TH − TC )2 > 0
2
Problem 3: Cooling Liquid Helium
a) Since the system is thermally isolated and no work is done in the process, the heat gained
by the salt must equal the heat lost by the liquid.
∆QS = −∆QL
Z
1/2
bT
−2
1/2
Z
aT 3 dT
dT = −
T0
1
1/2
1/2
a
−1
= −
T4
−b
T
4 1
T0
1
1a 1
1 a 15
1 128 15
2−
=
−1 =
=
= −2
T0
4 b 16
4 b 16
4 15 16
1/T0 = 4 → T0 = 1/4
b) Entropy is a state function, so one may compute its change by assuming the process was
quasi-static. We will use dS = d/Q/T = C dT /T .
∆S = ∆SS + ∆SL
Z
1/2
=
bT
−3
Z
T0
= −
b
2
1/2
dT +
aT 2 dT
1
1/2
1/4
T −2 +
a
3
1/2
T3
1
b
a 1
= 6b − (7/24)a
= − (4 − 16) +
−1
2 | {z } 3 8
| {z }
−12
−7/8
3
Problem 4: Torsional Pendulum
a) First find the Hamiltonian.
1
1
H = T + V + Iθ˙2 + K(θ − θ0 )2
2
2
Then use the canonical ensemble expression for the probability density.
˙ ∝ exp[− H ]
p(θ, θ)
kT
∝ exp[−
θ̇2
(θ − θ0 )2
] exp[−
]
2(kT /I)
2(kT /K)
˙ =
Notice two important features of this result. First, the probability density factors, p(θ, θ)
˙ so θ and θ˙ are statistically independent. Second, the dependence on both θ and θ˙
p(θ)p(θ),
has the Gaussian form. In particular, p(θ) is Gaussian with mean θ0 and variance σθ2 = kT /K.
Therefore,
r
kT
2
1/2
< (θ − θ0 ) > =
.
K
˙ is
b) Since θ and θ˙ are statistically independent, < θθ˙ >=< θ >< θ˙ >. By inspection, p(θ)
˙
˙
a zero-mean Gaussian, so < θ >= 0 which leads to < θθ >= 0.
4
Problem 5: The Hydrogen Atom
a)
A
|n, l, m >
n2
The lowest energy, −A, corresponds to |1, 0, 0 > and is non-degenerate. The next lowest
energy, −A/4, is four fold degenerate:
H|n, l, m >= −
|2, 0, 0 >,
|2, 1, 1 >,
|2, 1, 0 >,
and |2, 1, −1 > .
The ratio of the number of atoms in the first excited energy level to the number in the
ground state depends on both the energies and the degeneracies.
N (−A/4)
4 exp[A/4kT ]
=
= 4 exp[−(3/4)A/kT ]
N (−A)
exp[A/kT ]
Using the conversion factor 1meV = 11.6K we find that 13.6eV = 1.58 × 105 K. Evaluating
the above ratio gives 4.8 × 10−170 at 300K and 1.6 × 10−51 at 1000K.
b) The degeneracy of the nth energy level is
1 + 3 + 5 + · · · + (2n − 1) = n2 .
The partition function for a single atom, neglecting the unbound states, is
Z=
X
states
exp[−i /kT ] =
∞
X
n2 exp[α/n2 ],
n=1
i
where α ≡ A/kT . Since α > 0, it follows that exp[α/n2 ] > 1 for all n. Using this we can set
a lower bound for Z, but Z diverges since the lower bound diverges.
Z>
∞
X
n2
which diverges
n=1
c) The Coulomb potential is a mathematical oddity in that it produces an infinite number
of bound states with energies less than zero. This situation if modified in the real world by
the presence of walls (consider the energy levels of a particle in a box) or by the presence
of other atoms. The existence of hydrogen atoms in the interstellar medium, on the other
hand, probably has more to do with the absence of excitation mechanisms (non-equilibrium)
than with the presence of neighboring atoms.
5
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8.044 Statistical Physics I
Spring 2013
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