PHYSICS 112
Homework 7 Solutions
1. Since
n
L
where n = 1, 2, 3, the number of states with wavevector between k and k + dk is (L/ )dk , If
the corresponding (innitesimal) range of energy is d then we have
k=
L
() d = 2 dk,
(1)
where the factor of 2 comes
PHYSICS 112
Homework 1 Solutions
1. (a) From g (U ) = CU 3N/2 we have, with S = kB ln g ,
S
U
1
=
T
= kB
N
3N 1
,
2U
(1)
and hence
3
U = N kB T .
2
This is the form of the energy of an ideal gas, as we will discuss later in the course.
(b) Dierentiating E
PHYSICS 112
Homework 4 Solutions
1. The specic heat, per unit volume of a crystal at low temperature is given by
Cphonon
12 4 n
=
kB
5
T
D
3
,
while the specic heat per unit volume of a photon gas is
Cphoton
4 2
( kB T ) 3 ,
=
kB
15( c)3
h
which can be ob
PHYSICS 112
Homework 3 Solutions
1. As shown in class (see also handout) the allowed values of k in k -space form a regularly space grid
with spacing /L (where the system is of size L in each direction) and kx , ky etc. are taken to be
positive.
[Note: Th
PHYSICS 112
Homework 2 Solutions
1. (a) If the two states have energy 0 and , the partition function is given by
Z = 1 + e ,
where = 1/kB T , and so the free energy is given by
F = kB T ln Z = kB T ln 1 + e .
(b) The energy is given by
U=
1
(F ) =
.
e +1
PHYSICS 112
Homework 5 Solutions
1. We have
f ( + ) =
1
1
=
,
exp[ ( + )] + 1
exp[ ] + 1
and similarly (with replaced by )
f ( ) =
Now
1
.
exp[ ] + 1
1
ex
1
=
=1
,
x+1
x
e
1+e
1 + ex
and so, with x = , we have
f ( + ) = 1 f ( ) .
Hence the probability tha
PHYSICS 112
Homework 6 Solutions
1. (a) The thermodynamic identity is
T dS = dU + P dV.
If the volume changes by a small amount dV , and the temperatures by dT at constant entropy
we have 0 = dU + P dV . Now dU = CV dT since the energy of an ideal gas onl
PHYSICS 112
Homework 9 Solutions
1. As discussed in class the (Helmholtz) free energy of the van der Waals gas is given by
F = N kB T
n Q ( V N b)
N
+ 1 N a,
N
V
ln
where
mkB T
2 2
h
nQ =
3/2
.
(a) The entropy is given by
S=
F
T
V ,N
= N kB l n
=
N kB l n
PHYSICS 112
Homework 8 Solutions
1. (a) In the reaction
e + H + H
we have e = 1, H + = 1, H = 1 and so the law of mass action gives us
nQe nQH +
1
ne nH +
=
nH
nQH ZH (int)
(1)
where ZH (int) = e E is the internal partition function of the hydrogen atom (
Name: _
Date: _
Period: _
Renaissance and Reformation TIMELINE and VOCABULARY
Honors: Ch. 13 (pgs.223-247)
Regular: Ch. 9 and 10 (pgs. 160-187)
NGSSS Benchmarks
* SS.912.W.1.1 Use timelines to establish cause and effect relationships of historical events.