MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
8.044 Statistical Physics I
Spring Term 2013
Solutions to Problem Set #6
Problem 1: Sound Waves in a Solid
We need to find (∂T /∂P )∆Q=0 . To do this we will use in sequence the first law, the energy
derivative given in the statement of the problem, and the chain rule for partial derivatives.
d
/Q = dU − d/W = dU + P dV
∂U
∂U
=
dT +
+ P dV = 0
∂T V
∂V T
| {z }
|
{z
}
CV
A
A = T
∂P
∂T
0 = CV dT +
V
= T ∂T
∂V
−1
P
∂V
∂P
T
∂V
∂T P
= T ∂V
−
∂P T
=
αT
κT
αT
dV
κT
Now express dV in terms of dT and dP ,
∂V
∂V
dV =
dT +
dP = αV0 dT − κT V0 dP
∂T P
∂P T
and substitute in to the adiabatic condition
αT
(αV0 dT − κT V0 dP )
κT
α 2 T V0
αT V0 dP =
CV +
dT
κT
0 = CV dT +
∆T
∆P
αT
2 = CV
α T
+
V0
κT
1
� ∂P �
∂V
T
Problem 2: Energy of a Film
a) The best approach to take here is to find a general expression for CA and then show that
its derivative with respect to A is zero.
d/Q dT A
∂S
= T
∂T A
CA ≡
∂CA
∂A
by the second law
∂ 2S
∂A∂T
∂
∂S
= T
∂T
∂A T A
= T
T
interchanging order of the derivatives
We use a Maxwell relation to find (∂S/∂A)T . Note that S and S are different variables. I
would normally construct a magic square to find the equivalent derivatives, but for clarity I
will go through the more fundamental route here.
dE = T dS + S dA
dF = dE − d(T S) = −S dT + S dA
Since F is a state function, the cross derivatives must be equal.
∂S
∂S
Nk
−
=−
=
∂A T
∂T A
A−b
Substitute this result into the expression for the derivative of the heat capacity.
∂CA
∂
Nk
=T
=0
∂T A − b A
∂A T
This shows that the heat capacity at constant area does not depend on the area: CA (T, A) =
CA (T ).
2
b) Now we find the exact differential for the energy and integrate up.
dE
= T dS + S dA
∂S
∂S
= T
dT + T
+ S dA
∂T A
∂A T
| {z }
|
{z
}
N kT
CA (T )
−
+S =0
A−b
∂E
⇒
=0
∂A T
E(T, A) = E(T )
=
Z
T
CA (T 0 ) dT 0 + E(T = 0)
0
Problem 3: Bose-Einstein Gas
a) In this problem, we just follow the directions.
dE
= T dS − P dV
dF
∂S
= T
∂T
| {z
CV
∂P
∂T
V
}
∂S
− P dV
dT + T
∂V T
= dE − d(T S) = −S dT − P dV
⇒ −
∂S
∂V
T
= (5/2)aT
3/2
V
=−
∂P
∂T
2
+ 3bT =
V
∂S
∂V
T
∂S
T
−P
= (5/2)aT 5/2 + 3bT 3 − aT 5/2 − bT 3 − cV −2
∂V T
= (3/2)aT 5/2 + 2bT 3 − cV −2
Collecting this all together gives
dE = (dT 3/2 V + eT 2 V + f T 1/2 )dT + ((3/2)aT 5/2 + 2bT 3 − cV −2 )dV
3
b) Use the fact that the energy is a state function which requires that the cross derivatives
must be equal.
∂
∂E
∂
∂E
=
∂V
∂T V T
∂T
∂V T V
dT 3/2 + eT 2 = (15/4)aT 3/2 + 6bT 2
⇒ d = (15/4)a,
e = 6b
c) Use the results from b) to simplify the expression for dE in a).
dE = ((15/4)aT 3/2 V + 6bT 2 V + f T 1/2 )dT + ((3/2)aT 5/2 + 2bT 3 − cV −2 )dV
Integrate with respect to T first.
E
∂E
∂V
= (3/2)aT 5/2 V + 2bT 3 V + (2/3)f T 3/2 + F(V )
= (3/2)aT 5/2 + 2bT 3 + F 0 (V )
from above
= (3/2)aT 5/2 + 2bT 3 − cV −2
from dE
T
⇒ F 0 = −cV −2 ,
E
F = cV −1 + KE
= (3/2)aT 5/2 V + 2bT 3 V + (2/3)f T 3/2 + cV −1 + KE
d) Proceed just as we did above for E.
∂S
∂S
dS =
dT +
dV
∂T V
∂V T
| {z }
| {z }
CV /T
∂P
from a)
∂T V
= (dT 1/2 V + eT V + f T −1/2 )dT + ((5/2)aT 3/2 + 3bT 2 )dV
4
Integrate with respect to T first.
S
∂S
∂V
= (2/3)dV T 3/2 + (1/2)eV T 2 +2f T 1/2 + G(V )
|
{z
} | {z }
3/2
3bV T 2
(5/2)aV T
= (5/2)aT 3/2 + 3bT 2 + G 0 (V )
from above
= (5/2)aT 3/2 + 3bT 2
from dS
T
⇒ G 0 (V ) = 0,
G(V ) = KS
S(T, V ) = (5/2)aV T 3/2 + 3bV T 2 + 2f T 1/2 + KS
Problem 4: Paramagnet
a) This is virtually identical in approach to problem 2.
d/Q dT M
∂S
= T
∂T M
CM ≡
∂CM
∂M
T
by the second law
∂ 2S
= T
∂M ∂T
∂
∂S
= T
∂T
∂M T M
interchanging order of the derivatives
We will need H(T, M ) for what follows.
M=
A
H
T − T0
⇒
H=
M
(T − T0 )
A
We use a Maxwell relation to find (∂S/∂M )T .
dE = T dS + H dM
dF = dE − d(T S) = −S dT + H dM
Since F is a state function, the cross derivatives must be equal.
∂S
∂H
M
−
=
=
∂M T
∂T M
A
5
Substitute this result into the expression for the derivative of the heat capacity.
∂CM
∂
M
=T
−
=0
∂M T
∂T
A M
This shows that the heat capacity at constant magnetization does not depend on the magnetization: CM (T, M ) = CM (T ).
b)
dE = T dS + H dM
∂S
∂S
= T
dT +
T
+H
dM
∂T M
∂M T
| {z }
|
{z
}
−M T /A + H = −M T0 /A
CM (T )
Do the T integration first.
E(T, M ) =
Z
T
CM (T 0 ) dT 0 + f (M )
0
∂E
∂M
= f 0 (M )
from above
T
= −
M T0
A
from dE
⇒ f (M ) = −
E(T, M ) =
Z
M 2 T0
+ KE
2A
T
CM (T 0 ) dT 0 −
0
M 2 T0
+ KE
2A
c)
∂S
∂S
dS =
dT +
dM
∂T M
∂M T
| {z }
| {z }
CM (T )/T
−M/A from a)
Z T
CM (T 0 ) 0
S(T, M ) =
dT + g(M )
T0
0
6
∂S
∂M
= g 0 (M )
from above
T
= −
M
A
from dS
⇒ g(M ) = −
S(T, M ) =
Z
0
T
M2
+ KS
2A
CM (T 0 ) 0 M 2
dT −
+ KS
T0
2A
7
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8.044 Statistical Physics I
Spring 2013
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