Problem Set 2
Solutions
3.20 MIT
Professor Gerbrand Ceder
Fall 2003
Problem 1.1
S(U, V ) : dS =
1
P
dU + dV
T
T
Variables here are U and V and intensive variables are T1 and PT .
To go to T1 as a natural variables take the Legendre transform by subtracting
J =S−
U
T
from S
U
T
So,
� �
P
1
dJ = −U d
+ dV
T
T
Note that J is equal to − FT (F is the Helmholtz energy)
Similarly, go from S(U, V ) to B(U, PT ) by taking
P
V
T
� �
1
P
dB = dU − V d
T
T
B=S−
A Legendre transform with respect to both U and V defines the Planck function (Yes, you too
can be famous by defining your own Legendre transform)
P
U
V −
T
T
� �
� �
1
P
dY = −U d
−Vd
T
T
Y =S−
One can show that these functions have extrema properties just like the Legendre transforms of
energy. Massieu functions are maximum under constant value of their natural variables.
Problem 1.2
We know that Cp is defined as
�
Cp =
∂H
∂T
�
P
Enthalpy can be obtained from the Legendre transform of the internal energy as follows,
H = U + PV
We know that for an ideal gas U is a function of temperature only and the P V = RT (for one
mole), thus
�
� ∂U �
∂H
∂T
�
�
=
P
∂U
∂T
�
+
P
∂
(RT )
∂T
is only a function of temperature and this is not a function of pressure. Also, R is not a
∂T P
function of pressure. So Cp of an ideal gas must be independent of pressure.
Problem 1.3
F = F (T, V )
�
dF =
∂F
∂T
�
�
∂F
∂V
dT +
V
�
dV
T
Compare this with
F = U − TS
dF = −SdT − P dV
Therefore,
�
P =−
∂F
∂V
�
T
Using the given formula for F , solve for P by taking the derivative w.r.t V at constant T .
�
�
�
�
∂F
a
RT
∂f
= 2 −
+
∂V T
Vm Vm − b
∂V T
Since f(T ) is only a function of T , this term drops out and the solution is:
�
�
∂F
RT
a
P =−
=
− 2
∂V T
Vm − b Vm
Problem 1.4
(a) We can write the differential form of the entropy as a function of T and P
�
dS =
∂S
∂T
�
�
dT +
P
�
∂S
∂P
dP
T
Multiplying by T to get TdS (which is equal to dQ)
�
T dS = T
∂S
∂T
�
�
dT + T
P
∂S
∂P
�
dP
T
Since we are told the process is adiabatic, T dS = ∂q = 0, so,
�
−T
Or,
∂S
∂P
�
�
dP = T
T
∂S
∂T
�
dT
P
�
dT
dP
� ∂S �
�
=
�T
− � ∂P
∂S
∂T P
� ∂V �
=
∂T P
Cp
T
=
T αV
Cp
(b) Using the above relationship, we can get a value for the final temperature,
�
�
dT
αV
=
dP
T
Cp
� �
Tf
αV
ln
=
ΔP
Ti
Cp
�
�
αV
Tf = Ti exp
ΔP
Cp
For this case we can assume we have 1kg of material which means the volumes is 0.001 m3 .
Remember that αV = 3αL
�
3 · 19 × 10−5 K −1 ∗ 0.001m3 ∗ 999atm ∗ 1.01 × 105 P a/atm
Tf = 300 ∗ exp
2.3J/g · K ∗ 1000g
Tf = 300 ∗ 1.0254 = 307.6K
The only trick here is to be careful with your units and the definition of αL and αV .
Problem 1.5
(a) Start with the internal energy
dU = T dS − P dV + HdM
Using a Legendre transformation get G(T, P, H)
G = U − ST + P V − HM
dG = −SdT + V dP − M dH
And since we know that the order of differentiation does not matter
�
�
∂2G
=
∂P ∂H
�
�
�
�
� �
∂
∂G
∂
∂G
=
∂H ∂P H,T
∂P ∂H P,T
P,T
H,T
�
�
�
�
∂M
∂V
=−
∂H P,T
∂P H,T
∂ 2G
∂H∂P
� �
�
�
�
(b) One possibility is
�
∂S
∂H
�
�
=
T ,P
∂M
∂T
�
H,P
Problem 1.6
(a)
U = T dS + τ dL
G = U − TS
dG = SdT + τ dL − T dS + SdT
Which yields the solution:
dG = τ dL − SdT
(b)We also know if G=G(L,T) that
�
dG =
∂G
∂L
�
�
dL +
T
∂G
∂T
�
dT
L
Thus,
�
τ=
∂G
∂L
�
�
and S = −
T
∂G
∂T
�
L
Since we know that dG is a perfect differential, then the second derivatives must be equal
independent of the order which they are taken.
So,
�
� � 2 �
∂ 2G
∂ G
=
∂T ∂L
∂L∂T
�
�
� �
�
�
� �
∂ ∂G
∂G
∂
=
∂T ∂L T L
∂L ∂T L T
� �
� �
∂S
∂τ
=−
∂T L
∂L T
(c)
dU = T dS + τ dL
�
∂U
∂L
�
�
=T
T
∂S
∂L
�
+τ
T
Using the relation from part (b),
�
∂U
∂L
�
�
= −T
T
�
∂τ
∂T
+τ
L
(d) If U is only a function of temperature, then
�
∂U
∂L
�
�
T =0=τ −T
1
τ
�
∂τ
∂T
�
=
L
∂τ
∂T
�
L
1
T
LEVEL 2 PROBLEMS
Problem 2.1
(a) Start by constructing the differential form of S(T,P) [In Zemansky this is called the second
T dS equation]
�
�
�
∂S
∂S
dT +
dP
dS =
∂T P
∂P T
�
�
� �
∂S
∂S
dT + T
dP
T dS = T
∂T P
∂P T
�
But we know T
� ∂S �
∂T P
= Cp and from a Maxwell relation we know
�
T dS = Cp dT − T
∂V
∂T
� ∂S �
∂P T
=−
� ∂V �
∂T
�
dP = Cp dT − T V αdP
P
For an isothermal compression,
�
Q = T dS = −T
V αdP
In the case of a solid or liquid, neither V or α is very sensitive to pressure. So,
�
Q = −T V α
dP = −T V αΔP
Now we can use the data given to calculate Q (Replacing V with M/ρ)
Q=
298K ∗ 0.5kg ∗ 49.5 × 10−6 K −1 ∗ 5.05 × 108 P a
−T M αΔP
=−
ρ
8.96 × 103 kg/m3
Q = −416J
P
.So,
(b)We now need to calculate the work during the compression
�
W = −
� �
W =−
W =
∂V
∂P
�
�
�
W =
P dV
V βP dP = V β
P dP
T
P dP =
�
Vβ � 2
Pf − Pi2
2
�2
M β � 2 � 0.5kg ∗ 6.18 × 10−12 P a−1 �
8
∗
5.05
×
10
P
a
Pf =
2ρ
2 ∗ 8.96 × 103 kg/m3
W = 44J
(c) We know the first law,
ΔU = Q + W = −416J + 44J
ΔU = −372J
Thus it can be seen that the extra amount of energy in the form of heat comes from the storage
of internal energy.
(d) Now we can go back to the relationship we obtained for T dS (which we know is 0 since
the process is adiabatic and reversible)
T dS = 0 = Cp dT − T V αdP
�
�
dT
αV
=
dP
T
Cp
� �
Tf
αV
ln
=
ΔP
Cp
Ti
�
�
�
�
αV
αM
ΔP
Tf = Ti exp
ΔP = Ti exp
Cp
Cp ρ
�
0.5kg ∗ 49.5 × 10−6 K −1 ∗ 5000atm ∗ 1.01 × 105 P a/atm
Tf = 298 exp
8.96 × 103 kg/m3 ∗ 385J/kg · K ∗ 0.5kg
Tf = 300K
ΔT = 300 − 298 = 2K
Problem 2.2
�
In terms of T and P, dU can be written
�
�
�
∂U
∂U
dU =
dT +
dP
∂T P
∂P T
� �
� �
using U (S, V ) which we know
We can evaluate the two partials ∂U
and ∂U
∂T P
∂P T
�
dU = T dS − P dV
So
�
∂U
∂T
�
�
=T
P
∂S
∂T
�
�
−P
P
� ∂S �
Which can be simplified since we know Cp = T
�
∂U
∂T
∂V
∂T
�
P
and V α =
∂T P
� ∂V �
∂T
P
�
= Cp − P V α
P
Similarly
�
∂U
∂P
�
�
=T
T
∂S
∂P
�
�
−P
T
which can be simplified using the Maxwell relation
�
∂U
∂P
∂V
∂P
� ∂S �
∂P T
�
T
=−
� ∂V �
∂T
P
and V β = −
� ∂V �
∂P
T
�
= −T V α + P V β = V (βP − αT )
T
Putting this all together,
dU = (CP − P V α) dT + V (βP − αT ) dP
Problem 2.3
Given the definition of dH : dH ≡ T dS + V dP , we want first to determine how the enthalpy
varies with pressure, at constant temperature:
�
�
�
�
∂H
∂S
=T
+V
∂P T
∂P T
By using the Maxwell’s Relation
�
∂S
∂P
�
�
=−
T
∂V
∂T
�
P
(from the expression for the Gibbs Free Energy) and the definition
�
�
1 ∂V
α≡
V ∂T P
, we have:
�
∂H
∂P
�
= V (1 − αT )
T
What we are really interested in is to find how �Hevap = H vap − H liq changes with tempera­
ture:
�
�
�
�
� liq �
� ∂H vap �
∂ [H vap −H liq ]
∂ΔHevap
∂H
=
=
−
∂P
∂P
∂P
∂P
T
T
T
T
�
�
�
�
�
�
∂ΔHevap
vap vap
liq liq
vap
liq
− V α − V α T ≈ V vap (1 − αvap T )
= V
−V
∂P
T
In general, at relatively low pressures and densities, all gases behave ideally.
For an ideal gas, α ≡
1
V
� ∂V �
∂T
P
=
nR
PV
=
1
T
Therefore,
�
�
∂ΔHevap
≈0
∂P
T
Problem 2.4
Given
U (S, X) = X a exp(bS)
First find the equations of state
�
�
∂U
T =
= X a exp(bs)b = bU (S, X)
∂S X
�
�
∂U
aU (S, X)
f=
= aX a−1 exp(bs) =
∂X S
X
From these we can derive the following which will be useful later
1
S = ln
b
�
T
bX a
�
T
fX
=
b
a
U (S, X) =
(1)
(2)
T
b
(3)
�
�1−a
f
X=
exp (−bS)
a
Starting with the first function D(S, f ) we perform a Legendre transform as follows
(4)
D(S, f ) = U − f X = (1 − a)U (S, X)
Now using (4) to get this in terms of S and f only
�
f
D(S, f ) = (1 − a)
exp (−bS)
a
�(1−a)a
exp(bS)
Similarly for F(T,X)
F (T, X) = U − T S = U (S, X) − bU (s, x)S = (1 − bs)U (S, X)
Using (3) and (1) from above
�
�
��
T
T
F (T, X) = 1 − ln
a
bX
b
Now we do a transformation for to both T and f
K(T, f ) = U − T S − f X
Using (3), (1) and (2)
�
�
T
T
T
Ta
K(T, f ) = − ln
−
a
b
b
b
bX
⎡
⎛
⎞⎤
T
T
K(T, f ) = ⎣1 − a − ln ⎝ � �a ⎠⎦
b
b Ta
bf
Problem 2.5
(a) Start by using the chain rule on the left side of the equation
�
We know that
� ∂S �
∂T V
=
CV
T
∂S
∂P
�
�
=
V
∂S
∂T
� �
V
∂T
∂P
�
V
and we can expand the second derivative as follows
�
∂T
∂P
�
�
=−
V
∂T
∂V
� �
P
∂V
∂P
�
T
Putting those together,
CV
T
�
∂T
∂V
� �
P
∂V
∂P
�
T
CV
=−
∗
T
We can replace CV by CP using the relationship
�
1
Vα
�
∗ −V β
T V α2
β
CP − CV =
Yielding
�
∂S
∂P
�
�
�
β
T V α2
CP β
=
CP −
=
−Vα
Tα
β
Tα
V
(b) We can expand the derivative as follows
�
∂T
∂P
�
�
=−
S
∂T
∂S
� �
P
∂S
∂P
�
T
The first term is CTP and the second term can be transformed using a Maxwell relation
� ∂V �
− ∂T P . This gives
�
∂T
∂P
�
� ∂S �
∂P T
=
�
�
�
T
∂V
=
CP ∂T P
S
�
∂T
T αV
=
∂P S
CP
(c) Start with the left side
�
T
∂2P
∂T 2
�
V
��
∂
=T
∂T
∂P
∂T
� �
V
V
Using a Maxwell relation,
∂
T
∂T
��
∂P
∂T
� �
V
V
∂
=T
∂T
��
∂S
∂V
� �
=T
T
V
∂ 2S
∂TV ∂VT
Now the right side
�
�
�
� � �
�� � �
� �
∂
∂Cv
∂T
∂S
∂S
∂S
→
Cv = T
=
+T
∂T V
∂V T
∂V T ∂T V
∂V
∂T V T
The first term in the above is zero since the change in temperature at constant temperature is
zero. The second term can be written as
∂
T
∂V
��
∂S
∂T
� �
=T
V
T
∂ 2S
∂VT ∂TV
And since we know the order of differentiation doesn’t matter
T
∂ 2S
∂ 2S
=T
∂TV ∂VT
∂VT ∂TV
Problem 2.6
(a) Start with the differential form of G
dG = −SdT + V dP
�
�
∂G
∂T
�
�
= −S
�
V
�
∂T
∂T
�
+V
�
V
∂P
∂T
� �
�
V
�
∂G
∂P
∂V
= −S − V
∂T V
∂V T ∂T
�
�
�
�
∂G
−1
= −S − V
(V α)
∂T V
Vβ
�
�
∂G
Vα
=
−S
∂T V
β
P
(b) Start with the differential form of H
dH = T dS + V dP
�
∂H
∂T
�
�
P
�
�
�
∂S
∂P
=T
+V
∂T P
∂T P
�
�
∂H
T CP
=
∂T P
T
�
�
∂H
= CP
∂T P
Problem 2.7
(For this problem Maple is our friend)
(a) This part involves taking the indicated
� ∂U � derivatives of the given function. The two tricks are
knowing what β is and how to change ∂V T into something with derivatives of P w.r.t T or V.
−1
β=
V
�
∂V
∂P
�
T
and using the differential form of U and a Maxwell relation we get,
�
The solutions are then
∂U
∂V
�
�
=T
T
∂P
∂T
�
−P
V
�
�
β=
−1
V
��
∂P
∂T
�
� ∂P �
∂V
∂U
∂V
R
V −b
=
V
(V − b)2 V 2
=
RT V 3 − 2aV 2 + 4aV b − 2ab2
1
�
�
T
�
=
T
RT
a
−P = 2
V −b
V
(b)Remember
1
α=
V
�
∂V
∂T
�
P
R
VP
� ∂H �
α=
Expand H in differential form to try and get
∂P
T
into a form you can solve for
dH = T dS + V dP
�
∂H
∂P
�
�
=T
T
∂S
∂T
�
+V
T
�
�
�
∂V
∂H
= −T
+V
∂P
∂T P
�
�
∂H
−T R
=
+V =b
∂P T
P
�
Now for CP − CV
CP − CV =
CP − CV =
T V α2
β
R3 T 2
P 2 (V − b)2
Problem 2.8
1
U=
P V + Nf
k−1
�
PV k
Nk
dU = T dS − P dV
For a reversible, adiabatic process,
�
(dU )S = −P (dV )S
Along a reversible adiabatic process: P V k = −g(S) so
P = −g(s)V −k
�
∂U
∂V
�
= −P = g(s)V −k
S
Again for a reversible, adiabatic process
�
(dU )S =
∂U
∂V
�
dV =
S
g(s)
dV
Vk
Integrate from V0 → V at constant S
�
�
1
1
−g(S)
U (V, S) − U (V0 , S) =
−
(k − 1) V k−1 V0k−1
Moving things around,
�
U (V, S) =
U (V, S) =
−g(S) 1
k V k−1
−g(s) 1
k − 1 V k−1
�
�
g(S)
+ U (V0 , S) +
(k − 1)V0k−1
�
�
g(S)
+ U (V0 , S) +
(k − 1)V0k−1
�
��
�
�
The second term above is only a function of S, so we can call it some function w(S)
U (V, S) =
PV k 1
PV
+ w(S) =
+ w(S)
k−1
k−1V
(k − 1)
w(S) is an extensive function, therefore it can be written as N (the number of moles in the
system) times a molar quantity f (S)
� � � �
k
V
f P
N
which is independent of the system size
PV
U (V, S) =
+ Nf
(k − 1)
Problem 2.9
�
Vk
P k
N
�
The thermal expansivity is defined as
1
α=
V
�
�
∂V
∂T
P
Using a Maxwell relation,
�
∂V
∂T
�
�
=
P
∂S
∂P
�
T
And we are told in the problem that S is independent of crystallite structure and pressure, so
�
∂S
∂P
�
=0
T =0
And thus α = 0 at absolute zero.
Problem 2.10
(a) The key property is that
� ∂S �
∂H T
< 0. This can be proven from the Maxwell relation.
�
∂S
∂H
�
�
=
T
∂M
∂T
�
H
(this can be proven from the differential of dφ = d(U − T S + P V − M H) = −SdT + V dP −
M dH ) and since
�
∂M
∂T
�
�
< 0 =⇒
H
∂S
∂H
�
<0
T
S(T ) as constant H therefore has the following shape.
Adiabatic demagnetization consists of two steps:
I. Isothermal application of a field
II. insulate material and turn off the field → isotropic demagnetization. System moves
back to state with H = 0 , hence T decreases.
(b)
� ∂S �
� ∂M �
�
�
∂T
∂H T
= − � ∂S � = − ∂TcH H
∂H S
T
∂T H
� ∂M �
�
�
−T ∂T H
∂T
=
∂H S
cH
� ∂M �
We can evaluate ∂T H as follows
κV
M=
H =⇒
T
�
∂M
∂T
�
=−
H
κV
H
T2
Figure 1:
So,
� ∂T �
∂H S
=
κV H
cH T
LEVEL 3 PROBLEMS
Problem 3.1
Given:
� �
� �
θ
Rθ
2
U=
S −
V2
R
V02
(a) We want to make this the absolute U and not the molar quantity, so multiply by N.
�
U = NU =
Nθ
R
�
�
�
�
� � �2 �
� � �2
N Rθ
Nθ
S
N Rθ
V
2
S −
V =
−
2
2
V0
R
N
V0
N
� � �
�� �
Rθ
1
θ
2
U=
S −
V2
N
R
V02
2
The equations of state give us
�
T =
∂U
∂S
�
N,V
2
=
N
� �
θ
S
R
�
P =
�
µ=
∂U
∂N
�
P,S
1
=
−
2
N
∂U
∂V
�
N,S
� �
2 Rθ
=
V
N
V02
�� �
� � �
�� �
� � �
θ
Rθ
θ
Rθ
2
2
2
S −
V =
−
S −
V2
2
2
R
V0
R
V0
or
µ = −U
(b) Express µ as a function of T and P
� �
R
T
θ
� �
N V02
V =
−
P
2
Rθ
N
S=
2
Substitute in to the expression for µ
1
µ=− 2
N
�� �
� � 2 � 2 �2 �
� �2
Rθ N
θ N2 R
V0
T2 −
P2
2
R
4
θ
V0
4
Rθ
� � �
� � �
1
R
1
V02
2
µ=−
T −
P2
4
θ
4
Rθ
Problem 3.2
Start with the expression we used in class
� �
� �
� � � �
∂y
∂x
∂x
∂x
=
+
∂f z
∂f y
∂y f ∂f z
For this problem, this equation can relate the constant l heat capacity to the constant P heat
capacity as follows,
�
∂S
∂T
�
�
=
l
∂S
∂T
�
�
+
P
∂S
∂P
� �
T
∂P
∂T
�
l
To get heat capacity we need to multiply both sides by T, and when we do this we get
�
Cl =
Cp + T
∂S
∂P
� �
T
∂P
∂T
�
l
The first derivative can be rewritten using the Maxwell relation
simplifies to
� ∂S �
∂P T
=
−
� ∂V �
∂T
P
which
�
Cl = Cp − T
And since we know that
� ∂V �
∂T
P
∂V
∂T
� �
P
∂P
∂T
�
l
=
V α
�
Cl =
Cp − T V α
∂P
∂T
�
l
Now we need to expand the second derivative
�
�
�
� � �
∂P
∂P
∂l
=−
∂T l
∂l T ∂T P
� ∂l �
Since the material is isotropic, ∂T P =
13 V α which now leave us with
1
Cl =
Cp + T V 2 α2
3
�
∂P
∂l
�
T
One more partial to go....if we look at the relationship between stress and strain,
�
σ = Eε and thus
∂σ
∂ε
�
=
E
T
where E is the Young’s modulus for an isotropic material. To get this in a form we can use, we
need to multiply by l to switch from ∂ε to ∂l and then invert. This then will give us
1 T V 2 α2 E
Cl =
Cp +
3
l
Problem 3.3
The total system is at constant volume and temperature hence its Helmholtz free energy is
minimal with respect to the internal degrees of freedom.
Figure 2:
dF tot = dF α + dF β
β
dF tot = −S α dT α − pα dV α + µαA dnαA − S β dT β − pβ dV β + µβA dnA
since dT α = dT β = 0 and dV α = −dV β and dnαA = −dnβA we can write
�
�
�
�
dF tot = − pα − pβ dV α + µαA − µβA dnαA
If dV α and dnαA were independent, equilibrium would require that pα = pβ and µαA = µβA .
However dV α and dnαA are NOT independent!
dV α = KdnαA
�
dF tot =
µαA
�
−
K
β
µA
�
�
− pα − pβ dV α
�
��
�
The bracketed term must equal zero for equilibrium.
β
− Kpβ
µαA − Kpα = µA
Problem 3.4
The heat capacity cx is related to the temperature derivative of the entropy under the condition
x. (x being the conditions defined in the problem statement)
�
cx = T
∂S
∂T
�
x
Starting with S(T, P ) we write
�
dS =
∂S
∂T
�
�
dT +
P
∂S
∂P
�
dP
T
We can take the temperature derivative of this holding x constant
�� �
�
�
�
∂
∂
∂S
∂S
(dS)x =
dT +
dP
∂T
∂T
∂T P
∂P T
x
�
� �
�
� �
cp
∂S
∂P
∂S
=
+
∂T x
T
∂P T ∂T x
� �
� �
� � � �
∂y
∂x
∂x
(Note: The above can also be obtained using ∂f
= ∂f
+ ∂x
)
∂y
∂f
z
y
f
z
� ∂S �
� ∂V �
= −αv V. So now we need to
We know through a Maxwell relation that ∂P
=
−
∂T P
T
� ∂P �
find ∂T x . We start from the given information that V − aP = a constant, thus dV = adP . This
gives us
�
∂P
∂T
�
x
1
=
a
�
∂V
∂T
�
x
Continuing from there, we write out the differential form of V (T, P )
�
�
�
∂V
∂V
dV =
dT +
dP
∂T P
∂P T
� �
� �
But we know ∂V
= −βT V. We now have
= αv V and ∂V
∂T P
∂P T
�
dV = αv V dT + −βT V dP
So
� ∂V �
∂T
x
is given by
�
Replacing
� ∂P �
∂T
x
∂V
∂T
�
�
= αv V − βT V
x
∂P
∂T
�
x
we get
�
∂V
∂T
�
x
βT V
= αv V −
a
�
∂V
∂T
�
x
or
�
∂V
∂T
�
=
x
αv V a
a + βT V
Putting this all together...
�
�
� ��
�
∂S
cp
1
∂V
=
+ (−αv V )
∂T x
T
a
∂T x
� �
� ��
�
∂S
cp
1
αv V a
=
+ (−αv V )
∂T x
T
a
a + βT V
� �
� ��
�
∂S
1
αv V a
T
= cp + T (−αv V )
∂T x
a
a + βT V
So we get finally that
cx = cp −
T (αv V )2
a+βT V