BE.011/2.772 Problem Set 5 Due March 17, 2004 Dill 9.1, 9.3, 9.8 and 9.10 Continued on next page....
 ∂U 
 ∂U 
We are given that 
 = 0 , and we are asked to show that this implies that CV = 
 is

∂V T
 ∂T V
∂CV
∂  ∂U 
not a function of V. For this, we need only prove that
=
 = 0 . But we know that we can

∂
V 
∂T V
∂V
∂CV
∂
 ∂U 
∂  ∂U 
∂
(0) = 0 . Thus,
=
permute the order of the derivatives, so we get that

 =

 =
∂V
∂V  ∂T V ∂T  ∂V T ∂T
the constant volume heat capacity of an ideal gas is independent of volume.
9.3
 ∂f 
9.8 The first thing to note is that if we take the cross derivatives of dU, we end up getting 


∂p0  E ,S
 ∂p 
rather than  0  which we want. We need to first transform our equation to a new energy variable
 ∂f  E ,T
X(T,f,E).
dU = TdS +
fdl + Edp0
X
= U − TS − fl − Ep0
dX = SdT −
ldf − p0 dE
 ∂p 
Now if we take the cross derivatives we can get an expression for  0  .
 ∂f  E ,T
 ∂
p0 
 ∂l 

 =
 
 ∂f 
E ,T  ∂E 
f ,T
This is useful, because we are likely able measure how the length of the crystal changes with the field
 ∂l 

 .
 ∂E  f ,T
9.10
a) The first thing to note in this question is that we are considering T and p as variables (which we end
up holding constant). That means that we need to use the Gibbs free energy.
G
= H − TS =
U + pV − TS
dG = dU + pdV +
Vdp − TdS − SdT
= − SdT +
Vdp + fdl
Now we want to create a Maxwell relation between S and f in order to get a function for S.
 ∂ 2G   ∂ 2G 

 = 

 ∂l∂T   ∂T∂l 
∂  ∂G 
∂  ∂G 


 =

∂l  ∂T l , p ∂T  ∂l T , p
 ∂f 
 ∂S 
−  = 

 ∂l T , p  ∂T  p ,l
∂
 ∂S 
(aT (l − l0 ))
  =−
∂T
 ∂l T , p
= − a (l − l0 )
l
S (l ) = ∫ − a(l − l0 )dl
l0
a (l − l0 )
2
So the entropy is proportional to the square of the extension distance.
S (l ) = −
2
H = U + pV
dH = dU + pdV + Vdp
= TdS + Vdp + fdl
 dH 
 dS 

 = T  + f
 dl T , p
 dl T , p
= T (− a(l − l0 )) + aT (l − l0 )
=0
So H is constant for all extensions l.
b) Since we are told that it is an adiabatic process (δq=0, so dU=δw=fdl), we know that we need to start
with the expression for internal energy U. We also know that by definition, for constant volume
dU=CvdT.
dU = fdl
Cv dT = aT (l − l0 )dl
Cv
dT = a (l − l0 )dl
T
T2
l
C
∫T Tv dT = l∫ a(l − l0 )dl
1
0
 T2  a (l − l0 )2
Cv ln  =
2
 T1 
 a(l − l0 )2 
T2
= exp 

T1
 2Cv 