Exam 1 solutions - Fall 2002
Question 1.) We have 100g of water in a beaker, and it is heated to 100 C by placeing it in contact with a temperature reservoir.
a.) Calculate the entropy change of the beaker of water.
The solution to this problem will depend on whether or not you considered if the water would boil. We didnt expect you to
consider this but many people actually did.
Cp = 4.184 ⋅
ds =
δQ
T
J
∆Hevap = 41⋅
K⋅ g
= 100 ⋅ Cp⋅
kJ
mole
M
water
= 18⋅
g
mole
 T2 
 T1
 
∆Swater = 100 ⋅ Cp⋅ ln
dT
T
without boiling......
∆Swater := 100 ⋅ 4.184 ⋅ ln
373 
 293 
∆Swater = 101.004
J
K
with boiling......
 T2 
∆Hevap
+
 T1
T
evap
 
∆Swater := 100 ⋅ 4.184 ⋅ ln
∆Swater = 100 ⋅ Cp⋅ ln
373 
 293 
41000
+
373 ⋅ 18
⋅100
∆Swater = 711.668
J
K
b.) What is the entropy change of the universe?? First, we shall find the entropy change of the reservoir....
Q
Because the reservoir is at constant temperature, we know that
(
2
Qres = −100 Cp⋅ T − T
∆Sres :=
Qres
)
∆Sres = −89.737
373
and we know that the heat that flowed into
the beaker flowed out of the reservoir.
res
T
res
4
Qres := −100⋅4.184 ⋅ (100 − 20)
1
∆S =
Qres = −3.347 × 10
J
J
K
if you included boiling.......
∆Sres =
Qres
T
−
res
4
∆Hevap
T
∆Sres :=
−3.347 × 10
evap
373
−
41000
373 ⋅ 18
⋅100 ∆Sres = −700.396
J
K
now for the entropy change of the universe......
∆Suniv = ∆Swater + ∆Sres
without boiling......
∆Suniv :=
Qres
373
+ 100 ⋅ 4.184 ⋅ ln
373 
 293 
∆Suniv = 11.267
J
K
with boiling.....
∆Suniv :=
Qres
373
−
41000
373 ⋅ 18
⋅100 + 100 ⋅ 4.184 ⋅ ln
373 
 293 
+
41000
373 ⋅ 18
⋅100
∆Suniv = 11.267
J
K
You notice that the entropy change of the universe is positive, as you would expect. Also, note that boiling has no effect
on the total entropy change.
c.) A carnot engine gives reversible heat transfer, so the entropy change of the universe will be zero
2.
dU = TdS + V⋅ σ xx⋅ dεxx + V⋅ σ yy⋅ dεyy + V⋅ σ zz⋅ dεzz + E⋅ dP Note: The pressure is absorbed into the stresses.
a.) We want a function of T, εxx, σyy, σzz, and E.
φ = U − TS − V⋅ σ yy⋅ ε yy − V⋅ σ zz⋅ ε zz − EP dφ = −SdT + V⋅ σ xx⋅ dεxx − V⋅ ε yy⋅ dσyy − V⋅ ε zz⋅ dσzz − PdE
(
b.)
−S = f1 T , ε xx , σ yy , σ zz , E
)
(
)
(
)
(
)
V⋅ σ xx = f2 T , ε xx , σ yy , σ zz , E
−V⋅ ε yy = f3 T , ε xx , σ yy , σ zz , E
−V⋅ ε zz = f4 T , ε xx , σ yy , σ zz , E
(
−P = f5 T , ε xx , σ yy , σ zz , E
)
3. Start with the following expression......
dL =
 ∂ 
 ∂ 
 L ⋅ dσ +  L ⋅dT
 ∂σ  T
 ∂T  σ
we will use this to get
 ∂ 
1  ∂ 
1  ∂ 
 ∂ 
L = ⋅ L + ⋅ L ⋅ T
L  ∂σ 
L  ∂T   ∂σ 
L  ∂σ 
S
T
σ
S
1
⋅
1
 ∂ 
⋅ L =
E
L  ∂σ 
s
S
1
 ∂ 
⋅ L =
E
L  ∂σ 
T
T
1
1
 ∂ 
⋅ L
L  ∂σ 
S
1
We know all these terms except for the very last derivative
 ∂ 
⋅ L
= αL
L  ∂T 
σ
1
 ∂ 
S
 ∂σ  T
−
 ∂ 
 T =
 ∂σ  S  ∂ S 

 ∂T  σ
we can now write.......  ∂ 
 ∂ 
 S = A⋅  L = A⋅ L⋅ α L
 ∂σ  T
 ∂T  σ
  ∂ 

− S

 ∂σ  T 
1
1
=
+ α L⋅ 

E
E
s
T
  ∂ S 

  ∂T  σ 


we found this using the maxwell relation of the
following potential
dφ = −SdT − A⋅ L⋅ dσ
 ∂ 
Cσ
 S =
 ∂T  σ T
substituting all this stuff in, we get.....
1
E
=
S
1
E
−
T
(
α L)2⋅T⋅V
C
NOTE: This derivation is totally analogous to deriving Cp-Cv
σ
Now we need to estimate the relative difference between the two.
αL ~ 10
(
α L)2⋅T⋅V
C
σ
− 15
= 7.1⋅ 10
−5 1
⋅
K
C
σ
V
C
is roughly the same as
if we assumed a temperature of 300K
P
V
so then
C
P
V
=
6
4.2⋅ J
4.2⋅ 10 ⋅J
1 ⋅ mol 10g
⋅
=
=
3
K⋅ cc
mol⋅ K 60⋅ g cc
K⋅ m
25⋅ J
⋅
E −E
T
S
=−
E ⋅E
(α L)2⋅T
T S
C
assume that E ⋅E = <E>
S
σ
so then...
T
E −E
if <E> is roughly 100 GPa, then
E −E
2
T
S
<E>
T
S
<E>
=−
(α L)2⋅T
C
− 15
⋅<E> = −7.1⋅10
⋅<E>
σ
~ -1%
4. We have a piston attached to a generator.....
a.) what is the final temperature of the gas?
denote the environment pressure by P
o
work done by the gas: −P⋅dV
work done on environment:
work done on generator:
P ⋅dV
o
(P − Po)⋅dV
We know that the work done on the generator is fed back into the gas as heat, so....
dU
V
(
C ⋅ T −T
V
(
)
= C ⋅dT = P − P ⋅dV − P⋅ dV = −P ⋅dV we will solve this for the final temperature, and we know the final pressure is Po.
gas
2
)
1
o
(
o
= −P ⋅ V − V
o
2
1
)
 R⋅ T
= −P ⋅ 
o
2
 Po

R⋅ T
−
1
P
1

(
C
V
)
+ R ⋅T =
2
 Po

 ⋅R + C ⋅T
V
1
P
 1

 R + C 
V
 10
26
T =
⋅T =
⋅T
2
1
R+C
35 1
V 

b.) define the environment as everything except the piston/gas assembly and the generator. There was no heat flow to the environment so the entropy of the environment doesnt change.
c.) No. the entropy of the system increases, so the total entropy change is greater than zero and therefore irreversible.
5. find the metastable melting point of ε.
we want to know where ∆Gε−L ( T) = 0
(
)
∆Gε−L = ∆Gε − β + ∆Gβ −L = ∆Hε −β + ∆Hβ −L − T⋅ ∆Sε −
β + ∆Sβ −L
∆Hε −β = T
T
M
=
ε −β
⋅∆Sε − β
∆Hε − β + ∆Hβ −L
∆Sε − β + ∆Sβ −L
∆Hβ −L = T
⋅∆Sβ −L
β −L
= 1735.6 Kelvin