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ENGRD 221 –Prof. N. Zabaras
Prelim II
11/15/2007
Page 1 of 10
Thursday November 15, 7:30 pm – 9:30 pm
Closed books and notes.
The notation is all problems is the one used in class and further explanations are not
needed. Answer all questions. Make sure your answers are legible.
The TAs and instructor will not respond to any questions during the exam. If you think
that something is wrong with one of the problems below, please state your concern in
your exam books.
Problem 1
(20 points)
Answer each question carefully providing all necessary details.
1.
(3 points) For the following combined form of the first and second laws of
thermodynamics, dG=SdT+VdP, write down the corresponding Maxwell
equation as well as the two coefficient relations.
Solution: (a)

,
,
PT
T
TP
P
T
SV
G
G
dG
SdT
VdP
S
V
P
⎛
⎞⎛⎞
⎛⎞
∂∂
∂
∂
⎟⎟
⎟
⎟
⎜⎜
⎜
⎜
=
−
+
⇒
==
−
=
⎟
⎟
⎜
⎜
⎟
⎟
⎜
⎜
⎝
⎠⎝⎠
⎝⎠
∂
∂
2.
(5 point) Use the methodology of your choice to compute an expression for
T
V
U
∂
⎜⎟
∂
in terms of T, P,
α
and
β
.
Answer:
Since T is one of the independent variables we can write that
T
T
T
P
U
P
V
U
V
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
=
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
(1)
Solving for the resulting relations, we use equations for dU and dV
dP
V
dT
V
dV
β
α
−
=
(2)
The equation for dS is given as: dS= c
P
/T dT –
α
v dP (the Maxwell equation used here is
V
α
⎛
⎞
=−
⎜
⎟
⎝
⎠
derived from dG = S dT + V dP in part 1 above)
From the above equation using TdS = dU+ P dV, we derive the following:
()
( )
dP
T
P
V
dT
PV
C
dU
P
−
+
−
=
(3)
At constant T, the coefficients of
dP
are (from eqs 2 and 3, respectively):
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View Full DocumentENGRD 221 –Prof. N. Zabaras
Prelim II
11/15/2007
Page 2 of 10
β
V
P
V
T
−
=
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
(4)
()
α
T
P
V
P
U
T
−
=
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
(5)
Substituting 4 and 5 into 1 and reducing we find that
T
P
U
V
T
−
−
=
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
3.
(5 points) From Eqs. C
P
= (
∂
h/
∂
T)
P
= T (
∂
s/
∂
T)
P
and C
v
= (
∂
u/
∂
T)
v
= T (
∂
s/
∂
T)
v
and the knowledge that Cp > Cv what can you conclude about the slopes of
constant v and constant P curves in a Ts diagram? Notice that we are looking at
functions T(s) with P or v given.
Solution
The slopes of constant v or constant P curves in a Ts diagram are T lines as a function of
s with either v or P held constant.
Therefore, the slopes of these lines are (
∂
T/
∂
s)
v
and
(
∂
T/
∂
s)
P
.
From the definitions of c
v
and c
p
:
c
v
= (
∂
u/
∂
T)
v
= T (
∂
s/
∂
T)
v
(for the derivation on the right we used: du = Tds – Pdv)
c
p
= (
∂
h/
∂
T)
P
= T (
∂
s/
∂
T)
P
(for the derivation on the right we used: dh = Tds + vdP)
we get using the inversion relation (e.g. (
∂
T/
∂
s)
v
= 1 / (
∂
s/
∂
T)
v ,
etc.).
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