PHYS 253: Assignment 3
Problem 1
For this hydrostatic system, the first law is
dU
=
dQ

PdV
dU
=
emf
·
I
·
dt

p
0
·
dV
Integrating, we get
Δ
U
=
emf
·
I
·
t

p
0
·
(
nv
o

V
)
Problem 2
a) We write the 1st law as
¯
dQ
=
dU
+
PdV
Keeping in mind that
U
=
U
(
P, T
), we can rewrite
dU
as
dU
=
∂U
∂T
P
dT
+
∂U
∂P
T
dP
and from the EOS, we write
PdV
as
PdV
=
P
∂V
∂T
P
dT
+
∂V
∂P
T
dP
Regrouping the terms, we get
¯
dQ
=
∂U
∂T
P
+
P
∂V
∂T
P
dT
+
∂U
∂P
T
+
P
∂V
∂P
T
dP
b) setting
P
to be a constant, we rewrite the expression from a) as
¯
dQ
dT
P
=
∂U
∂T
P
+
P
∂V
∂T
P
Rewriting the individual terms:
¯
dQ
dT
P
=
C
P
and
P
∂V
∂T
P
=
PV β
we finally get
∂U
∂T
P
=
C
P

PV β
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– 2 –
c) Starting from the expression in a), at constant
V
, we first use the result from b) to rewrite the
∂U
∂T

P
term
¯
dQ
=
C
P
dT
+
∂U
∂P
T
+
P
∂V
∂P
T
dP
and now rewriting the
∂V
∂P

T
using the definition of
κ
, and dividing by
dT
, we get:
¯
dQ
dT
V
=
C
P
+
∂U
∂P
T

PV κ
dP
dT
V
(1)
We can rewrite
dP
dT

V
using
dP
=
∂P
∂T
V
dT
+
∂P
∂V
T
dV
However, since
dV
= 0 then
dP
dT
V
=
∂P
∂T
V
Using equation (2.6) from the textbook:
∂P
∂V
T
∂V
∂T
P
∂T
∂P
V
=

1
we rewrite
∂P
∂T

V
as
∂P
∂T
V
=

∂P
∂V
T
∂V
∂T
P
=


1
κV
(
βV
) =
β
κ
Putting everything together, we can rewrite equation (1) as
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 Fall '10
 PeterGrutter
 Thermodynamics, Expression, Trigraph, SEPTA Regional Rail, Rewrite engine

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