Show that entropy is a state function
Answers
AnswerEssentially, this shows a derivation of entropy and that a state function can be written as a total derivative, dF(x,y)=(∂F∂x)ydx+(∂F∂y)xdy . where q is the heat flow, w is the work (which we define as −∫PdV ), and δ indicates that heat flow and work are inexact differentials (path functions).
Essentially, this shows a derivation of entropy and that a state function can be written as a total derivative,
d
F
(
x
,
y
)
=
(
∂
F
∂
x
)
y
d
x
+
(
∂
F
∂
y
)
x
d
y
.
From the first law of thermodynamics:
d
U
=
δ
q
rev
+
δ
w
rev
,
where
q
is the heat flow,
w
is the work (which we define as
−
∫
P
d
V
), and
δ
indicates that heat flow and work are inexact differentials (path functions).
Solving for
δ
q
rev
gives:
δ
q
rev
=
d
U
−
∂
w
rev
=
C
V
(
T
)
d
T
+
P
d
V
,
since
(
∂
U
∂
T
)
V
=
C
V
, the constant-volume heat capacity. For an ideal gas, we'd get:
δ
q
rev
(
T
,
V
)
=
C
V
(
T
)
d
T
+
n
R
T
V
d
V
It can be shown that this is an inexact total derivative, indicative of a path function. Euler's reciprocity relation states that for the total derivative
d
F
(
x
,
y
)
=
M
(
x
)
d
x
+
N
(
y
)
d
y
,
where
M
(
x
)
=
(
∂
F
∂
x
)
y
and
N
(
y
)
=
(
∂
F
∂
y
)
x
,
a differential is exact if
(
∂
M
∂
y
)
x
=
(
∂
N
∂
x
)
y
. If this is the case, this would indicate that we have a state function.
Let
M
(
T
)
=
(
∂
q
rev
∂
T
)
V
=
C
V
(
T
)
,
N
(
V
)
=
(
∂
q
rev
∂
V
)
T
=
n
R
T
V
,
x
=
T
, and
y
=
V
. If we use our current expression for
δ
q
rev
, we obtain:
(
∂
C
V
(
T
)
∂
V
)
T
?
=
(
∂
(
n
R
T
/
V
)
∂
T
)
V
But since
C
V
(
T
)
is only a function of
T
for an ideal gas, we have:
0
≠
n
R
V
However, if we multiply through by
1
T
, called an integrating factor, we would get a new function of
T
and
V
which is an exact differential:
δ
q
rev
(
T
,
V
)
T
=
C
V
(
T
)
T
d
T
+
n
R
V
d
V
Now, Euler's reciprocity relation works:
(
∂
[
C
V
(
T
)
/
T
]
∂
V
)
T
?
=
(
∂
(
n
R
/
V
)
∂
T
)
V
0
=
0
√
Therefore, this new function,
q
rev
(
T
,
V
)
T
can be defined as the state function
S
, entropy, which in this case is a function of
T
and
V
:
d
S
(
T
,
V
)
=
δ
q
rev
T
and it can be shown that for the definition of the total derivative of
S
:
d
S
=
(
∂
S
∂
T
)
V
d
T
+
(
∂
S
∂
V
)
T
d
V
=
(
∂
S
∂
T
)
V
d
T
+
(
∂
P
∂
T
)
V
d
V