1
Note 5.
Equilibrium I: Free Energy
5.1 Helmholtz free energy
5.1.1 Derivation
In the previous section, I alluded to a new state function which will help us sort out the
total entropy change. This new state function is called the
free energy
.
When we created the enthalpy state function, we said that we wanted something to
reflect the change in heat at constant pressure. Here, we have a very different goal. In the
previous section, we saw that using just entropy alone required us to consider the entropy
changes of both the system and its surroundings. This is often difficult to do and best to
avoid if possible. But how can we avoid this?
One way is to build a state function which reflects the nature of equilibrium. To find
such a state function, we return to the idea of maximum work. In Note 2, we learned that
a system can do maximum work in a reversible case (i.e. equilibrium along the path). We
will now introduce thermodynamic functions which can indicate whether one can do the
maximum amount of work and thus tell us whether we are at equilibrium. In this section,
we only consider systems at constant temperature.
For a reversible reaction, we have
rev
rev
dq
dw
dU
+
=
Since we can relate the change in reversible heat to entropy by
T
dq
dS
rev
=
We can write
TdS
dw
dU
rev
+
=
rearranging, we get
const
T
at
TS
U
d
TdS
dU
dw
rev
=
−
=
−
=
)
(
Thus, it is natural to define a new state function, called the
Helmholtz free energy
A
TS
U
A
−
=
since (at constant temperature)
rev
dw
TS
U
d
dA
=
−
=
)
(
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One way to think of the free energy is that
it is a measure of the maximum work the
system can do on its surroundings
. Just as mechanical systems work to decrease
U
,
molecular systems (at constant
T
and
V
) try to reduce
A
to reach equilibrium. What does
this mean? In mechanical systems, we expect spontaneous processes to occur only if
dU
< 0
. Apples spontaneously fall from trees, but they never spontaneously leap from the
ground into the sky. Similarly, spontaneous processes in molecular systems occur only if
dA
< 0
. If there is a state with lower free energy, then the system will eventually get there.
What happens at equilibrium? At equilibrium, we expect that the system will do no
work and thus
dA
= 0. This naturally parallels
dU
= 0
in mechanical systems.
5.2 Another way to look at free energy and equilibria
In the Note 3, we showed that the 2
nd
law is
T
dq
dS
≥
and said that
dS > dq/T
for spontaneous processes and
dS = dq/T
for reversible processes.
Now, we’ll see what the interpretations of this are for other state functions. Let’s take
dU
for example. From the first law, we have
dU = dq + dw
and thus
dq=dU  dw
. For a
gas, we have
dw = PdV
and thus
dq = dU + PdV
. Putting this into the inequality above,
we get
TdS
≥
dU + PdV
. Rearranging terms, we get
PdV
TdS
dU
−
≤
Thus, we see that at constant
S
and
V
(
i.e.
dS
= 0
and
dV
= 0)
0
≤
dU
Thus, under constant
S
and
V
, spontaneous processes must have
dU
< 0
. Note that this is
what we expect for the mechanical examples – you only have to monitor
U
to determine
the equilibrium.
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 Spring '10
 LIM
 Thermodynamics, Entropy

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