1. Helmholtz Free Energy (A).
A
≡
U – TS
(combine 1
st
and 2
nd
Law at const T and V)
Δ
A =
Δ
U  T
Δ
S < 0
(spontaneous process at const T and V)
Δ
A = w
rev
=maximum work
(isothermal, reversible, i.e. T
Δ
S=q
rev
)
2. Gibbs Free Energy (G).
TS
H
G
−
≡
(combine 1
st
and 2
nd
Law at const T and P)
Δ
G =
Δ
H – T
Δ
S < 0
(spontaneous process at const T and P)
Δ
G = w
nonPV
(isothermal, reversible, i.e. T
Δ
S=q
rev
)
3. Fuel Cell Extracts Electrical Work (
Δ
G=w
nonPV
) from H
2
(g) + O
2
(g)
→
H
2
O(l).
Δ
Hº = 285 kJ/mol and T
Δ
Sº = 48.7 kJ/mol at T=298 K, w
nonPV
= nFE =
Δ
G
Δ
G =
Δ
H T
Δ
S = 237 kJ/mol
⇒
efficiency = w/q =
Δ
G/
Δ
H = 83% vs 30% gas engine
H
2
2 e

2H
+
½O
2
O
2
H
2
O
H
2
+ O
2
h
ν
Solar
Fuel Cell
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What if Fuel Cell reaction is run in a piston and cylinder and does only PV work?
H
2
(
g
)
+
1
2
O
2
(
g
)
⎯ →
⎯
H
2
O
(
l
)
(
)
(
)
J
w
V
V
nRT
V
dV
nRT
dV
V
nRT
PdV
w
PV
PV
3
1
2
2
1
2
1
2
1
10
33
.
28
7
.
36
018
.
0
ln
15
.
298
314
.
8
5
.
1
ln
×
=
⎟
⎠
⎞
⎜
⎝
⎛
−
=
−
=
∫
−
=
∫
−
=
∫
−
=
mol
kJ
G
/
1
.
237
−
=
Δ
KJ
H
8
.
285
−
=
Δ
(
)
KJ
RT
n
PV
gas
717
.
3
−
=
Δ
≅
Δ
(
)
KJ
PV
H
U
1
.
282
−
=
Δ
−
Δ
=
Δ
(
)
KJ
S
T
TS
7
.
48
−
=
Δ
=
Δ
kJ
q
4
.
310
1
.
282
33
.
28
−
=
−
−
=
Efficiency = w
PV
/q
(q = w
PV
+
Δ
U)
Efficiency of PV work = w
PV
/q
= 28.33/310 = 9%
Efficiency of Fuel Cell (w
nonPV
/q) =
Δ
G/
Δ
H = 237/285 = 83%
W
nonPV
= 0