Entropy of mixing
[tln25]
Consider two dilute gases in a rigid and insulat-
ing box separated by a mobile conducting wall:
n
1
=
x
1
n
(helium) and
n
2
=
x
2
n
(neon) with
x
1
+
x
2
= 1. Thermal equilibrium:
p, T, V
1
=
x
1
V, V
2
=
x
2
V.
p
n
p T
T
V
n
1
1
2
V
2
When the interior wall is removed, the gases mix spontaneously (irreversible
process). By how much does the entropy increase?
To calculate ∆
S
, consider a reversible mixing process involving an isothermal
expansion of semipermeable walls:
U
=
U
1
+
U
2
with
U
i
(
T
) = const
⇒
dU
i
=
T dS
i
−
p
i
dV
i
= 0.
p
i
=
n
i
RT
V
i
are the partial pressures exerted on the semipermeable walls.
⇒
dS
i
=
p
i
T
dV
i
=
n
i
R
V
i
dV
i
=
nRx
i
dV
i
V
i
.
Entropy of mixing: ∆
S
=
nR
X
i
x
i
Z
V
x
i
V
dV
i
V
i
=
−
nR
X
i
x
i
ln
x
i
>
0
.
Gibbs paradox: No entropy increase should result if the two gases happen
to be of the same kind. This distinction is not reflected in the above deriva-
tion. The paradox is resolved by quantum mechanics, which requires that
distinguishable and indistinguishable particles are counted differently.
To calculate the change in Gibbs free energy during mixing, we use the
result for
G
(
T, p, N
) of an ideal gas (see [tex15]) rewritten as
G
(
T, p, n
) =
−
nRT
[ln(
T /T
0
)
α
+1
−
ln(
p/p
0
)].
Initially:
p
1
=
p
2
=
p
⇒
G
ini
=
−
X
i
nx
i
RT
"
ln
T
T
0
α
+1
−
ln
p
p
0
#
.
Finally:
p
i
=
px
i
⇒
G
f in
=
−
X
i
nx
i
RT
"
ln
T
T
0
α
+1
−
ln
px
i
p
0
#
.
Change in Gibbs free energy: ∆
G
=
nRT
X
i
x
i
ln
x
i
.
Use
S
=
−
∂G
∂T
p
to recover ∆
S
=
−
nR
X
i
x
i
ln
x
i
.
Change in chemical potential: use
G
=
X
i
n
i
µ
i
⇒
∆
µ
i
=
RT
ln
x
i
.