9.12 Entropy Maximum and Energy Minimum
357
If we had a mixture of ideal gases, all acting independently, we would have for
each:
μ i (T , p, {N i }) = RT ln
p i
p o
+ μ
o
i (T , {N i }) ,
(9.84)
where p i = n i RT /V = (n i /n)p = x i p is the partial pressure for the gas
component labeled by the index ‘i’, and x i is the ‘mole fraction’ of that component.
As a result, we have the alternative expression:
μ i (T , p, {x i }) = RT ln x i + RT ln
p
p o
+ μ
o
i (T , {x i }) .
(9.85)
9.12 Entropy Maximum and Energy Minimum
Consider a system characterized by an internal energy E (S, V ) dependent only on
its entropy and volume. If we solve for the value of the entropy S, we have the
identity
E = E (S (E, V ) , V )
(9.86)
By differentiation, small deviations from the equilibrium volume will give
0 =
∂E
∂S
δS
δV
+
∂E
∂V
,
(9.87)
which gives ∂E/∂V , the slope of the energy function as the volume deviates from
equilibrium. The curvature of the function can be found by differentiating with
respect to V again and rearranging to get
∂ 2 E
∂V 2 = −
∂ 2 E
∂S 2
δS
δV
−
∂E
∂S
δ 2 S
δV 2
= −
∂ 2 E
∂S 2
∂S
∂V
− T
∂ 2 S
∂V 2
But at equilibrium, we know that S has reached a minimum as a function of volume,
so
δS
δV = 0, and
δ 2 S
δV 2 < 0. Thus, since T > 0, we have
∂ 2 E
∂V 2 > 0. (See Fig. 9.6.) This
means the energy as a function volume is at the bottom of a parabola when near
equilibrium. The energy of the system reaches equilibrium when it is at a minimum
as a function of volume, holding entropy fixed.