2.6 The Second Law and Stability
83
so that the entropy of the isolated thermodynamic system has increased in passing
from initial state i to final state f via a spontaneous irreversible process. By
considering the universe as a closed system and recognizing that naturally-occurring
processes are typically spontaneous irreversible processes, this argument provides
the source of the statement that the entropy of the universe always increases or,
equivalently, tends to a maximum. It will be useful to consider two examples that
illustrate the nature of a spontaneous irreversible change.
Example 2.6 Irreversible vs. reversible changes.
We examine in this example the (spontaneous irreversible) expansion of an ideal
gas of volume V i into a connected empty chamber of volume V c to occupy a final
volume V f = V i + V c . To obtain an expression for the change S = S f − S i in
entropy for this expansion, we shall utilize a reversible path (as the change in any
state function for a process connecting two thermodynamic states is independent of
the path and depends only upon its values for the two states).
In particular, we shall employ a reversible isothermal expansion of the ideal gas
(for which U = 0) into the empty chamber. The First Law then necessitates that
δQ rev = −δW rev and, as δW rev = −P dV with P given by P = Nk B T /V for an
ideal gas, we see that ((S) rev
sys is thus given as
((S)
rev
sys ≡
V f
V i
δQ rev
T
= −
V f
V i
δW rev
T
= Nk B
V f
V i
dV
V
.
Because ((S) irr
sys = ((S) rev
sys for this expansion, we have
((S) sys ≡ ((S)
irr
sys = Nk B ln
V f
V i
> 0 ,
and the entropy of the system increases due to the spontaneous irreversible
expansion of the ideal gas.
To see how to distinguish the way in which an irreversible expansion differs from
a reversible expansion of an ideal gas system, we shall examine the corresponding
entropy changes for the surroundings.
(a) To evaluate S for a reversible isothermal expansion, we may use the fact that
because U = 0 the expanding gas must have acquired heat (energy) Q rev =
−W rev = Nk B T ln(V f /V i ) from its surroundings. This, in turn, means that the
entropy of the surroundings will have decreased by an amount Q rev /T , i.e.,
((S)
rev
surr = −
Q rev
T
= −Nk B ln
V f
V i
.
However, as the statement S ≥ 0 applies only to an isolated system, it will
be clear that we must examine ((S) rev for the relevant isolated system, namely,
the ideal gas plus its surroundings (i.e., effectively the entire universe). Thus,
the appropriate entropy change to be considered is given by
83
so that the entropy of the isolated thermodynamic system has increased in passing
from initial state i to final state f via a spontaneous irreversible process. By
considering the universe as a closed system and recognizing that naturally-occurring
processes are typically spontaneous irreversible processes, this argument provides
the source of the statement that the entropy of the universe always increases or,
equivalently, tends to a maximum. It will be useful to consider two examples that
illustrate the nature of a spontaneous irreversible change.
Example 2.6 Irreversible vs. reversible changes.
We examine in this example the (spontaneous irreversible) expansion of an ideal
gas of volume V i into a connected empty chamber of volume V c to occupy a final
volume V f = V i + V c . To obtain an expression for the change S = S f − S i in
entropy for this expansion, we shall utilize a reversible path (as the change in any
state function for a process connecting two thermodynamic states is independent of
the path and depends only upon its values for the two states).
In particular, we shall employ a reversible isothermal expansion of the ideal gas
(for which U = 0) into the empty chamber. The First Law then necessitates that
δQ rev = −δW rev and, as δW rev = −P dV with P given by P = Nk B T /V for an
ideal gas, we see that ((S) rev
sys is thus given as
((S)
rev
sys ≡
V f
V i
δQ rev
T
= −
V f
V i
δW rev
T
= Nk B
V f
V i
dV
V
.
Because ((S) irr
sys = ((S) rev
sys for this expansion, we have
((S) sys ≡ ((S)
irr
sys = Nk B ln
V f
V i
> 0 ,
and the entropy of the system increases due to the spontaneous irreversible
expansion of the ideal gas.
To see how to distinguish the way in which an irreversible expansion differs from
a reversible expansion of an ideal gas system, we shall examine the corresponding
entropy changes for the surroundings.
(a) To evaluate S for a reversible isothermal expansion, we may use the fact that
because U = 0 the expanding gas must have acquired heat (energy) Q rev =
−W rev = Nk B T ln(V f /V i ) from its surroundings. This, in turn, means that the
entropy of the surroundings will have decreased by an amount Q rev /T , i.e.,
((S)
rev
surr = −
Q rev
T
= −Nk B ln
V f
V i
.
However, as the statement S ≥ 0 applies only to an isolated system, it will
be clear that we must examine ((S) rev for the relevant isolated system, namely,
the ideal gas plus its surroundings (i.e., effectively the entire universe). Thus,
the appropriate entropy change to be considered is given by