84
2 Macroscopic Thermodynamics
((S)
rev
total = ((S)
rev
sys + ((S)
rev
surr .
Because ((S) rev
surr is the negative of ((S) rev
sys , we see that ((S) rev
total = 0 for the
reversible isothermal expansion of an ideal gas.
(b) For the irreversible (isothermal) expansion of the system (the ideal gas), we
have U ≡ 0 and W irr = 0, as no work is done during this expansion. By the
First Law we therefore have Q irr = 0, i.e., no heat (energy) is delivered to the
system from its surroundings. However, we note that as ((S) irr
sys is still equal to
((S) rev
sys , ((S) irr
surr = 0 for the irreversible expansion of an ideal gas. The total
entropy change is therefore
((S)
irr
total = ((S)
irr
sys + ((S)
irr
surr = Nk B ln
V f
V i
,
and thus ((S) irr
total > 0.
Example 2.7 The mixing of two ideal gases.
We shall now consider two ideal gases A and B each initially in chambers of
volume V A and V B but separated by a removable impermeable membrane. Each gas
is at temperature T and under pressure P . Removal of the impermeable membrane
separating the two gases then enables them to mix (by expanding spontaneously and
independently of one another), with each gas thereby occupying the total volume
V f = V A + V B . To examine the entropy change for the resultant (spontaneous
irreversible) mixing process, we may once again make use of the fact that S is a
thermodynamic state function in order to employ a reversible isothermal expansion
to evaluate the entropy change ((S) irr ≡ ((S) mix resulting from the mixing
process.
For gas A we obtain (from Eq. (2.2.9b), for example)
S A = N A k B ln
V A + V B
V A
,
while for gas B we obtain
S B = N B k B ln
V A + V B
V B
,
with N A and N B the numbers of molecules A and B in the two ideal gas subsystems.
We may employ the ideal gas equation of state (with T A = T B ≡ T and P A = P B ≡
P ) to express S A and S B as
S A = N A k B ln
N
N A
and S B = N B k B ln
N
N B
,
with N = N A + N B the total number of molecules in the gas mixture. The total
entropy change for mixing of the two gases is thus
2 Macroscopic Thermodynamics
((S)
rev
total = ((S)
rev
sys + ((S)
rev
surr .
Because ((S) rev
surr is the negative of ((S) rev
sys , we see that ((S) rev
total = 0 for the
reversible isothermal expansion of an ideal gas.
(b) For the irreversible (isothermal) expansion of the system (the ideal gas), we
have U ≡ 0 and W irr = 0, as no work is done during this expansion. By the
First Law we therefore have Q irr = 0, i.e., no heat (energy) is delivered to the
system from its surroundings. However, we note that as ((S) irr
sys is still equal to
((S) rev
sys , ((S) irr
surr = 0 for the irreversible expansion of an ideal gas. The total
entropy change is therefore
((S)
irr
total = ((S)
irr
sys + ((S)
irr
surr = Nk B ln
V f
V i
,
and thus ((S) irr
total > 0.
Example 2.7 The mixing of two ideal gases.
We shall now consider two ideal gases A and B each initially in chambers of
volume V A and V B but separated by a removable impermeable membrane. Each gas
is at temperature T and under pressure P . Removal of the impermeable membrane
separating the two gases then enables them to mix (by expanding spontaneously and
independently of one another), with each gas thereby occupying the total volume
V f = V A + V B . To examine the entropy change for the resultant (spontaneous
irreversible) mixing process, we may once again make use of the fact that S is a
thermodynamic state function in order to employ a reversible isothermal expansion
to evaluate the entropy change ((S) irr ≡ ((S) mix resulting from the mixing
process.
For gas A we obtain (from Eq. (2.2.9b), for example)
S A = N A k B ln
V A + V B
V A
,
while for gas B we obtain
S B = N B k B ln
V A + V B
V B
,
with N A and N B the numbers of molecules A and B in the two ideal gas subsystems.
We may employ the ideal gas equation of state (with T A = T B ≡ T and P A = P B ≡
P ) to express S A and S B as
S A = N A k B ln
N
N A
and S B = N B k B ln
N
N B
,
with N = N A + N B the total number of molecules in the gas mixture. The total
entropy change for mixing of the two gases is thus