From Equation 5.11, for n moles of an ideal gas the entropy is S=nc, InT+nRln + So n (5.13) Now nд moles of an ideal gas A of volume VA and temperature T are separated from ng moles of another ideal gas B of volume V, at the same temperature T (see Figure 5-13a). The partition is removed so that the gases mix isothermally at the temperature T, the mixture then occupying the volume V + V, (see Figure 5-13b). (a) Use Gibbs's theo- rem, introduced in the preceding problem, to show that the entropy change occurring in this mixing is ASmising Rn, In VA+VB VA +ng In VA+VB VB (b) Suppose that the gases are identical. Clearly, on removing the partition, there can now be no entropy change, because the physical system is unchanged. However, the result just proved in (a) gives AS mixing 0! This is known as the Gibbs paradox. Is the result given in (a) valid for identical gases and if not, why not? (b) (a) VA VB V₁+V₁₂ ПА Ив T T T Mixture A + B Gas A Gas B Figure 5-13 (Hint: Consider how Gibbs's theorem was proved.) (c) Obtain the correct expression ASmining (na+n)Rln V₁+V nА+n B VB Rin -ngRin nA Пв for the entropy of mixing of identical gases by applying Equation 5.11 to the three volumes V, V, and V₁+ VB, all containing the same gas. (d) By using the fact that, for identical gases, VA+VB VA VB = = nА +nB ПА NB show that the entropy of mixing given in (c) is indeed zero.
From Equation 5.11, for n moles of an ideal gas the entropy is S=nc, InT+nRln + So n (5.13) Now nд moles of an ideal gas A of volume VA and temperature T are separated from ng moles of another ideal gas B of volume V, at the same temperature T (see Figure 5-13a). The partition is removed so that the gases mix isothermally at the temperature T, the mixture then occupying the volume V + V, (see Figure 5-13b). (a) Use Gibbs's theo- rem, introduced in the preceding problem, to show that the entropy change occurring in this mixing is ASmising Rn, In VA+VB VA +ng In VA+VB VB (b) Suppose that the gases are identical. Clearly, on removing the partition, there can now be no entropy change, because the physical system is unchanged. However, the result just proved in (a) gives AS mixing 0! This is known as the Gibbs paradox. Is the result given in (a) valid for identical gases and if not, why not? (b) (a) VA VB V₁+V₁₂ ПА Ив T T T Mixture A + B Gas A Gas B Figure 5-13 (Hint: Consider how Gibbs's theorem was proved.) (c) Obtain the correct expression ASmining (na+n)Rln V₁+V nА+n B VB Rin -ngRin nA Пв for the entropy of mixing of identical gases by applying Equation 5.11 to the three volumes V, V, and V₁+ VB, all containing the same gas. (d) By using the fact that, for identical gases, VA+VB VA VB = = nА +nB ПА NB show that the entropy of mixing given in (c) is indeed zero.
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Transcribed Image Text:From Equation 5.11, for n moles of an ideal gas the entropy is
S=nc, InT+nRln
+ So
n
(5.13)
Now nд moles of an ideal gas A of volume VA and temperature T are
separated from ng moles of another ideal gas B of volume V, at the
same temperature T (see Figure 5-13a). The partition is removed so
that the gases mix isothermally at the temperature T, the mixture then
occupying the volume V + V, (see Figure 5-13b). (a) Use Gibbs's theo-
rem, introduced in the preceding problem, to show that the entropy
change occurring in this mixing is
ASmising Rn, In
VA+VB
VA
+ng In
VA+VB
VB
(b) Suppose that the gases are identical. Clearly, on removing
the partition, there can now be no entropy change, because the
physical system is unchanged. However, the result just proved in
(a) gives AS mixing 0! This is known as the Gibbs paradox. Is the
result given in (a) valid for identical gases and if not, why not?
(b)
(a)
VA
VB
V₁+V₁₂
ПА
Ив
T
T
T
Mixture A + B
Gas A
Gas B
Figure 5-13
(Hint: Consider how Gibbs's theorem was proved.) (c) Obtain the
correct expression
ASmining (na+n)Rln
V₁+V
nА+n B
VB
Rin
-ngRin
nA
Пв
for the entropy of mixing of identical gases by applying Equation 5.11
to the three volumes V, V, and V₁+ VB, all containing the same gas.
(d) By using the fact that, for identical gases,
VA+VB VA VB
=
=
nА +nB
ПА
NB
show that the entropy of mixing given in (c) is indeed zero.
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